Absolute Maxima and Minima

Concept of Absolute Maxima and Minima

While local maxima and minima tell us about the peaks and valleys of a function in a small neighborhood, the absolute maximum (or global maximum) and absolute minimum (or global minimum) refer to the highest and lowest values of the function over its entire given domain.

Let $f$ be a continuous function defined on an interval $I = [a, b]$.

Absolute Maximum

A point $c \in I$ is a point where $f$ attains an absolute maximum if $f(c) \ge f(x)$ for all $x \in I$. The value $f(c)$ is called the absolute maximum value.

Absolute Minimum

A point $d \in I$ is a point where $f$ attains an absolute minimum if $f(d) \le f(x)$ for all $x \in I$. The value $f(d)$ is called the absolute minimum value.

Extreme Value Theorem

Every continuous function defined on a closed interval $[a, b]$ must have both an absolute maximum and an absolute minimum value on that interval.

Important Note: Unlike local extrema, absolute extrema are judged using the entire interval, not just nearby points. Also, the absolute maximum or minimum may occur at an endpoint.

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Working Rule for Finding Absolute Maxima and Minima

To find the absolute maximum and minimum values of a continuous function $f$ on a closed interval $[a, b]$, use the following standard procedure:

Step 1: Find all critical points of $f$ in the open interval $(a, b)$. These are points where $f'(x) = 0$ or where $f'(x)$ is not defined.

Step 2: Take the endpoints of the interval, namely $x = a$ and $x = b$.

Step 3: Evaluate $f(x)$ at all critical points found in Step 1 and at the endpoints found in Step 2.

Step 4: Compare all these values.

• The greatest value is the absolute maximum value.
• The smallest value is the absolute minimum value.

This method works because, on a closed interval, absolute extrema can occur only at interior critical points or at endpoints.

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Open Intervals and Optimization

If the interval is open, say $(a, b)$, or if the domain is the set of all real numbers $\mathbb{R}$, an absolute maximum or minimum may fail to exist. For example, $f(x) = x$ on $(0, 1)$ has neither an absolute minimum nor an absolute maximum, because the values approach 0 and 1 but never actually attain them.

However, in many real-world optimization problems, physical restrictions naturally create a closed interval. For example, when cutting squares from the corners of a sheet, the cut size cannot be negative and cannot exceed half the side length. In such cases, the Extreme Value Theorem applies, and absolute extrema are guaranteed.

If the domain is not closed, then endpoint behavior must be checked separately, and it is possible that no absolute extremum exists.

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Solved Examples

Example 1: Absolute Extrema of a Polynomial

Find the absolute maximum and absolute minimum values of the function $f(x) = 2x^3 - 15x^2 + 36x + 1$ on the interval $[1, 5]$.

Solution: Step 1: Find the derivative. $$f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 1) = 6x^2 - 30x + 36.$$

Step 2: Find the critical points by setting $f'(x)=0$. $$6x^2 - 30x + 36 = 0$$ $$6(x^2 - 5x + 6)=0$$ $$6(x-2)(x-3)=0$$ So the critical points are $x=2$ and $x=3$, both of which lie in $[1,5]$.

Step 3: Evaluate $f(x)$ at the critical points and the endpoints.

• At $x=1$: $$f(1)=2(1)^3-15(1)^2+36(1)+1=2-15+36+1=24.$$
• At $x=2$: $$f(2)=2(8)-15(4)+36(2)+1=16-60+72+1=29.$$
• At $x=3$: $$f(3)=2(27)-15(9)+36(3)+1=54-135+108+1=28.$$
• At $x=5$: $$f(5)=2(125)-15(25)+36(5)+1=250-375+180+1=56.$$

Step 4: Compare the values $24, 29, 28, 56$. The greatest value is $56$ and the smallest value is $24$.

Answer: Absolute maximum value is $56$ at $x=5$. Absolute minimum value is $24$ at $x=1$.

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Example 2: Absolute Extrema of a Trigonometric Function

Find the absolute maximum and minimum values of $f(x) = \sin x + \cos x$ on the interval $[0, \pi]$.

Solution: Step 1: Find the derivative. $$f'(x)=\cos x-\sin x.$$

Step 2: Find the critical points by setting $f'(x)=0$. $$\cos x-\sin x=0 \implies \sin x=\cos x \implies \tan x=1.$$ In the interval $[0,\pi]$, the only solution is $$x=\frac{\pi}{4}.$$

Step 3: Evaluate $f(x)$ at the critical point and the endpoints.

• At $x=0$: $$f(0)=\sin 0+\cos 0=0+1=1.$$
• At $x=\frac{\pi}{4}$: $$f\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}.$$
• At $x=\pi$: $$f(\pi)=\sin \pi+\cos \pi=0-1=-1.$$

Step 4: Compare the values $1, \sqrt{2}, -1$. The largest value is $\sqrt{2}$ and the smallest value is $-1$.

Answer: Absolute maximum is $\sqrt{2}$ at $x=\frac{\pi}{4}$. Absolute minimum is $-1$ at $x=\pi$.

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Example 3: Function with Fractional Powers

Find the absolute maximum and absolute minimum values of $f(x) = 12x^{4/3} - 6x^{1/3}$ on the interval $[-1, 1]$.

Solution: Step 1: Find the derivative. $$f'(x)=12\left(\frac{4}{3}\right)x^{1/3}-6\left(\frac{1}{3}\right)x^{-2/3}=16x^{1/3}-2x^{-2/3}.$$ Writing it as a single fraction: $$f'(x)=\frac{16x-2}{x^{2/3}}.$$

Step 2: Find the critical points. A critical point occurs where $f'(x)=0$ or $f'(x)$ is not defined.

• For $f'(x)=0$: $$16x-2=0 \implies x=\frac{1}{8}.$$
• For $f'(x)$ undefined: $$x^{2/3}=0 \implies x=0.$$ Both lie in the interval $[-1,1]$.

Step 3: Evaluate the function at $x=-1,0,\frac18,1$.

• At $x=-1$: $$f(-1)=12(-1)^{4/3}-6(-1)^{1/3}=12(1)-6(-1)=18.$$
• At $x=0$: $$f(0)=0.$$
• At $x=\frac18$: Since $\left(\frac18\right)^{1/3}=\frac12$ and $\left(\frac18\right)^{4/3}=\left(\frac12\right)^4=\frac1{16}$, $$f\left(\frac18\right)=12\left(\frac1{16}\right)-6\left(\frac12\right)=\frac34-3=-\frac94.$$
• At $x=1$: $$f(1)=12-6=6.$$

Step 4: Compare the values $18,0,-\frac94,6$. The largest value is $18$ and the smallest value is $-\frac94$.

Answer: Absolute maximum is $18$ at $x=-1$. Absolute minimum is $-\frac94$ at $x=\frac18$.

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Example 4: Application - Optimization Problem

A square piece of tin of side $18$ cm is to be made into an open box by cutting a square from each corner and folding up the flaps. What should be the side of the square cut off so that the volume of the box is maximum?

Solution: Step 1: Form the volume function. Let the side of each square cut from the corners be $x$ cm. Then the dimensions of the box are:

• Length $=18-2x$
• Breadth $=18-2x$
• Height $=x$

So the volume is $$V(x)=x(18-2x)^2.$$ Since $x$ cannot be negative and cannot exceed $9$, the physical domain is $$0 \le x \le 9.$$

Step 2: Differentiate the volume function. Using the product rule, $$V'(x)=(18-2x)^2+x\cdot 2(18-2x)(-2).$$ Factor out $(18-2x)$: $$V'(x)=(18-2x)[(18-2x)-4x]=(18-2x)(18-6x).$$

Step 3: Find the critical points. $$V'(x)=0 \implies (18-2x)(18-6x)=0.$$ So, $$x=9 \quad \text{or} \quad x=3.$$ Here $x=9$ is also an endpoint of the interval.

Step 4: Evaluate $V(x)$ at the endpoints and the interior critical point.

• At $x=0$: $$V(0)=0.$$
• At $x=3$: $$V(3)=3(18-6)^2=3(12)^2=432.$$
• At $x=9$: $$V(9)=9(0)^2=0.$$

Step 5: Compare the values. The greatest value is $432$.

Answer: The side of the square to be cut off should be $3$ cm.

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Example 5: Absolute Extrema of a Rational Function

Find the absolute maximum and minimum values of $f(x) = \frac{x^2 - 1}{x^2 + 1}$ on the interval $[-2, 2]$.

Solution: Step 1: Differentiate using the quotient rule. $$f'(x)=\frac{(x^2+1)(2x)-(x^2-1)(2x)}{(x^2+1)^2}.$$ Simplify the numerator: $$f'(x)=\frac{2x^3+2x-2x^3+2x}{(x^2+1)^2}=\frac{4x}{(x^2+1)^2}.$$

Step 2: Find the critical points. Since the denominator is always positive, $f'(x)=0$ only when $$4x=0 \implies x=0.$$ This lies in $[-2,2]$.

Step 3: Evaluate the function at $x=-2,0,2$.

• At $x=-2$: $$f(-2)=\frac{4-1}{4+1}=\frac35.$$
• At $x=0$: $$f(0)=\frac{-1}{1}=-1.$$
• At $x=2$: $$f(2)=\frac{4-1}{4+1}=\frac35.$$

Step 4: Compare the values. The largest value is $\frac35$ and the smallest value is $-1$.

Answer: Absolute maximum is $\frac35$ at $x=-2$ and $x=2$. Absolute minimum is $-1$ at $x=0$.