Maxima and Minima

Introduction

Optimization is a fundamental concept in mathematics and engineering, concerned with determining the maximum or minimum values of a function. This process is essential across various fields, including physics, economics, and engineering design, where one often seeks to minimize costs, maximize efficiency, or optimize performance parameters.

At its core, the problem of optimization begins with a function \( f(x) \) defined on some interval, and the goal is to identify values of \( x \) at which \( f(x) \) attains either its highest or lowest value. Such points are called extrema. If a function reaches its highest value at some point within a given domain, that point is called a maximum; if it attains its lowest value, it is called a minimum. Depending on the context, one may be interested in either local extrema—where the function reaches a peak or valley within a small neighborhood—or global extrema, where the function attains its absolute highest or lowest value over the entire domain.

Optimization problems appear naturally in real-world applications. Engineers optimize material usage in construction, economists seek to maximize profit functions, and physicists analyze equilibrium conditions by finding extrema of energy functions. Understanding how to systematically identify these points and characterize their nature is an essential step before progressing to more advanced optimization techniques.

The goal of optimization is to determine the maximum and minimum values of a function over a given interval. These values, called the global maximum and global minimum, represent the highest and lowest function values within the interval and are essential in many mathematical and real-world applications.

One way to find these extrema is by drawing the graph of the function and visually identifying the highest and lowest points. While this can sometimes be helpful, it is not always practical. For many functions, sketching an accurate graph is difficult or even impossible without prior knowledge of the function’s behavior. Therefore, we seek a structured mathematical approach that allows us to determine maximum and minimum values without relying on graphing.

In some cases, it is possible to find extrema using algebraic methods, such as factoring or completing the square. However, these methods work only for specific types of functions and are not always applicable. To develop a more general and systematic approach, we rely on calculus, which provides powerful techniques for finding local and global extrema of a function.

Our approach begins with identifying all local maxima and minima, which are points where the function reaches a peak or a valley in a small neighborhood. Once we determine these points, we use these with some other information to find global maximum and global minimum values of the function.

Local Maxima and Minima

A function \( f \) is said to have a local extremum at \( x = a \) if there exists \( \delta > 0 \) such that:

• \( f \) has a local maximum at \( x = a \) if for all \( x \in (a - \delta, a) \) and \( y \in (a, a + \delta) \),

  \[ f(x) \leq f(a) \geq f(y). \]

  This means that within a small interval around \( a \), the function attains its highest value at \( x = a \), making it a local peak.

Similarly,

• \( f \) has a local minimum at \( x = a \) if for all \( x \in (a - \delta, a) \) and \( y \in (a, a + \delta) \),

  \[ f(x) \geq f(a) \leq f(y). \]

  This means that within a small neighborhood around \( a \), the function attains its lowest value at \( x = a \), forming a local valley.

These definitions formalize the idea that a local maximum is a peak, while a local minimum is a valley, both occurring within a small region of the function's domain.

In this definition it does not matter whether the function is continuous or distcontinuous at \(a\). This is valid for any possibility.

Note

• The plural of maximum is maxima, and the plural of minimum is minima.
• The term extremum (plural: extrema) refers to either a maximum or a minimum.
• When discussing multiple peaks and valleys of a function, we use the terms local maxima and local minima to describe multiple high and low points, respectively.
• A function may have several local extrema but only one global maximum or global minimum over a given domain, unless the same value is attained at multiple points.

To check whether a point \( x = a \) is a local maximum, we compare \( f(a) \) with \( f(a+\delta) \) and \( f(a-\delta) \) for a small positive number \( \delta \). This is quite a practical approach. Our goal is to show that:

\[ f(a-\delta) \leq f(a) \geq f(a+\delta) \]

for every sufficiently small \( \delta \). This ensures that as \( \delta \) gets arbitrarily small, \( f(a) \) remains greater than or equal to the values of the function at nearby points.

Similarly, for a local minimum, we must show:

\[ f(a-\delta) \geq f(a) \leq f(a+\delta) \]

for all sufficiently small \( \delta \), ensuring that \( f(a) \) is the smallest function value in its neighborhood.

Consider the function

\[ f(x) = x(2-x). \]

We check whether it has a local maximum at \( x = 1 \).

First, we evaluate:

\[ f(1) = 1(2-1) = 1. \]

Now, for a small \( \delta > 0 \), compute \( f(1+\delta) \):

\[ f(1+\delta) = (1+\delta)(2-(1+\delta)) = (1+\delta)(1-\delta). \]

Expanding,

\[ f(1+\delta) = 1 + \delta - \delta - \delta^2 = 1 - \delta^2. \]

Since \( \delta^2 > 0 \) for any \( \delta \neq 0 \), we see that:

\[ f(1+\delta) = 1 - \delta^2 < 1 = f(1). \]

Similarly, compute \( f(1-\delta) \):

\[ f(1-\delta) = (1-\delta)(2-(1-\delta)) = (1-\delta)(1+\delta). \]

Expanding,

\[ f(1-\delta) = 1 - \delta + \delta - \delta^2 = 1 - \delta^2. \]

Again, since \( \delta^2 > 0 \),

\[ f(1-\delta) = 1 - \delta^2 < 1 = f(1). \]

Since both \( f(1+\delta) \) and \( f(1-\delta) \) are less than \( f(1) \) for every sufficiently small \( \delta \), we conclude that \( x=1 \) is a local maximum of \( f(x) \).

Using Derivatives

The above method to check whether a point is a local maximum or minimum is valid for any function. Let us now consider a smaller class of functions: those that are continuous but may be non-differentiable at some points. For this class of functions, we can use derivatives to quickly identify all local maxima and minima.

Let \( f \) be a continuous function, possibly non-differentiable at a countable number of points, defined on an interval \( (a, b) \). It turns out that if a function has a local extremum at some point, then either its derivative there vanishes or does not exist. However, the converse is not true—a point where \( f'(x) = 0 \) or does not exist may not necessarily be a local extremum.

At a local maximum, the function changes its behavior from increasing to decreasing. This means that \( f' \) must change from positive to negative. At the maximum itself, \( f' \) may be zero or may not exist.

Similarly, at a local minimum, the function changes from decreasing to increasing, meaning \( f' \) must change from negative to positive. Again, at the minimum itself, \( f' \) may be zero or may not exist.

Example 1: \( f(x) = |x| \)

The function \( f(x) = |x| \) is continuous but not differentiable at \( x = 0 \). Its derivative is:

\[ f'(x) = \begin{cases} 1, & x > 0 \\ -1, & x < 0 \end{cases} \]

At \( x = 0 \), \( f' \) does not exist. Checking the behavior of \( f' \), we see that as \( x \) increases past 0, \( f' \) changes from negative to positive, indicating that \( f(x) \) is transitioning from decreasing to increasing. Thus, \( x = 0 \) is a local minimum, even though \( f' \) does not exist there.

Example 2: \( g(x) = x^2 \)

For \( g(x) = x^2 \), the derivative is:

\[ g'(x) = 2x. \]

At \( x = 0 \), we have \( g'(0) = 0 \).
For \( x < 0 \), \( g'(x) < 0 \), so the function is decreasing.
For \( x > 0 \), \( g'(x) > 0 \), so the function is increasing.
Since \( g'(x) \) changes from negative to positive at \( x = 0 \), we conclude that \( x = 0 \) is a local minimum.

From these two examples, we understand that at a local extremum, the derivative either vanishes or does not exist. To determine whether a critical point is an extremum, we just need to check whether \( f' \) changes sign around that point.

Now consider the function \( f(x) = x^3 \). Its derivative is

\[ f'(x) = 3x^2. \]

At \( x = 0 \), we compute

\[ f'(0) = 3(0)^2 = 0. \]

However, \( x = 0 \) is not a local extremum. To see why, observe the sign of \( f'(x) \) on both sides of \( x = 0 \):

• For \( x > 0 \), we have \( f'(x) = 3x^2 > 0 \).
• For \( x < 0 \), we also have \( f'(x) = 3x^2 > 0 \).

Since \( f'(x) \) is positive on both sides of \( x = 0 \), the function does not change from increasing to decreasing or vice versa. Instead, \( f(x) \) is strictly increasing for all \( x \), meaning \( x = 0 \) is neither a local maximum nor a local minimum.

This example illustrates that having \( f'(x) = 0 \) at a point does not necessarily mean the function has a local extremum there. The behavior of the derivative around that point must also be examined.

Necessary Condition for Local Extrema

Let \( f \) be a continuous function, possibly non-differentiable at a countable number of points, defined on an open interval \( (a, b) \). If \( f \) has a local extremum at \( x = c \), then either

\[ f'(c) = 0 \quad \text{or} \quad f'(c) \text{ does not exist}. \]

However, the converse is not true—a point where \( f'(c) = 0 \) or does not exist may not necessarily be a local extremum.

Moreover,

• If \( f \) has a local maximum at \( x = c \), then \( f' \) changes sign from positive to negative around \( c \).
• If \( f \) has a local minimum at \( x = c \), then \( f' \) changes sign from negative to positive around \( c \).

At the extremum itself, \( f' \) may either be zero or may not exist.

Local Extremum are not defined at Endpoints

In the definition of local maxima and minima, notice that we do not define local extrema at the endpoints of the domain of the function. Suppose the function \( f \) is defined on the closed interval \( [a, b] \). According to our definition of local maximum and local minimum, a point \( x = c \) is a local extremum if there exists a small neighborhood around \( c \) such that \( f(c) \) is either greater than or less than all function values in that neighborhood.

However, at the endpoints \( x = a \) and \( x = b \), a neighborhood on both sides does not exist. The left endpoint \( x = a \) has points only to its right, and the right endpoint \( x = b \) has points only to its left. This means our standard definition of local extrema does not apply at endpoints.

One possible way to extend our definition to endpoints is by considering one-sided neighborhoods:

• At \( x = a \), we could say \( f(a) \) is a local maximum if \( f(a) \geq f(x) \) for all \( x \) in a small right-hand neighborhood \( (a, a + \delta) \), and a local minimum if \( f(a) \leq f(x) \) for all such \( x \).
• Similarly, at \( x = b \), we could define a local maximum or minimum using a left-hand neighborhood \( (b - \delta, b) \).

However, if we adopt this modified definition, the behavior of derivatives at the endpoints is fundamentally different from that at interior extrema. If \( f \) is differentiable at an interior local extremum, we necessarily have:

\[ f'(c) = 0. \]

But at an endpoint, if \( f \) is differentiable there, the derivative can be any finite number. It does not necessarily vanish. This distinction arises because, at an interior extremum, the function changes from increasing to decreasing (or vice versa), forcing the derivative to be zero. At an endpoint, there is no function behavior on one side to compare, so the derivative does not have the same constraint.

Thus, we treat endpoints separately in optimization problems. While they may contribute to global extrema (as we will learn later), they are not considered local extrema under the standard definition.

Tests to Find all points of Maxima and Minima

To find all local maxima and minima of a function in a given interval, there are three primary methods: the first derivative test, the second derivative test, and the higher-order derivative test. We begin with the first derivative test, which we have already described conceptually. Here, we present it in a more algorithmic manner for systematic application.

First Derivative Test

Let \( f \) be a continuous function on an open interval \( (a, b) \), possibly non-differentiable at some points. The first derivative test consists of the following steps:

1. Compute the derivative and find all critical points

   • Recall that a critical point of \( f \) is any point \( c \) in \( (a, b) \) where either

     \[ f'(c) = 0 \quad \text{or} \quad f'(c) \text{ does not exist}. \]

   • Compute \( f'(x) \) and solve for \( f'(x) = 0 \).

   • Identify points where \( f'(x) \) does not exist.

2. Plot these critical points in ascending order on the real number line and analyze the sign of \( f'(x) \) around them

   • Mark the critical points on a number line.
   • Choose test points in each subinterval formed by these critical points.
   • Determine the sign of \( f'(x) \) at these test points and construct a sign graph for \( f'(x) \).

3. Make conclusions based on the sign changes of \( f'(x) \):

   • If \( f'(x) \) changes sign from positive to negative at \( x = c \) (i.e., \( f' > 0 \) for \( x < c \) and \( f' < 0 \) for \( x > c \)), then \( f \) has a local maximum at \( x = c \). This corresponds to the function changing from increasing to decreasing.
   • If \( f'(x) \) changes sign from negative to positive at \( x = c \) (i.e., \( f' < 0 \) for \( x < c \) and \( f' > 0 \) for \( x > c \)), then \( f \) has a local minimum at \( x = c \). This corresponds to the function changing from decreasing to increasing.
   • If \( f'(x) \) does not change sign at \( x = c \) (i.e., \( f' > 0 \) on both sides or \( f' < 0 \) on both sides), then \( f \) is monotonic near \( x = c \), and there is no local extremum at \( x = c \).

Example

Find Local Maxima and Minima of \( f(x) = (x+3)^2 (x-2)^3 \)

Solution:

The function

\[ f(x) = (x+3)^2 (x-2)^3 \]

is differentiable everywhere. To find critical points, we compute its derivative:

\[ f'(x) = 2(x+3)(x-2)^3 + 3(x+3)^2(x-2)^2. \]

Factoring common terms,

\[ f'(x) = (x+3)(x-2)^2(2(x-2) + 3(x+3)). \]

Simplifying,

\[ = (x+3)(x-2)^2(2x - 4 + 3x + 9). \]

\[ = (x+3)(x-2)^2(5x + 5). \]

Factoring further,

\[ = 5(x+3)(x-2)^2(x+1). \]

Setting \( f'(x) = 0 \), we get critical points:

\[ x+3 = 0 \Rightarrow x = -3, \quad (x-2)^2 = 0 \Rightarrow x = 2, \quad x+1 = 0 \Rightarrow x = -1. \]

Thus, the critical points are \( x = -3, -1, 2 \).

Sign Analysis of \( f'(x) \)

To determine the nature of these critical points, we analyze the sign of \( f'(x) \) in the intervals divided by the critical points \( x = -3, -1, 2 \):

• For \( x < -3 \), choose \( x = -4 \):

  \[ f'(-4) = 5(-4+3)(-4-2)^2(-4+1) = 5(-1)(36)(-3) > 0. \]

  So, \( f'(x) > 0 \) (positive).

• For \( -3 < x < -1 \), choose \( x = -2 \):

  \[ f'(-2) = 5(-2+3)(-2-2)^2(-2+1) = 5(1)(16)(-1) < 0. \]

  So, \( f'(x) < 0 \) (negative).

• For \( -1 < x < 2 \), choose \( x = 0 \):

  \[ f'(0) = 5(0+3)(0-2)^2(0+1) = 5(3)(4)(1) > 0. \]

  So, \( f'(x) > 0 \) (positive).

• For \( x > 2 \), choose \( x = 3 \):

  \[ f'(3) = 5(3+3)(3-2)^2(3+1) = 5(6)(1)(4) > 0. \]

  So, \( f'(x) > 0 \) (positive).

Conclusion:

• At \( x = -3 \), \( f'(x) \) changes from positive to negative \( \Rightarrow \) local maximum.
• At \( x = -1 \), \( f'(x) \) changes from negative to positive \( \Rightarrow \) local minimum.
• At \( x = 2 \), \( f'(x) \) does not change sign (remains positive), so it is not a local extremum.

Final Answer:

• Local Maximum at \( x = -3 \).
• Local Minimum at \( x = -1 \).

Example

Find all points of local maxima and minima of the function

\[ f(x) = \begin{cases} (2 + x)^3, & -3 < x \leq -1 \\ x^{2/3}, & -1 < x < 2 \end{cases} \]

Solution:

Step 1: Check Continuity

Both pieces are always continuous. Although, we must check the continuity ar \(x=-1\)

For continuity at \( x = -1 \), we compute:

\[ \lim_{x \to -1^-} f(x) = (-1 + 2)^3 = 1, \quad f(-1) = 1, \quad \lim_{x \to -1^+} f(x) = (-1)^{2/3} = 1. \]

Since \( \lim\limits_{x \to -1^-} f(x) = \lim\limits_{x \to -1^+} f(x) = f(-1) \), the function is continuous at \( x = -1 \).

Now we can differentiate:

Step 2: Compute Derivatives

For \( -3 < x < -1 \),

\[ f'(x) = 3(x+2)^2. \]

For \( -1 < x < 2 \),

\[ f'(x) = \frac{2}{3} x^{-1/3} = \frac{2}{3} \frac{1}{\sqrt[3]{x}}. \]

At \(x=-1\) we remove the equal sign because we are unsure of differentiability at this point.

Step 3: Check Differentiability at \( x = -1 \)

For \( x \to -1^- \),

\[ \lim_{x \to -1^-} f'(x) = 3(-1+2)^2 = 3. \]

For \( x \to -1^+ \),

\[ \lim_{x \to -1^+} f'(x) = \frac{2}{3} (-1)^{-1/3} = -\frac{2}{3}. \]

Since \( \lim\limits_{x \to -1^-} f'(x) \neq \lim\limits_{x \to -1^+} f'(x) \), the derivative does not exist at \( x = -1 \). Thus, \( x = -1 \) is a critical point.

Step 4: Find Other Critical Points

Setting \( f'(x) = 0 \) in each piece:

For \( -3 < x < -1 \):

\[ 3(x+2)^2 = 0 \implies x = -2. \]

Since \(-2 \in (-3, -1)\), it is a critial point.

For \( -1 < x < 2 \), the derivative is \( f'(x) = \frac{2}{3} x^{-1/3} \), which does not become zero for any finite \( x \). However, \( f'(x) \) does not exist at \( x = 0 \), making \( x = 0 \) another critical point.

Thus, the critical points are \( x = -2, -1, 0 \).

Step 5: Sign Analysis for \( f'(x) \) to Determine Local Maxima and Minima

We analyze the sign of \( f'(x) \) in different intervals determined by the critical points \( x = -2, -1, 0 \).

For \( x < -2 \):

\[ f'(x) = 3(x+2)^2 > 0 \]

\( f(x) \) is increasing.

For \( -2 < x < -1 \):

\[ f'(x) = 3(x+2)^2 > 0 \]

\( f(x) \) is still increasing. Since \( f'(x) \) does not change sign at \( x = -2 \), \( x = -2 \) is not a local extremum.

For \( -1 < x < 0 \):

\[ f'(x) = \frac{2}{3} x^{-1/3} < 0 \]

\( f(x) \) is decreasing. Since \( f'(x) \) changes from positive to negative at \( x = -1 \), we conclude \( x = -1 \) is a local maximum.

For \( 0 < x < 2 \):

\[ f'(x) = \frac{2}{3} x^{-1/3} > 0 \]

\( f(x) \) is increasing. Since \( f'(x) \) changes from negative to positive at \( x = 0 \), we conclude \( x = 0 \) is a local minimum.

Final Answer:

• Local Maximum: \( x = -1 \).
• Local Minimum: \( x = 0 \).

Second Derivative Test

In the previous discussion, \( f \) was assumed to be a continuous function, possibly non-differentiable at some points. The second derivative test, however, applies to a smaller class of functions—those that are differentiable in an interval.

Let \( f \) be a differentiable function on an open interval \( (a, b) \). The idea is the same as before:

1. First, we compute \( f'(x) \) and find all points where \( f'(x) = 0 \). These points are called critical points.
2. Unlike the first derivative test, where we examined sign changes in \( f' \), we now differentiate \( f' \) again to obtain the second derivative \( f''(x) \).
3. We check the sign of \( f''(x) \) at each stationary point (a critical point where \( f'(x) = 0 \)).

Classification Using \( f''(x) \):

• If \( f''(s) > 0 \) at a stationary point \( s \), then \( f \) is concave upward at \( s \), indicating a local minimum.
• If \( f''(s) < 0 \) at a stationary point \( s \), then \( f \) is concave downward at \( s \), indicating a local maximum.
• If \( f''(s) = 0 \), the test is inconclusive—the point may be a local extremum, or it may not. Further analysis (such as higher derivatives or another method) is needed.

Rationale Behind the Test:

At a local minimum, the function transitions from decreasing to increasing, implying an upward curvature—which means \( f''(x) > 0 \). Similarly, at a local maximum, the function transitions from increasing to decreasing, implying a downward curvature—which means \( f''(x) < 0 \). However, when \( f''(x) = 0 \), the curvature test fails, and further investigation is required.

Example

Find the local maximum and local minimum values of the function

\[ f(x) = 3x^4 + 4x^3 - 12x^2 + 12. \]

Solution:

To find the local extrema, we first compute the first derivative:

\[ f'(x) = \frac{d}{dx} (3x^4 + 4x^3 - 12x^2 + 12). \]

\[ = 12x^3 + 12x^2 - 24x. \]

Factoring,

\[ f'(x) = 12x (x^2 + x - 2). \]

Further factorizing,

\[ f'(x) = 12x (x - 1)(x + 2). \]

Setting \( f'(x) = 0 \), we get the critical points:

\[ x = 0, \quad x = 1, \quad x = -2. \]

Next, we compute the second derivative:

\[ f''(x) = \frac{d}{dx} (12x^3 + 12x^2 - 24x). \]

\[ = 36x^2 + 24x - 24. \]

Evaluating at the critical points:

\[ f''(0) = 36(0)^2 + 24(0) - 24 = -24 < 0. \]

\[ f''(1) = 36(1)^2 + 24(1) - 24 = 36 > 0. \]

\[ f''(-2) = 36(-2)^2 + 24(-2) - 24 = 72 > 0. \]

By the second derivative test: - Since \( f''(0) < 0 \), \( x = 0 \) is a local maximum. - Since \( f''(1) > 0 \), \( x = 1 \) is a local minimum. - Since \( f''(-2) > 0 \), \( x = -2 \) is a local minimum.

Now, we compute the function values:

\[ f(0) = 12, \quad f(1) = 7, \quad f(-2) = -20. \]

Final Answer:

• Local Maximum at \( x = 0 \) with \( f(0) = 12 \).
• Local Minima at \( x = 1 \) and \( x = -2 \) with \( f(1) = 7 \) and \( f(-2) = -20 \).

Alternatively, we could have directly applied the first derivative test. Once we obtained

\[ f'(x) = 12x (x - 1)(x + 2), \]

we identified the critical points as \( x = 0, 1, -2 \). Instead of using the second derivative, we analyze the sign changes of \( f'(x) \) around these points.

To do this, we draw a sign chart for \( f'(x) \) and examine its behavior in the intervals determined by the critical points:

• For \( x < -2 \), choose \( x = -3 \):

  \[ f'(-3) = 12(-3)(-3 - 1)(-3 + 2) = 12(-3)(-4)(-1) < 0. \]

  So, \( f'(x) < 0 \) (negative).

• For \( -2 < x < 0 \), choose \( x = -1 \):

  \[ f'(-1) = 12(-1)(-1 - 1)(-1 + 2) = 12(-1)(-2)(1) > 0. \]

  So, \( f'(x) > 0 \) (positive).

• For \( 0 < x < 1 \), choose \( x = 0.5 \):

  \[ f'(0.5) = 12(0.5)(0.5 - 1)(0.5 + 2) = 12(0.5)(-0.5)(2.5) < 0. \]

  So, \( f'(x) < 0 \) (negative).

• For \( x > 1 \), choose \( x = 2 \):

  \[ f'(2) = 12(2)(2 - 1)(2 + 2) = 12(2)(1)(4) > 0. \]

  So, \( f'(x) > 0 \) (positive).

From the sign changes:

• \( f'(x) \) changes from negative to positive at \( x = -2 \), so \( x = -2 \) is a local minimum.
• \( f'(x) \) changes from positive to negative at \( x = 0 \), so \( x = 0 \) is a local maximum.
• \( f'(x) \) changes from negative to positive at \( x = 1 \), so \( x = 1 \) is a local minimum.

Thus, using the first derivative test, we arrive at the same conclusion:

• Local Maximum at \( x = 0 \) with \( f(0) = 12 \).
• Local Minima at \( x = -2 \) and \( x = 1 \) with \( f(-2) = -20 \) and \( f(1) = 7 \).

Handling the Case When \( f''(s) = 0 \) at a Stationary Point

If \( f''(s) = 0 \) at a stationary point \( s \), the second derivative test becomes inconclusive. In such cases, we need an alternative method to determine whether \( s \) is a local maximum, a local minimum, or neither.

Consider the function:

\[ f(x) = x^4. \]

We compute its derivatives:

\[ f'(x) = 4x^3. \]

Setting \( f'(x) = 0 \), we find the stationary point:

\[ 4x^3 = 0 \quad \Rightarrow \quad x = 0. \]

Now, compute the second derivative:

\[ f''(x) = 12x^2. \]

At \( x = 0 \):

\[ f''(0) = 12(0)^2 = 0. \]

Since \( f''(0) = 0 \), the second derivative test is inconclusive.

Alternative Approaches

1. Fallback to the First Derivative Test

   • Checking the sign of \( f'(x) \) on either side of \( x = 0 \), we see that:

     • For \( x < 0 \), \( f'(x) < 0 \) (decreasing).
     • For \( x > 0 \), \( f'(x) > 0 \) (increasing).

   • Since \( f(x) \) changes from decreasing to increasing, \( x = 0 \) is a local minimum.

2. Checking the Sign of \( f''(x) \) Around \( s \)

   • Even though \( f''(0) = 0 \), we examine the sign of \( f''(x) \) for \( x \) near \( 0 \):

     • For \( x > 0 \), \( f''(x) = 12x^2 > 0 \).
     • For \( x < 0 \), \( f''(x) = 12x^2 > 0 \).

   • Since \( f''(x) \) is positive on both sides, the function is concave upwards, confirming \( x = 0 \) as a local minimum.

General Rule When \( f''(s) = 0 \):

• If \( f''(x) \) is positive on both sides of \( s \), then \( s \) is a local minimum (concave up).
• If \( f''(x) \) is negative on both sides, then \( s \) is a local maximum (concave down).
• If \( f''(x) \) changes sign (from \( + \) to \( - \) or vice versa), then \( s \) is neither a maximum nor a minimum, but an inflection point—indicating a change in concavity.

Thus, when the second derivative test is inconclusive (i.e., \( f''(s) = 0 \)), checking the sign of \( f''(x) \) in a small neighborhood of \( s \) is an effective alternative approach.

Another Approach: nth Derivative Test

When the second derivative test fails for a stationary point \( s \), meaning we obtain \( f''(s) = 0 \), we may apply a more general method known as the \( n \)th derivative test. This test can help classify \( s \) as a local maximum, local minimum, or neither when lower-order derivatives provide no information.

\( n \)th Derivative Test

We assume that \( f \) is differentiable at least \( n \) times in a neighborhood of \( s \). If the second derivative test fails at \( s \), meaning \( f''(s) = 0 \), we continue differentiating higher-order derivatives until we find the first nonzero derivative.

1. Compute the third derivative \( f'''(x) \) and evaluate at \( x = s \):

   \[ f'''(s). \]
   • If \( f'''(s) = 0 \), we differentiate again.

2. Compute the fourth derivative \( f^{(4)}(x) \) and evaluate at \( x = s \):

   \[ f^{(4)}(s). \]
   • If \( f^{(4)}(s) = 0 \), we differentiate again.

3. Continue this process until we obtain the first nonzero derivative at some order \( n \), i.e.,

   \[ f^{(n)}(s) \neq 0. \]

Conclusion Based on \( n \):

• If \( n \) is odd, then \( s \) is a point of inflection (not a local extremum). The function does not attain a local maximum or minimum at \( s \), but instead undergoes a change in concavity.

• If \( n \) is even, then the sign of \( f^{(n)}(s) \) determines the nature of the extremum:

  • If \( f^{(n)}(s) > 0 \), then \( s \) is a local minimum.
  • If \( f^{(n)}(s) < 0 \), then \( s \) is a local maximum.

Example: \( f(x) = x^4 \)

Consider \( f(x) = x^4 \), for which we already established that the second derivative test fails at \( x = 0 \). Applying the \( n \)th derivative test:

1. Compute derivatives:

   \[ f'(x) = 4x^3, \quad f''(x) = 12x^2, \quad f'''(x) = 24x, \quad f^{(4)}(x) = 24. \]

2. Evaluate at \( x = 0 \):

   \[ f'(0) = 0, \quad f''(0) = 0, \quad f'''(0) = 0, \quad f^{(4)}(0) = 24. \]

3. Since \( f^{(4)}(0) \neq 0 \) and \( n = 4 \) (even), we check the sign:

   \[ f^{(4)}(0) = 24 > 0. \]

   Since \( n \) is even and \( f^{(n)}(0) > 0 \), \( x = 0 \) is a local minimum.

Example: \( f(x) = x^5 \)

Consider \( f(x) = x^5 \):

1. Compute derivatives:

   \[ f'(x) = 5x^4, \quad f''(x) = 20x^3, \quad f'''(x) = 60x^2, \quad f^{(4)}(x) = 120x, \quad f^{(5)}(x) = 120. \]

2. Evaluate at \( x = 0 \):

   \[ f'(0) = 0, \quad f''(0) = 0, \quad f'''(0) = 0, \quad f^{(4)}(0) = 0, \quad f^{(5)}(0) = 120. \]

3. Since \( f^{(5)}(0) \neq 0 \) and \( n = 5 \) (odd), we conclude that \( x = 0 \) is a point of inflection, not a local extremum.

Conclusion

The \( n \)th derivative test is a more general tool that applies when the second derivative test fails. It provides a systematic way to determine whether a stationary point is a local extremum or a point of inflection. If the first nonzero derivative beyond the second is:

• Of even order (\( n \) is even), we use its sign to classify the point as a local minimum or maximum.
• Of odd order (\( n \) is odd), the point is an inflection point, not an extremum.

Example

Check whether \( g(x) = \cos x - 1 + \frac{x^2}{2!} \) has a local extremum at \(x=0\)

Solution:

We analyze whether \( x = 0 \) is a local extremum for the function

\[ g(x) = \cos x - 1 + \frac{x^2}{2}. \]

First, we compute the first derivative:

\[ g'(x) = \frac{d}{dx} (\cos x - 1 + \frac{x^2}{2}) = -\sin x + x. \]

Setting \( g'(x) = 0 \) to find stationary points, we get

\[ -\sin x + x = 0 \quad \Rightarrow \quad \sin x = x. \]

At \( x = 0 \), clearly \( \sin(0) = 0 \), so \( g'(0) = 0 \), making \( x = 0 \) a stationary point.

Next, we compute the second derivative:

\[ g''(x) = \frac{d}{dx} (-\sin x + x) = -\cos x + 1. \]

Evaluating at \( x = 0 \):

\[ g''(0) = -\cos(0) + 1 = -1 + 1 = 0. \]

Since \( g''(0) = 0 \), the second derivative test is inconclusive. We proceed by computing higher derivatives until we obtain the first nonzero value.

The third derivative is

\[ g'''(x) = \frac{d}{dx} (-\cos x + 1) = \sin x. \]

Evaluating at \( x = 0 \), we get

\[ g'''(0) = \sin(0) = 0. \]

Differentiating again, the fourth derivative is

\[ g^{(4)}(x) = \frac{d}{dx} (\sin x) = \cos x. \]

Evaluating at \( x = 0 \),

\[ g^{(4)}(0) = \cos(0) = 1. \]

Since the first nonzero derivative is the fourth derivative and \( g^{(4)}(0) > 0 \), we apply the \( n \)th derivative test. Since \( n = 4 \) is even and \( g^{(4)}(0) > 0 \), this confirms that \( x = 0 \) is a local minimum.

Thus, \( g(x) = \cos x - 1 + \frac{x^2}{2!} \) has a local minimum at \( x = 0 \).

Global (Absolute) Maximum and Minimum

Let \( f \) be a function defined on a domain \( D \).

• \( f(x) \) has a global (absolute) maximum at \( x = c \) if

  \[ f(c) \geq f(x) \quad \text{for all } x \in D. \]

  The value \( f(c) \) is called the global maximum value of \( f \).

• \( f(x) \) has a global (absolute) minimum at \( x = d \) if

  \[ f(d) \leq f(x) \quad \text{for all } x \in D. \]

  The value \( f(d) \) is called the global minimum value of \( f \).

A function may have local extrema that are not global, meaning they are extreme only within a small neighborhood but not over the entire domain. The global extrema are the true highest and lowest values of the function across its entire domain.

To find the global extremum, we have two approaches: one through algebraic methods, which apply to many functions, and the other through derivatives, which we will focus on first.

We distinguish two cases:
1. When \( f \) is defined on a closed interval.
2. When \( f \) is defined on an open interval.

Case 1: Function Defined on a Closed Interval

Let \( f \) be a continuous function defined on a closed interval \([a, b]\). The process to find the global extrema is straightforward:

1. Find all points of local extrema by solving \( f'(x) = 0 \) or where \( f'(x) \) does not exist. Let these points be \( c_1, c_2, \dots, c_n \) inside \( (a, b) \).
2. Evaluate \( f(x) \) at these local extrema and at the endpoints \( a \) and \( b \).
3. The global maximum occurs at either one of the local maxima or one of the endpoints.
4. The global minimum occurs at either one of the local minima or one of the endpoints.

Thus, the global maximum value is given by:

\[ M = \max \{ f(a), f(c_1), f(c_2), \dots, f(c_n), f(b) \}. \]

Similarly, the global minimum value is:

\[ m = \min \{ f(a), f(c_1), f(c_2), \dots, f(c_n), f(b) \}. \]

Since \( f \) is continuous on a closed interval, the Extreme Value Theorem guarantees that \( f \) attains both a global maximum and a global minimum at some points in \([a, b]\).

Case 2: Function Defined on a Open Interval

For an open interval, we may not have a global minimum or maximum. The absence of closed endpoints means the function may approach extreme values without actually attaining them. To see this, consider an example.

Let

\[ f(x) = x^3, \quad x \in (-1, 2). \]

What are the maximum and minimum values of \( f \) in this interval?

From our previous knowledge, we recognize that \( f(x) = x^3 \) is a strictly increasing function. This means that as \( x \) increases, \( f(x) \) also increases. The function takes values in the range:

\[ (f(-1), f(2)) = (-1, 8). \]

This is an open interval, meaning the function values approach \(-1\) and \( 8 \) but never actually attain them.

Is 8 the maximum value of \( f \)? No, because the interval is open at \( x = 2 \), meaning \( x = 2 \) is not included in the domain. Thus, the function never actually attains the value \( 8 \) for any \( c \in (-1,2) \).

Similarly, the function does not attain \(-1\) as a minimum value, because \( x = -1 \) is also not included in the domain.

Thus, \( f(x) = x^3 \) on \( (-1,2) \) has no global maximum value and no global minimum value.

In contrast consider the following example:

Consider the function

\[ f(x) = x^2 - 2x, \quad x \in (-1,2). \]

What are the maximum and minimum values of \( f \) in this interval?

Since the function is defined on an open interval, we must carefully analyze whether the function attains a maximum or minimum.

From the graph of \( f(x) = x^2 - 2x \), we observe that the function achieves a minimum value at \( x = 1 \). Substituting \( x = 1 \),

\[ f(1) = (1)^2 - 2(1) = 1 - 2 = -1. \]

Thus, the global minimum value is \( -1 \) at \( x = 1 \).

Now, does the function attain a global maximum? The function is increasing for \( x > 1 \), and if the left endpoint \( x = -1 \) were included, it would serve as the maximum point. However, since \( x = -1 \) is not part of the domain, the function does not actually attain this value. Instead, we analyze its limit:

\[ \lim\limits_{x \to -1^+} f(x) = (-1)^2 - 2(-1) = 1 + 2 = 3. \]

Thus, the function approaches 3 but never actually attains it. This means that there is no global maximum value.

Procedure to Calculate Global Extrema for Functions on Open Intervals

For functions defined on open intervals, we must be extra cautious at the endpoints, as the function may not attain extreme values there. The process follows these steps:

1. Find all local extrema by solving \( f'(x) = 0 \) or checking where \( f'(x) \) does not exist. Suppose the local maxima occur at points \( c_1, c_2, \dots, c_n \) and the local minima occur at points \( d_1, d_2, \dots, d_m \).

2. Analyze the function's behavior at the endpoints by computing

   \[ p = \lim\limits_{x \to a^+} f(x), \quad q = \lim\limits_{x \to b^-} f(x). \]

3. Compare all function values at the local extrema and the endpoint limits:

   • The global maximum is determined by

     \[ M = \sup \{ f(c_1), f(c_2), \dots, f(c_n), p, q \}. \]

     If this maximum is attained at some \( x \) in \( (a, b) \), we declare it as the global maximum. If it is never attained, there is no global maximum.

   • The global minimum is determined by

     \[ m = \inf \{ f(d_1), f(d_2), \dots, f(d_m), p, q \}. \]

     If this minimum is attained at some \( x \) in \( (a, b) \), we declare it as the global minimum. Otherwise, no global minimum exists.

Since the function is not defined at \( x = a \) and \( x = b \), the values \( p \) and \( q \) provide only approaching behavior, not actual function values. If the function diverges at an endpoint, meaning \( p = \pm \infty \) or \( q = \pm \infty \), no finite global extremum can exist at that endpoint.

Unlike the closed interval case, where the Extreme Value Theorem guarantees the existence of global extrema, in an open interval, global extrema may fail to exist. Thus, we must carefully consider the endpoint limits when evaluating the function's overall behavior.

Example

Find the maximum and minimum values of

\[ f(x) = x^3 - 3x + 2, \quad x \in [-2,2]. \]

Solution:

The function \( f(x) \) is differentiable on \( [-2,2] \), so we proceed by finding its critical points.

Computing the first derivative:

\[ f'(x) = \frac{d}{dx} (x^3 - 3x + 2) = 3x^2 - 3. \]

Setting \( f'(x) = 0 \) to find critical points:

\[ 3x^2 - 3 = 0. \]

\[ 3(x^2 - 1) = 0. \]

\[ 3(x - 1)(x + 1) = 0. \]

Thus, the critical points are \( x = -1, 1 \).

Now, applying the second derivative test to classify these points:

\[ f''(x) = \frac{d}{dx} (3x^2 - 3) = 6x. \]

Evaluating at \( x = -1 \):

\[ f''(-1) = 6(-1) = -6 < 0. \]

Since \( f''(-1) < 0 \), \( x = -1 \) is a local maximum.

Evaluating at \( x = 1 \):

\[ f''(1) = 6(1) = 6 > 0. \]

Since \( f''(1) > 0 \), \( x = 1 \) is a local minimum.

Now, to find the global maximum and minimum, we compare function values at critical points and endpoints.

Computing function values:

\[ f(-2) = (-2)^3 - 3(-2) + 2 = -8 + 6 + 2 = 0. \]

\[ f(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4. \]

\[ f(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0. \]

\[ f(2) = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4. \]

Thus, the maximum value is:

\[ \max \{ f(-2), f(-1), f(2) \} = \max \{ 0, 4, 4 \} = 4. \]

And the minimum value is:

\[ \min \{ f(-2), f(1), f(2) \} = \min \{ 0, 0, 4 \} = 0. \]

Thus,

• Maximum value = 4, occurring at \( x = -1 \) and \( x = 2 \).
• Minimum value = 0, occurring at \( x = -2 \) and \( x = 1 \).

Example

Find the maximum and minimum values of

\[ g(x) = x^x, \quad x > 0. \]

Solution:

The function \( g(x) = x^x \) is differentiable for \( x > 0 \). Expressing it as \( g(x) = e^{x \ln x} \), differentiation gives

\[ g'(x) = e^{x \ln x} \cdot (1 + \ln x) = x^x (1 + \ln x). \]

Setting \( g'(x) = 0 \) gives \( x^x (1 + \ln x) = 0 \), which is possible only if \( 1 + \ln x = 0 \), yielding \( x = e^{-1} = \frac{1}{e} \). This is the only critical point.

To determine its nature, we check the sign of \( g'(x) \). For \( x > \frac{1}{e} \), choosing \( x = 1 \) gives \( g'(1) = 1^1(1 + \ln 1) = 1(1+0) = 1 > 0 \), so \( g(x) \) is increasing for \( x > \frac{1}{e} \). For \( 0 < x < \frac{1}{e} \), choosing \( x = e^{-2} \) gives \( g'(e^{-2}) = e^{-2e^{-2}}(1 + \ln e^{-2}) = e^{-2e^{-2}}(1-2) < 0 \), so \( g(x) \) is decreasing for \( 0 < x < \frac{1}{e} \). Since \( g'(x) \) changes from negative to positive at \( x = \frac{1}{e} \), it is a local minimum.

For global behavior, consider the limits. As \( x \to 0^+ \),

\[ \lim\limits_{x \to 0^+} x^x = \lim\limits_{x \to 0^+} e^{x \ln x}. \]

Since \( x \ln x = \frac{\ln x}{1/x} \) is of the form \( \frac{-\infty}{\infty} \), applying L’Hôpital’s Rule gives

\[ \lim\limits_{x \to 0^+} \frac{\ln x}{1/x} = \lim\limits_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim\limits_{x \to 0^+} -x = 0. \]

Thus, \( \lim\limits_{x \to 0^+} x^x = e^0 = 1 \).

As \( x \to \infty \), we observe \( \lim\limits_{x \to \infty} x^x = \infty \), so no global maximum exists. The global minimum must be determined by

\[ \min \left\{ g\left(\frac{1}{e}\right), \lim\limits_{x \to 0^+} g(x) \right\} = \min \left\{ e^{-1/e}, 1 \right\}. \]

Since \( e^{-1/e} < 1 \), the global minimum is attained at \( x = \frac{1}{e} \) with minimum value

\[ g\left(\frac{1}{e}\right) = e^{-1/e}. \]

Single Critical Point Theorem

The Single Critical Point Theorem states that if a function has exactly one critical point in an interval, and that point is a local extremum, then it must also be the global extremum.

Theorem Statement

Let \( f \) be a differentiable function on an interval \( I \), and suppose that \( f \) has exactly one critical point \( c \) in \( I \) (i.e., \( f'(c) = 0 \) and there are no other critical points in \( I \)). Then:

• If \( f \) has a local maximum at \( c \), then \( f(c) \) is the global maximum of \( f \) over \( I \).
• If \( f \) has a local minimum at \( c \), then \( f(c) \) is the global minimum of \( f \) over \( I \).

Proof

Since \( f \) has only one critical point, it must be monotonic on both sides of \( c \). This means that the derivative does not vanish or change direction anywhere else in \( I \). If \( f \) is increasing before \( c \) and decreasing after \( c \), then \( c \) is the highest point in \( I \), making it the global maximum. Similarly, if \( f \) is decreasing before \( c \) and increasing after \( c \), then \( c \) is the global minimum.

This conclusion follows from the First Derivative Test: since \( f' \) has only one zero, \( f(x) \) cannot attain a greater or smaller value at any other point in \( I \). Thus, the local extremum at \( c \) must also be the absolute extremum.

\(\blacksquare\)

This theorem is particularly useful in situations where checking endpoints or analyzing the global behavior of the function might be tedious. If we establish that a function has exactly one local extremum within an interval, the theorem guarantees that this extremum is also the global extremum, eliminating the need to verify function values at endpoints or evaluate additional limits.

For example, consider the function \( f(x) = x^x \) for \( x > 0 \). Previously, we found that the only critical point occurs at \( x = \frac{1}{e} \), obtained by solving \( f'(x) = x^x(1 + \ln x) = 0 \), which leads to \( \ln x = -1 \) and hence \( x = e^{-1} \). By analyzing the sign of \( f'(x) \), we concluded that \( x = \frac{1}{e} \) is a local minimum.

Since this function is defined on an open interval \( (0, \infty) \), we might typically check the behavior of \( f(x) \) as \( x \to 0^+ \) and \( x \to \infty \) to determine whether \( x = \frac{1}{e} \) is indeed the global minimum. However, applying the Single Critical Point Theorem, we recognize that \( x = \frac{1}{e} \) is the only critical point in \( (0, \infty) \). Since it is a local minimum and no other extrema exist, it must also be the global minimum by virtue of the theorem.

Thus, we do not need to explicitly evaluate the limiting values \( \lim\limits_{x \to 0^+} x^x \) or \( \lim\limits_{x \to \infty} x^x \), nor do we need to compare function values across the interval. The theorem allows us to immediately conclude that

\[ \min_{x > 0} f(x) = f(1/e) = e^{-1/e}. \]

This application of the Single Critical Point Theorem not only confirms the correctness of our conclusion but also eliminates unnecessary computations that would otherwise be required to establish the global minimum.

Using Algebra to Find Global Maximum and Minimum

We use several algebraic techniques to evaluate the maximum and minimum values of functions, depending on the nature of the function itself. One common technique is transforming the expression through algebraic manipulation into a form where only a single term depends on \( x \), whose range is either known or easy to compute. By isolating this term, we can directly determine the range of the entire function without differentiation.

Consider \( f(x) = \frac{1}{2 + \sin x}, \quad x \in \mathbb{R} \), let us find the maximum and minimum value of this fucntion.

The function depends on \( x \) only through \( \sin x \). Since the well-known range of the sine function is

\[ -1 \leq \sin x \leq 1, \]

we transform the denominator:

\[ 2 + \sin x. \]

Adding 2 throughout the inequality gives:

\[ 1 \leq 2 + \sin x \leq 3. \]

Now, reciprocating (which reverses the inequality because all terms are positive):

\[ \frac{1}{3} \leq \frac{1}{2 + \sin x} \leq 1. \]

Thus, the range of \( f(x) \) is:

\[ \left[ \frac{1}{3}, 1 \right]. \]

To determine when these values occur, we recall that the function \( f(x) = \frac{1}{2 + \sin x} \) reaches its global maximum when its denominator is smallest, which happens when \( \sin x = -1 \). The sine function achieves \( \sin x = -1 \) at

\[ x = 2n\pi - \frac{\pi}{2}, \quad n \in \mathbb{Z}. \]

Substituting \( \sin x = -1 \) into \( f(x) \), we get:

\[ f(x) = \frac{1}{2 + (-1)} = \frac{1}{1} = 1. \]

Thus, the global maximum value is \( 1 \), occurring at:

\[ x = 2n\pi - \frac{\pi}{2}, \quad n \in \mathbb{Z}. \]

Similarly, the function reaches its global minimum when its denominator is largest, which happens when \( \sin x = 1 \). The sine function achieves \( \sin x = 1 \) at

\[ x = 2n\pi + \frac{\pi}{2}, \quad n \in \mathbb{Z}. \]

Substituting \( \sin x = 1 \) into \( f(x) \), we get:

\[ f(x) = \frac{1}{2 + (1)} = \frac{1}{3}. \]

Thus, the global minimum value is \( \frac{1}{3} \), occurring at:

\[ x = 2n\pi + \frac{\pi}{2}, \quad n \in \mathbb{Z}. \]

This confirms that without using derivatives, we can algebraically determine the global extrema of the function solely by analyzing the range of \( \sin x \) and applying transformations.

Example

Let \( f(x) = \frac{\cos x}{2 + \cos x}, \quad x \in \mathbb{R} \)

We aim to determine the global maximum and minimum values of \( f(x) \) using algebraic manipulation. The given function contains two terms that depend on \( x \), namely \( \cos x \) in both the numerator and denominator. To simplify the expression, we rewrite it as follows:

\[ f(x) = \frac{\cos x + 2 - 2}{2 + \cos x} = 1 - \frac{2}{2 + \cos x}. \]

Now, the expression consists of the constant term 1 and a fraction that depends only on \( \cos x \). This allows us to analyze the range of \( f(x) \) using the known range of \( \cos x \), which is

\[ -1 \leq \cos x \leq 1. \]

Thus, the denominator satisfies

\[ 2 - 1 \leq 2 + \cos x \leq 2 + 1, \]

which simplifies to

\[ 1 \leq 2 + \cos x \leq 3. \]

Taking reciprocals (which reverses the inequality because all terms are positive), we obtain

\[ \frac{1}{3} \leq \frac{1}{2 + \cos x} \leq 1. \]

Multiplying by \(-2\) (which reverses the inequality again),

\[ -2 \leq -\frac{2}{2 + \cos x} \leq -\frac{2}{3}. \]

Adding 1 throughout the inequality,

\[ 1 - 2 \leq 1 - \frac{2}{2 + \cos x} \leq 1 - \frac{2}{3}, \]

which simplifies to

\[ -1 \leq f(x) \leq \frac{1}{3}. \]

Thus, the global maximum of \( f(x) \) is \( \frac{1}{3} \), and the global minimum is \( -1 \).

To determine where these extrema occur, recall that \( f(x) = 1 - \frac{2}{2 + \cos x} \) attains its maximum when the fraction \( \frac{2}{2 + \cos x} \) is smallest, which happens when \( \cos x \) is largest. Since \( \cos x = 1 \) at

\[ x = 2n\pi, \quad n \in \mathbb{Z}, \]

substituting \( \cos x = 1 \) gives

\[ f(x) = 1 - \frac{2}{2 + 1} = 1 - \frac{2}{3} = \frac{1}{3}. \]

Thus, the global maximum value \( \frac{1}{3} \) occurs at

\[ x = 2n\pi, \quad n \in \mathbb{Z}. \]

Similarly, \( f(x) \) attains its minimum when the fraction \( \frac{2}{2 + \cos x} \) is largest, which happens when \( \cos x \) is smallest at \( \cos x = -1 \). This occurs at

\[ x = (2n+1)\pi, \quad n \in \mathbb{Z}. \]

Substituting \( \cos x = -1 \) gives

\[ f(x) = 1 - \frac{2}{2 - 1} = 1 - 2 = -1. \]

Thus, the global minimum value \( -1 \) occurs at

\[ x = (2n+1)\pi, \quad n \in \mathbb{Z}. \]

This example illustrates how algebraic manipulation allows us to determine global extrema efficiently by reducing the function to a form where only a single term depends on \( x \).

Example

Find the global maximum and minimum values of

\[ f(x) = \frac{x^2}{1 + 2x^2}, \quad x \in \mathbb{R}. \]

Solution:

The function contains two terms that depend on \( x \), namely \( x^2 \) in both the numerator and denominator. To simplify, we express \( f(x) \) in a form where only one term depends on \( x \). Rewriting,

\[ f(x) = \frac{x^2}{1+2x^2}. \]

Factorizing the numerator in terms of the denominator,

\[ f(x) = \frac{1}{2} \cdot \frac{2x^2}{1 + 2x^2}. \]

Recognizing that we can write \( 2x^2 \) in terms of \( 1 + 2x^2 \),

\[ f(x) = \frac{1}{2} \cdot \frac{1 + 2x^2 - 1}{1 + 2x^2} = \frac{1}{2} \left(1 - \frac{1}{1+2x^2} \right). \]

Now, only one term in the expression depends on \( x \), namely \( \frac{1}{1+2x^2} \).

Since \( x^2 \geq 0 \) for all \( x \in \mathbb{R} \), we analyze its effect on the expression:

\[ 0 \leq x^2 < \infty \quad \Rightarrow \quad 0 \leq 2x^2 < \infty. \]

Adding 1 throughout,

\[ 1 \leq 1 + 2x^2 < \infty. \]

Taking reciprocals (which reverses the inequality because all terms are positive),

\[ 0 < \frac{1}{1+2x^2} \leq 1. \]

Multiplying -1 thoughout,

\[ -1 \leq -\frac{1}{1+2x^2} < 0. \]

Adding 1 throughout,

\[ 0 \leq 1 - \frac{1}{1+2x^2} < 1. \]

Multiplying by \( \frac{1}{2} \),

\[ 0 \leq \frac{1}{2} \left(1 - \frac{1}{1+2x^2} \right) < \frac{1}{2}. \]

Thus, the range of \( f(x) \) is:

\[ \left[0, \frac{1}{2} \right). \]

The minimum value is clearly 0, which occurs at \( x = 0 \), since substituting \( x = 0 \) gives

\[ f(0) = \frac{0^2}{1+2(0)^2} = 0. \]

For the maximum, observe that \( f(x) \to \frac{1}{2} \) as \( |x| \to \infty \), but the function never actually attains this value. This is because as \( |x| \to \infty \), we get

\[ \frac{1}{1+2x^2} \to 0 \quad \Rightarrow \quad f(x) \to \frac{1}{2}. \]

However, no finite \( x \) satisfies \( f(x) = \frac{1}{2} \). Thus, there is no global maximum, since \( f(x) \) approaches but never reaches \( \frac{1}{2} \).

Hence,

• Global minimum: \( 0 \) at \( x = 0 \).
• No global maximum (supremum is \( \frac{1}{2} \), but it is not attained).

Example

Find the maximum and minimum values of the function

\[ f(x) = \sin^2 x - 3\sin x + 2, \quad x \in \mathbb{R}. \]

Solution:

To determine the maximum and minimum values, we utilize the fact that \( f(x) \) is a quadratic expression in \( \sin x \). We transform it using the completing the square method to rewrite it in a form where only one term depends on \( x \).

Rewriting the function,

\[ f(x) = \sin^2 x - 3\sin x + 2. \]

Completing the square in terms of \( \sin x \),

\[ f(x) = \sin^2 x - 2 \cdot \frac{3}{2} \sin x + \left( \frac{3}{2} \right)^2 - \left( \frac{3}{2} \right)^2 + 2. \]

This simplifies to

\[ f(x) = \left( \sin x - \frac{3}{2} \right)^2 - \frac{9}{4} + 2. \]

\[ f(x) = \left( \sin x - \frac{3}{2} \right)^2 - \frac{1}{4}. \]

Now, only one term depends on \( x \) through \( \sin x \). Using the range of \( \sin x \),

\[ -1 \leq \sin x \leq 1. \]

Subtracting \( \frac{3}{2} \) throughout,

\[ -1 - \frac{3}{2} \leq \sin x - \frac{3}{2} \leq 1 - \frac{3}{2}. \]

\[ -\frac{5}{2} \leq \sin x - \frac{3}{2} \leq -\frac{1}{2}. \]

Squaring both sides,

\[ \left( -\frac{5}{2} \right)^2 \geq \left( \sin x - \frac{3}{2} \right)^2 \geq \left( -\frac{1}{2} \right)^2. \]

\[ \frac{25}{4} \geq \left( \sin x - \frac{3}{2} \right)^2 \geq \frac{1}{4}. \]

Subtracting \( \frac{1}{4} \) throughout,

\[ \frac{25}{4} - \frac{1}{4} \geq \left( \sin x - \frac{3}{2} \right)^2 - \frac{1}{4} \geq \frac{1}{4} - \frac{1}{4}. \]

\[ \frac{24}{4} \geq f(x) \geq 0. \]

\[ 6 \geq f(x) \geq 0. \]

Thus, the maximum value of \( f(x) \) is 6, and the minimum value is 0. These occur at values of \( x \) where the squared term attains its extreme values:

• The maximum value occurs when \( \sin x = -1 \), which gives \( x = 2n\pi - \frac{\pi}{2}, n \in \mathbb{Z} \).
• The minimum value occurs when \( \sin x = 1 \), which gives \( x = 2n\pi + \frac{\pi}{2}, n \in \mathbb{Z} \).

Thus, the function attains all values in the interval \( [0,6] \), completing the determination of its range.

Example

Find the range of the function

\[ f(x) = \cos^2 x - \cos x, \quad x \in \mathbb{R}. \]

Solution:

Since the function is quadratic in \( \cos x \), we use the completing the square method to rewrite it in a form where only one term depends on \( x \).

Rewriting,

\[ f(x) = \cos^2 x - \cos x. \]

Completing the square in terms of \( \cos x \),

\[ f(x) = \cos^2 x - 2 \cdot \frac{1}{2} \cos x + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2. \]

\[ f(x) = (\cos x - \frac{1}{2})^2 - \frac{1}{4}. \]

Now, only one term depends on \( x \) through \( \cos x \). Using the known range of \( \cos x \),

\[ -1 \leq \cos x \leq 1. \]

Subtracting \( \frac{1}{2} \) throughout,

\[ -1 - \frac{1}{2} \leq \cos x - \frac{1}{2} \leq 1 - \frac{1}{2}. \]

\[ -\frac{3}{2} \leq \cos x - \frac{1}{2} \leq \frac{1}{2}. \]

Since the interval contains both negative and positive values, squaring reverses the sign of negative values, and the squared term satisfies

\[ 0 \leq (\cos x - \frac{1}{2})^2 \leq \max \left( \left(-\frac{3}{2} \right)^2, \left(\frac{1}{2} \right)^2 \right). \]

\[ 0 \leq (\cos x - \frac{1}{2})^2 \leq \max \left(\frac{9}{4}, \frac{1}{4} \right) = \frac{9}{4}. \]

Subtracting \( \frac{1}{4} \) throughout,

\[ -\frac{1}{4} \leq (\cos x - \frac{1}{2})^2 - \frac{1}{4} \leq \frac{9}{4} - \frac{1}{4}. \]

\[ -\frac{1}{4} \leq f(x) \leq 2. \]

Thus, the range of \( f(x) \) is:

\[ \left[ -\frac{1}{4}, 2 \right]. \]

The function attains its maximum value \( 2 \) when \( (\cos x - \frac{1}{2})^2 = \frac{9}{4} \), which happens at \( \cos x = -1 \), i.e.,

\[ x = (2n+1)\pi, \quad n \in \mathbb{Z}. \]

The function attains its minimum value \( -\frac{1}{4} \) when \( (\cos x - \frac{1}{2})^2 = 0 \), which happens at \( \cos x = \frac{1}{2} \), i.e.,

\[ x = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z}. \]

Thus, the function takes all values in the interval \( \left[ -\frac{1}{4}, 2 \right] \), completing the determination of its range.

Application of the AM-GM Inequality in Finding Maximum and Minimum Values

The Arithmetic Mean - Geometric Mean (AM-GM) Inequality states that for any set of non-negative real numbers \( a_1, a_2, \dots, a_n \),

\[ \frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \dots a_n}, \]

with equality holding if and only if \( a_1 = a_2 = \dots = a_n \).

This inequality is particularly useful in optimization problems where the function consists of non-negative terms whose product is constant or has a simple bound. By applying AM-GM, we can establish lower or upper bounds on the function without using derivatives.

Using \(AM\ge GM\)

Find the maximum and minimum values of

\[ f(x) = 3\tan^2 x + 2\cot^2 x, \quad x \in \mathbb{R}. \]

Solution:

We observe that \( f(x) \) consists of two non-negative terms, and their product is constant. This suggests that we can apply the Arithmetic Mean - Geometric Mean (AM-GM) Inequality, which states that for any non-negative numbers \( a \) and \( b \),

\[ \frac{a + b}{2} \geq \sqrt{ab}. \]

Applying this to our function, setting \( a = 3\tan^2 x \) and \( b = 2\cot^2 x \),

\[ \frac{3\tan^2 x + 2\cot^2 x}{2} \geq \sqrt{(3\tan^2 x) (2\cot^2 x)}. \]

Since \( \tan^2 x \cot^2 x = 1 \), we get

\[ \sqrt{6 \tan^2 x \cot^2 x} = \sqrt{6}. \]

Multiplying both sides by 2,

\[ f(x) = 3\tan^2 x + 2\cot^2 x \geq 2\sqrt{6}. \]

Thus, the minimum value of \( f(x) \) is \( 2\sqrt{6} \).

To check if this minimum is attained, equality in AM-GM holds when

\[ 3\tan^2 x = 2\cot^2 x. \]

Rearranging,

\[ 3\tan^2 x \cot^2 x = 2. \]

Substituting \( \tan^2 x \cot^2 x = 1 \),

\[ 3 = 2\cot^2 x. \]

\[ \cot^2 x = \frac{3}{2}. \]

Taking square roots,

\[ \cot x = \pm\sqrt{\frac{3}{2}}. \]

Since \( \tan x = \frac{1}{\cot x} \),

\[ \tan x = \pm\sqrt{\frac{2}{3}}. \]

Thus, the minimum value is attained when \( x \) satisfies

\[ \tan x = \pm\sqrt{\frac{2}{3}}. \]

For the maximum value, we analyze the behavior as \( x \to \frac{\pi}{2} \). As \( x \to \frac{\pi}{2} \),

• \( \tan^2 x \to \infty \),
• \( \cot^2 x \to 0 \),

which implies

\[ f(x) = 3\tan^2 x + 2\cot^2 x \to \infty. \]

Since \( f(x) \) is unbounded above, it has no global maximum.

Conclusion:

• Minimum value: \( 2\sqrt{6} \), attained when \( \tan x = \pm\sqrt{\frac{2}{3}} \).
• No global maximum, since \( f(x) \to \infty \) as \( x \to \frac{\pi}{2} \).

This example demonstrates how AM-GM inequality provides an efficient method for some kind of fucntions to determine extrema without differentiation.

Example

If \( x > 0 \) and \( y > 0 \) satisfy \( x + y = 10 \), find the maximum value of \( xy \).

Solution:

Applying the AM-GM inequality,

\[ \frac{x + y}{2} \geq \sqrt{xy}. \]

Substituting \( x + y = 10 \),

\[ \frac{10}{2} \geq \sqrt{xy}. \]

\[ 5 \geq \sqrt{xy}. \]

Squaring both sides,

\[ 25 \geq xy. \]

Thus, the maximum possible value of \( xy \) is 25.

To check if this value is attainable, equality in AM-GM holds when \( x = y \). Since \( x + y = 10 \), setting \( x = y \) gives

\[ 2x = 10 \quad \Rightarrow \quad x = 5, y = 5. \]

Substituting,

\[ xy = 5 \times 5 = 25. \]

Thus, the maximum value of \( xy \) is 25, which occurs when \( x = 5, y = 5 \).

Example

If \( x > 0 \) and \( y > 0 \) satisfy \( x + y = 10 \), find the maximum value of \( x^2 y \).

Solution:

Split \( x \) into two equal parts as \( \frac{x}{2} \) and \( \frac{x}{2} \) in the sum \( x + y \), so that

\[ x + y = \frac{x}{2} + \frac{x}{2} + y. \]

Applying the AM-GM inequality to the three non-negative terms \( \frac{x}{2}, \frac{x}{2}, y \), we obtain

\[ \frac{\frac{x}{2} + \frac{x}{2} + y}{3} \geq \sqrt[3]{\frac{x}{2} \cdot \frac{x}{2} \cdot y}. \]

Since \( x + y = 10 \), this simplifies to

\[ \frac{10}{3} \geq \sqrt[3]{\frac{x^2 y}{4}}. \]

Cubing both sides,

\[ \left( \frac{10}{3} \right)^3 \geq \frac{x^2 y}{4}. \]

\[ \frac{1000}{27} \geq \frac{x^2 y}{4}. \]

Multiplying by 4,

\[ \frac{4000}{27} \geq x^2 y. \]

Thus, the maximum value of \( x^2 y \) is \( \frac{4000}{27} \).

To check when equality holds, in AM-GM, equality occurs when all terms are equal, i.e.,

\[ \frac{x}{2} = \frac{x}{2} = y. \]

This implies

\[ x = 2y. \]

Substituting into \( x + y = 10 \),

\[ 2y + y = 10 \quad \Rightarrow \quad 3y = 10 \quad \Rightarrow \quad y = \frac{10}{3}, \quad x = \frac{20}{3}. \]

Thus, the maximum value \( \frac{4000}{27} \) occurs when \( x = \frac{20}{3}, y = \frac{10}{3} \).

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Alternate Solution Using Calculus:

The given condition \( x + y = 10 \) implies \( y = 10 - x \). Substituting into the function,

\[ x^2 y = x^2 (10 - x). \]

Defining

\[ f(x) = 10x^2 - x^3. \]

Since \( f(x) \) is differentiable, we differentiate to find the critical points:

\[ f'(x) = 20x - 3x^2. \]

Setting \( f'(x) = 0 \),

\[ 20x - 3x^2 = 0. \]

Factoring,

\[ x(20 - 3x) = 0. \]

Solving for \( x \), we get

\[ x = 0 \quad \text{or} \quad x = \frac{20}{3}. \]

Since \( x > 0 \), we discard \( x = 0 \), leaving only one critical point, \( x = \frac{20}{3} \).

Applying the second derivative test,

\[ f''(x) = 20 - 6x. \]

Evaluating at \( x = \frac{20}{3} \),

\[ f''\left(\frac{20}{3}\right) = 20 - 6 \times \frac{20}{3} = 20 - 40 = -20 < 0. \]

Since \( f''(x) < 0 \), \( x = \frac{20}{3} \) is a local maximum. By the Single Critical Point Theorem, since there is only one critical point in the domain, it must also be the global maximum.

Computing the maximum value,

\[ f\left(\frac{20}{3}\right) = 10 \left(\frac{20}{3}\right)^2 - \left(\frac{20}{3}\right)^3. \]

\[ = 10 \times \frac{400}{9} - \frac{8000}{27} = \frac{4000}{27}. \]

Thus, the maximum value of \( x^2 y \) is \( \frac{4000}{27} \), attained at \( x = \frac{20}{3}, y = \frac{10}{3} \).

Cauchy-Schwarz Inequality and Its Lemma

The Cauchy-Schwarz inequality and its lemma are fundamental results in inequalities that can be ingeniously applied to functions to determine their maximum and minimum values. These results allow us to bound expressions involving sums and products, making them powerful tools in optimization problems without requiring calculus.

Cauchy-Schwarz Inequality

For any real numbers \( a_1, a_2, \dots, a_n \) and \( b_1, b_2, \dots, b_n \), the following inequality holds:

\[ (a_1b_1 + a_2b_2 + \dots + a_n b_n)^2 \leq (a_1^2 + a_2^2 + \dots + a_n^2)(b_1^2 + b_2^2 + \dots + b_n^2). \]

Equality holds if and only if there exists a constant \( k \) such that

\[ \frac{a_1}{b_1} = \frac{a_2}{b_2} = \dots = \frac{a_n}{b_n} = k, \]

for all indices where \( b_i \neq 0 \). If any \( b_i = 0 \), then the corresponding \( a_i \) must also be 0.

Cauchy-Schwarz Lemma

For any real numbers \( a_1, a_2, \dots, a_n \) and positive real numbers \( b_1, b_2, \dots, b_n \), we have:

\[ \frac{a_1^2}{b_1} + \frac{a_2^2}{b_2} + \dots + \frac{a_n^2}{b_n} \geq \frac{(a_1 + a_2 + \dots + a_n)^2}{b_1 + b_2 + \dots + b_n}. \]

Equality holds if and only if there exists a constant \( k \) such that

\[ \frac{a_1}{b_1} = \frac{a_2}{b_2} = \dots = \frac{a_n}{b_n} = k. \]

Example

Find the minimum value of

\[ g(x) = \frac{5}{\sin^2 x} + \frac{2}{\cos^2 x}, \quad x \in \mathbb{R}, x \neq n\pi, x \neq \frac{\pi}{2} + n\pi. \]

Solution:

We express the given function in a form suitable for applying Cauchy-Schwarz Lemma:

\[ g(x) = \frac{(\sqrt{5})^2}{\sin^2 x} + \frac{(\sqrt{2})^2}{\cos^2 x}. \]

By Cauchy-Schwarz Lemma,

\[ \frac{a_1^2}{b_1} + \frac{a_2^2}{b_2} \geq \frac{(a_1 + a_2)^2}{b_1 + b_2}. \]

Applying this to our function with \( a_1 = \sqrt{5} \), \( a_2 = \sqrt{2} \), \( b_1 = \sin^2 x \), and \( b_2 = \cos^2 x \), we get

\[ \frac{(\sqrt{5})^2}{\sin^2 x} + \frac{(\sqrt{2})^2}{\cos^2 x} \geq \frac{(\sqrt{5} + \sqrt{2})^2}{\sin^2 x + \cos^2 x}. \]

Since \( \sin^2 x + \cos^2 x = 1 \), this simplifies to

\[ g(x) \geq (\sqrt{5} + \sqrt{2})^2. \]

Thus, the minimum value of \( g(x) \) is \( (\sqrt{5} + \sqrt{2})^2 \).

To check when equality holds, we use the equality condition from Cauchy-Schwarz Lemma, which states that

\[ \frac{\sqrt{5}}{\sin^2 x} = \frac{\sqrt{2}}{\cos^2 x}. \]

Rearranging,

\[ \frac{\sin^2 x}{\cos^2 x} = \frac{\sqrt{5}}{\sqrt{2}}. \]

\[ \tan^2 x = \sqrt{\frac{5}{2}}. \]

Thus, the minimum value \( (\sqrt{5} + \sqrt{2})^2 \) occurs when

\[ \tan^2 x = \sqrt{\frac{5}{2}}. \]

Example

Let \( x, y, z \in \mathbb{R} \) satisfy

\[ x^2 + y^2 + z^2 = 5. \]

Find the maximum value of

\[ 3x + 4y + 5z. \]

Solution:

Applying the Cauchy-Schwarz inequality,

\[ (3x + 4y + 5z)^2 \leq (3^2 + 4^2 + 5^2)(x^2 + y^2 + z^2). \]

Since

\[ 9 + 16 + 25 = 50 \quad \text{and} \quad x^2 + y^2 + z^2 = 5, \]

we obtain

\[ (3x + 4y + 5z)^2 \leq 50 \times 5 = 250. \]

Taking square roots,

\[ -5\sqrt{10} \leq 3x + 4y + 5z \leq 5\sqrt{10}. \]

Thus, the maximum value is \( 5\sqrt{10} \).

To check when equality holds, we use the equality condition in Cauchy-Schwarz, which states that

\[ \frac{x}{3} = \frac{y}{4} = \frac{z}{5}. \]

Let this common ratio be \( k \), so that

\[ x = 3k, \quad y = 4k, \quad z = 5k. \]

Substituting into the given constraint,

\[ (3k)^2 + (4k)^2 + (5k)^2 = 5. \]

\[ 9k^2 + 16k^2 + 25k^2 = 5. \]

\[ 50k^2 = 5. \]

Solving for \( k \),

\[ k = \frac{1}{\sqrt{10}}. \]

Thus, the values of \( x, y, z \) at which the maximum occurs are:

\[ x = \frac{3}{\sqrt{10}}, \quad y = \frac{4}{\sqrt{10}}, \quad z = \frac{5}{\sqrt{10}}. \]

Geometrical Application of Maxima and Minima

In engineering design, optimizing material usage is crucial for cost efficiency. Suppose you are tasked with designing a closed cylindrical tin box intended to store powdered milk. The box must have a fixed volume of \( V \) cubic units, but you have the freedom to choose its radius and height. The goal is to minimize the amount of tin used in constructing the box, as more material results in higher manufacturing costs.

Since the amount of tin required is directly proportional to the surface area of the cylinder, minimizing the tin usage translates to minimizing the total surface area while ensuring that the volume remains \( V \). Thus, the problem reduces to determining the optimal values of the radius and height that achieve this minimum.

To proceed with the optimization, let us define the necessary mathematical expressions. Let the radius of the cylindrical box be \( r \) and the height be \( h \). The volume of the box is given by

\[ V = \pi r^2 h \]

Since the volume is fixed, the variables \( r \) and \( h \) are not independent; they must satisfy this equation. We refer to such an equation as a constraint because it restricts our choice of \( r \) and \( h \).

Now, let \( S \) denote the total surface area of the cylindrical box. The surface area consists of two circular ends and the lateral curved surface. Hence,

\[ S = 2\pi r h + 2\pi r^2 \]

Here, \( S \) is a function of both \( r \) and \( h \), but we can express it in terms of a single variable using the constraint equation. The choice of which variable to eliminate is arbitrary; here, we eliminate \( h \). Solving for \( h \) from the volume equation,

\[ h = \frac{V}{\pi r^2} \]

Substituting this into the expression for \( S \),

\[ S = 2\pi r \cdot \frac{V}{\pi r^2} + 2\pi r^2 \]

Simplifying,

\[ S = \frac{2V}{r} + 2\pi r^2 \]

Now, \( S \) is expressed as a function of \( r \) alone. The problem reduces to finding the value of \( r \) that minimizes this function, ensuring the optimal design of the cylindrical box using the least amount of tin.

To find the optimal value of \( r \) that minimizes the surface area \( S \), we differentiate \( S \) with respect to \( r \):

\[ \frac{dS}{dr} = \frac{d}{dr} \left( \frac{2V}{r} + 2\pi r^2 \right) \]

Differentiating term by term,

\[ \frac{dS}{dr} = -\frac{2V}{r^2} + 4\pi r \]

To find the critical points, we set \( \frac{dS}{dr} = 0 \):

\[ -\frac{2V}{r^2} + 4\pi r = 0 \]

Solving for \( r \),

\[ r^3 = \frac{2V}{4\pi} = \frac{V}{2\pi} \]

\[ r = \left( \frac{V}{2\pi} \right)^{\frac{1}{3}} \]

To confirm that the critical point we found corresponds to a minimum, we apply the first or second derivative test. We apply second derivative test here for its simplicity in this problem:

We compute the second derivative of \( S \):

\[ \frac{d^2S}{dr^2} = \frac{d}{dr} \left( -\frac{2V}{r^2} + 4\pi r \right). \]

Differentiating term by term,

\[ \frac{d^2S}{dr^2} = \frac{4V}{r^3} + 4\pi. \]

Since both terms are positive for all \( r > 0 \), we conclude that

\[ \frac{d^2S}{dr^2} > 0 \quad \text{for all } r > 0. \]

Thus, \( S \) is concave upward at the critical point, confirming that it is a local minimum. By Single Critical Point Theorem we can claim that it is also the global minimum.

Once \( r \) is determined, we substitute it back into the volume constraint equation to find \( h \):

\[ h = \frac{V}{\pi r^2} \]

Substituting \( r = \left( \frac{V}{2\pi} \right)^{\frac{1}{3}} \),

\[ h = \frac{V}{\pi \left( \frac{V}{2\pi} \right)^{\frac{2}{3}}} \]

Simplifying,

\[ h = 2 \left( \frac{V}{2\pi} \right)^{\frac{1}{3}} \]

Thus, the optimal dimensions for minimizing the tin used are:

\[ r = \left( \frac{V}{2\pi} \right)^{\frac{1}{3}}, \quad h = 2 \left( \frac{V}{2\pi} \right)^{\frac{1}{3}} \]

This result indicates that the height of the optimal cylinder is exactly twice its radius. Hence, for a fixed volume, the most efficient cylindrical tin box is one where its height is twice the radius, ensuring minimal material usage.

General Procedure of Optimization

Optimization is the process of finding the best possible value of a quantity while satisfying given conditions or constraints. In mathematics and engineering, this typically involves maximizing or minimizing a function subject to certain constraints. Problems in optimization arise in a wide range of applications, such as minimizing material costs in manufacturing, maximizing efficiency in transportation, or even determining the optimal shape for a physical structure.

The general procedure to solve an optimization problem follows these steps:

1. Identify the quantity to be optimized:

   • Determine what needs to be maximized or minimized (e.g., cost, surface area, time, distance).
   • Assign a variable to represent this quantity.

2. Express the quantity as a function:

   • Identify parameters relevant to the problem and express the quantity to be optimized as a function of these parameters.
   • This function may depend on multiple variables.

3. Find the constraints:

   • Constraints are given conditions that restrict the values of the parameters.
   • These may come from physical, geometrical, or problem-specific conditions.
   • If the function depends on two variables, we need one constraint to express one variable in terms of the other. If it depends on three variables, we need two constraints, and so on.

4. Eliminate extra parameters:

   • Use the constraints to express the function in terms of a single variable by eliminating extra parameters.

5. Differentiate and find critical points:

   • Compute the derivative of the function with respect to the remaining variable.
   • Solve for the critical points by setting the derivative equal to zero.

6. Confirm whether the critical point is a minimum or maximum:

   • Use either the first derivative test (checking sign changes) or the second derivative test (checking concavity) to confirm whether the critical point corresponds to a minimum or maximum.

7. Interpret the result:

   • Verify that the solution satisfies all constraints and makes sense in the context of the problem.

Example

A gardener has 20 meters of wire available to fence off a flowerbed in the shape of a circular sector. The wire is used to enclose the curved boundary as well as the two straight radii of the sector. Determine the maximum possible area of the flowerbed and find the corresponding values of the radius and central angle.

Solution:

A total of 20 meters of wire is available to fence a flowerbed in the shape of a circular sector. The objective is to determine the maximum possible area of this sector.

Let \( r \) be the radius of the sector and \( \theta \) be its central angle in radians. The total fencing consists of the two radii and the arc length. Hence, the constraint equation is

\[ 2r + r\theta = 20. \]

Rearranging,

\[ \theta = \frac{20 - 2r}{r}. \]

The area of a circular sector is given by

\[ A = \frac{1}{2} r^2 \theta. \]

Substituting the value of \( \theta \),

\[ A = \frac{1}{2} r^2 \cdot \frac{20 - 2r}{r} \]

\[ A = \frac{1}{2} r (20 - 2r) \]

\[ A = \frac{1}{2} (20r - 2r^2) \]

\[ A = 10r - r^2. \]

To find the critical points, differentiate \( A \) with respect to \( r \):

\[ \frac{dA}{dr} = 10 - 2r. \]

Setting \( \frac{dA}{dr} = 0 \) gives

\[ 10 - 2r = 0 \]

\[ r = 5. \]

To confirm that this corresponds to a maximum, compute the second derivative:

\[ \frac{d^2A}{dr^2} = -2. \]

Since \( \frac{d^2A}{dr^2} \) is negative, the function is concave downward, ensuring that \( r = 5 \) is a point of maximum.

Substituting \( r = 5 \) into the equation for \( \theta \),

\[ \theta = \frac{20 - 2(5)}{5} = \frac{10}{5} = 2. \]

The maximum area is therefore

\[ A_{\max} = \frac{1}{2} (5)^2 (2) \]

\[ = \frac{1}{2} \times 25 \times 2 = 25 \text{ sq m}. \]

Thus, the flowerbed attains its maximum possible area when the radius is 5 meters and the central angle is 2 radians, yielding an area of 25 square meters.

Optimization of a Right Circular Cone's Volume

A right circular cone has a fixed slant height of 3 meters. Determine the maximum possible volume of such a cone and find the corresponding height.

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Solution:

Let \( h \) be the height of the cone and \( r \) be the radius of its circular base. The given slant height is 3 meters. Using the Pythagorean theorem in the right triangle formed by the height, radius, and slant height,

\[ l^2 = h^2 + r^2. \]

Substituting \( l = 3 \),

\[ 3^2 = h^2 + r^2. \]

\[ 9 = h^2 + r^2. \]

Solving for \( r \),

\[ r = \sqrt{9 - h^2}. \]

The volume of a cone is given by

\[ V = \frac{1}{3} \pi r^2 h. \]

Substituting \( r^2 = 9 - h^2 \),

\[ V(h) = \frac{1}{3} \pi (9 - h^2) h. \]

\[ V(h) = \frac{1}{3} \pi (9h - h^3). \]

To find the maximum volume, differentiate with respect to \( h \):

\[ V'(h) = \frac{1}{3} \pi (9 - 3h^2). \]

Setting \( V'(h) = 0 \) to find critical points,

\[ 9 - 3h^2 = 0. \]

\[ h^2 = 3. \]

\[ h = \sqrt{3}. \]

Since height cannot be negative, we take \( h = \sqrt{3} \).

To confirm that this critical point corresponds to a maximum, compute the second derivative:

\[ V''(h) = \frac{1}{3} \pi (-6h). \]

Substituting \( h = \sqrt{3} \),

\[ V''(\sqrt{3}) = \frac{1}{3} \pi (-6\sqrt{3}). \]

\[ = -2\pi\sqrt{3}. \]

Since \( V''(h) < 0 \), the function is concave downward, confirming a maximum at \( h = \sqrt{3} \).

Now, substituting \( h = \sqrt{3} \) into the volume equation,

\[ V(\sqrt{3}) = \frac{1}{3} \pi (9\sqrt{3} - (\sqrt{3})^3). \]

\[ = \frac{1}{3} \pi (9\sqrt{3} - 3\sqrt{3}). \]

\[ = \frac{1}{3} \pi (6\sqrt{3}). \]

\[ = 2\sqrt{3} \pi. \]

Thus, the maximum possible volume of the cone is \( 2\sqrt{3} \pi \) cubic meters, which occurs when the cone's height is \( \sqrt{3} \) meters.

Example

A right circular cylinder is inscribed in a sphere of radius 3 meters. Determine the height of the cylinder that maximizes its volume.

Solution:

Let \( r \) be the radius and \( h \) be the height of the cylinder. Since the cylinder is symmetrically placed inside the sphere, its axis passes through the center of the sphere. The radius of the sphere is 3, meaning each half of the cylinder's height forms a right-angled triangle with the radius of the cylinder and the radius of the sphere.

Let \( \theta \) be the angle, as shown in the figure, between the height and the radius of sphere. Since the cylinder is inscribed within the sphere, \( \theta \) lies in the interval \( (0, \frac{\pi}{2}) \).

From the right-angled triangle,

\[ \frac{h}{2} = 3 \cos \theta \]

\[ h = 6 \cos \theta. \]

Similarly, the base radius of the cylinder is given by

\[ r = 3 \sin \theta. \]

The volume of the cylinder is given by

\[ V = \pi r^2 h. \]

Substituting \( r = 3 \sin \theta \) and \( h = 6 \cos \theta \),

\[ V = \pi (3 \sin \theta)^2 (6 \cos \theta). \]

\[ V = \pi (9 \sin^2 \theta) (6 \cos \theta). \]

\[ V = 54\pi \sin^2 \theta \cos \theta. \]

To find the critical points, differentiate \( V \) with respect to \( \theta \):

\[ \frac{dV}{d\theta} = 54\pi [2 \sin \theta \cos \theta \cos \theta - \sin^3 \theta]. \]

Taking \( \cos^2 \theta \) common,

\[ \frac{dV}{d\theta} = 54\pi \sin \theta \cos^2 \theta (2 - \tan^2 \theta). \]

For maxima or minima, set \( \frac{dV}{d\theta} = 0 \):

\[ \sin \theta \cos^2 \theta (2 - \tan^2 \theta) = 0. \]

Since \( \theta \in (0, \frac{\pi}{2}) \), neither \( \sin \theta \) nor \( \cos^2 \theta \) can be zero. Therefore,

\[ 2 - \tan^2 \theta = 0. \]

\[ \tan^2 \theta = 2. \]

\[ \tan \theta = \sqrt{2}. \]

Since \( \theta \) is in \( (0, \frac{\pi}{2}) \), we take the positive root:

\[ \theta = \tan^{-1} \sqrt{2}. \]

To confirm that this critical point corresponds to a maximum, apply the first derivative test. Consider \( \theta \) slightly less than \( \tan^{-1} \sqrt{2} \). In that case, \( 2 - \tan^2 \theta > 0 \), making \( \frac{dV}{d\theta} > 0 \), indicating that \( V \) is increasing. Similarly, for \( \theta \) slightly greater than \( \tan^{-1} \sqrt{2} \), \( 2 - \tan^2 \theta < 0 \), making \( \frac{dV}{d\theta} < 0 \), showing that \( V \) is decreasing. Since \( V \) changes from increasing to decreasing at \( \theta = \tan^{-1} \sqrt{2} \), it corresponds to a local maximum.

Now, substituting \( \theta = \tan^{-1} \sqrt{2} \) into the expression for \( h \):

\[ h = 6 \cos \theta = 6 \times \frac{1}{\sqrt{3}}. \]

\[ h = 2 \sqrt{3}. \]

Thus, the height of the right circular cylinder that maximizes the volume is \( 2\sqrt{3} \) meters.

In this example, the fact that \( \theta = \tan^{-1} \sqrt{2} \) gives the maximum volume can also be understood conceptually, without relying on the first derivative test.

Consider the extreme values of \( \theta \):

• When \( \theta = 0 \), the cylinder degenerates into a vertical line with zero radius, meaning the volume is zero.
• When \( \theta = \frac{\pi}{2} \), the cylinder flattens into a disc with height zero, again resulting in zero volume.

Since the volume function smoothly varies between these two extremes, there must exist some intermediate value of \( \theta \) where the volume reaches a maximum.

Now, since we found \( \theta = \tan^{-1} \sqrt{2} \) as a critical point, we can confidently conclude that it must correspond to the maximum volume—without explicitly verifying it through the second derivative or first derivative test. The reasoning is that a function transitioning from zero volume to zero volume across an interval must attain a maximum somewhere in between, and the critical point we found is precisely where this happens.