최댓값·최솟값과 최적화

Two examples are given below, but many are possible.

If $f(x)=-x^2$ or $f(x)=-\sqrt{|x|}$, then $f(x)$ has a global maximum at $x=0$. Since $f(x)$ keeps getting more and more strongly negative as $x$ gets farther and farther from 0, $f(x)$ has no global minimum.

Two examples are given below, but many are possible.

If $f(x)=e^x$, then $f(x) > 0$ for all $x$. As we move left along the $x$-axis, $f(x)$ gets smaller and smaller, approaching 0 but never reaching it. Since $f(x)$ gets smaller and smaller as we move left, there is no global minimum. Likewise, $f(x)$ increases more and more as we move right, so there is no maximum.

If $f(x)=\arctan(x)+2$, then $f(x) >\left(-\frac{\pi}{2}\right)+2>0$ for all $x$.

As we move left along the $x$-axis, $f(x)$ gets smaller and smaller, approaching $\left(-\frac{\pi}{2}+2\right)$ but never reaching it. Since $f(x)$ gets smaller and smaller as we move left, there is no global minimum.

Likewise, as we move right along the $x$-axis, $f(x)$ gets bigger and bigger, approaching $\left(\frac{\pi}{2}+2\right)$ but never reaching it. Since $f(x)$ gets bigger and bigger as we move right, there is no global maximum.

Since $f(5)$ is a global minimum, $f(5) \leq f(x)$ for all $x$, and so in particular $f(5) \leq f(-5)$.

Similarly, $f(-5) \leq f(x)$ for all $x$, so in particular $f(-5) \leq f(5)$.

Since $f(-5) \leq f(5)$ AND $f(5) \leq f(-5)$, it must be true that $f(-5)=f(5)$.

A sketch of one such graph is below.

Global extrema will occur at critical or singular points in the interval $(-5,5)$ or at the endpoints $x=5,\,x=-5$.

$f'(x)=2x+6$. Since this is defined for all real numbers, there are no singular points. The only time $f'(x)=0$ is when $x=-3$. This is inside the interval $[-5,5]$. So, our points to check are $x=-3,\,x=-5,$ and $x=5$.

$$\begin{array}{|c||c|c|c|}\hline c & -3 & -5 & 5 \\ \hline \text{type} & \text{critical point} & \text{endpoint} & \text{endpoint} \\ \hline f(c) & -19 & -15 & 45 \\ \hline \end{array}$$

The global maximum is 45 at $x=5$ and the global minimum is $-19$ at $x=-3$.

Global extrema will occur at the endpoints of the interval, $x=-4$ and $x=0$, or at singular or critical points inside the interval. Since $f(x)$ is a polynomial, it is differentiable everywhere, so there are no singular points. To find the critical points, we set the derivative equal to zero.

$$\begin{aligned} f'(x)&=2x^2-4x-30\\ 0&=2x^2-4x-30 &=(2x-10)(x+3)\\ x&=5,\,-3 \end{aligned}$$

The only critical point inside the interval is $x=-3$.

$$\begin{array}{|c||c|c|c|}\hline c & -3 & -4 & 0 \\ \hline \text{type} & \text{critical point} & \text{endpoint} & \text{endpoint} \\ \hline f(c) & 61 & \frac{157}{3}=52+\frac{1}{3} & 7 \\ \hline \end{array}$$

The global maximum over the interval is $61$ at $x=-3$, and the global minimum is $7$ at $x=0$.

Call the length of the wire $L$ units and suppose that it is cut $\ell$ units from one end. Make the square from the piece of length $\ell$, and make the circle from the remaining piece of length $L-\ell$.

The square has perimeter $\ell$, so its side length is $\ell/4$ and its area is $\left(\dfrac{\ell}{4}\right)^2$. The circle has circumference $L-\ell$, so its radius is $\dfrac{L-\ell}{2\pi}$ and its area is $\pi\left(\dfrac{L-\ell}{2\pi}\right)^2=\dfrac{(L-\ell)^2}{4\pi}$.

The area enclosed by the shapes, when the square is made from a length of size $\ell$, is

$$\begin{aligned} A(\ell)&=\dfrac{\ell^2}{16}+\dfrac{(L-\ell)^2}{4\pi} \\ \text{We want to find the global max and min for this function, given the constraint $0 \leq \ell \leq L$, so we find its derivative:} \\ A'(\ell)&=\dfrac{\ell}{8} -\dfrac{L-\ell}{2\pi} =\dfrac{\pi+4}{8\pi}\ell-\dfrac{L}{2\pi} \\ \text{Now, we find the critical point.} \\ A'(\ell)&=0\\ \frac{\pi+4}{8\pi}\ell&=\frac{L}{2\pi}\\ \ell&=\frac{4L}{\pi+4} \end{aligned}$$

$$\begin{array}{|c||c|c|c|}\hline \ell & 0 & L & \dfrac{.^.}{.}\frac{4L}{\pi+4} \\ \hline \text{type} & \text{endpoint} & \text{endpoint} & \text{critical point} \\ \hline A(\ell) & \dfrac{.^.}{.}\frac{L^2}{4\pi} & \frac{L^2}{16}_{} & A\left(\frac{4L}{\pi+4}\right) \\ \hline \end{array}$$

It seems obnoxious to evaluate $A\left(\frac{4L}{\pi+4}\right)$, and the problem doesn't ask for it–but we still have to figure out whether it is a global max or min.

When $\ell<\frac{4L}{\pi+4}$, $A'(\ell)<0$, and when $\ell>\frac{4L}{\pi+4}$, $A'(\ell)>0$. So, $A(\ell)$ is decreasing until $\ell=\frac{4L}{\pi+4}$, then increasing. That means our critical point $\ell=\frac{4L}{\pi+4}$ is a local minimum.

So, the minimum occurs at the only critical point, which is $\ell=\left(\dfrac{4}{4+\pi}\right)L$. This corresponds to $\dfrac{\ell}{L}=\dfrac{4}{4+\pi}$: the proportion of the wire that is cut is $\dfrac{4}{4+\pi}$.

The maximum has to be either at $\ell=0$ or at $\ell=L$. As $A(0)=\dfrac{L^2}{4\pi} > A(L)=\dfrac{L^2}{16}$, the maximum has $\ell=0$ (that is, no square).

(a) It's not possible to say without more information: for instance, the functions $f(x) = e^{-x}$, $f(x) = e^{-x}$, and $f(x) = x^2$ are each always concave up, but only one of them has a global minimum. We can say for sure that by being always concave up, $f$ cannot have a global maximum, and it also cannot have both a global maximum and a global minimum. It might have a global minimum, or it might have neither. (b) By being twice differentiable, we know that $g'(x) = 0$ any time $x$ is a critical number of $g$. Thus, given the signs of $g'(x)$ on the stated intervals, $g$ is decreasing before $x = a$ and decreasing on $a < x < b$, with a horizontal tangent line at $x = a$. In addition, $g$ is increasing for $x > b$. Thus, we know that $g$ has a global minimum at $x = b$. (c) Since $h$ is always decreasing for $x < a$, always increasing for $a < x < b$, and always decreasing for $x > b$, we know that $h$ has a local minimum at $x = a$ and a local maximum at $x = b$. Moreover, these local extremes are both global extremes since $h(x) to 0$ as $x \to \pm \infty$. In particular, since $h(x)$ decreases to $0$ for $x > b$ and increases on $a < x < b$, it follows that $h(b)$ is the largest value of $h(x)$ on $(a, \infty)$, and that $h(b) > 0$. Similarly, since $h(x)$ decreases on $(-\infty, a)$ and approaches $0$ as $x \to -\infty$, and $h(x)$ increases on $a < x < b$, it must be the case that $h(a)$ is the least value of $h(x)$ on $(-\infty, b)$ and that $h(b) < 0$. Hence, $h(x)$ is always negative on $x < a$ and $h(x)$ is always positive on $x > b$. From all of these observations, it follows that $h$ has a global minimum at $x = a$ and a global maximum at $x = b$. (d) First, observe that $p$ is defined for all real numbers by assumption, so $p(a)$ has some finite value. However, $p$ is not differentiable at $x = a$ and $p$ is concave up to the left of $x = a$ and concave down to the right of $x = a$. This indicates that $p$ has a relative maximum at $x = a$. Not enough information is provided, however, to determine if this relative maximum is a global maximum, nor if $p$ happens to have a global minimum at some other point.

(a) Given $p(x) = x^3 - a^2x$ on $[0,a]$, we first find the critical number(s) of $p$ that lie in $[0,a]$. Since $p'(x) = 3x^2 - a^2 = 3\left(x^2 - \frac{a^2}{3} \right)$, we know $p$ has critical numbers at $x = \pm \frac{a}{\sqrt{3}}$. The negative critical number lies outside the interval, while the positive one lies within the interval since $0 < \frac{1}{\sqrt{3}} < 1$. Evaluating $p$ at the endpoints and the critical number within the interval, we find that $p(0) = 0$, $p(a) = a^3 - a^2 \cdot a = 0$, and $p\left( \frac{a}{\sqrt{3}} \right) = \left( \frac{a}{\sqrt{3}} \right)^3 - a^2 \cdot \left( \frac{a}{\sqrt{3}} \right) = \frac{a^3}{3\sqrt{3}} - \frac{a^3}{\sqrt{3}} = -\frac{2a^3}{3\sqrt{3}}$. Hence, the absolute maximum of $p$ on $[0,a]$ is $p(0) = p(a) = 0$ and the absolute minimum is $p\left( \frac{a}{\sqrt{3}} \right) = -\frac{2a^3}{3\sqrt{3}}$. (b) Given $r(x) = axe^{-bx}$ on $\left[\frac{1}{2b}, \frac{2}{b}\right]$, we first find the critical numbers of $r$ that lie in the given interval. Note that by the product rule, $r'(x) = ax e^{-bx} (-b) + e^{-bx} a$. Factoring, $r'(x) = ae^{-bx}(-bx + 1)$, and thus the only critical number of $r$ occurs where $x = \frac{1}{b}$. Note that $\frac{1}{2b} < \frac{1}{b} < b$ since $b > 1$, and thus the critical number lies within the interval. Evaluating $r$ at the endpoints and the critical number, we find

• $r\left( \frac{1}{2b} \right) = a \cdot \frac{1}{2b} \cdot e^{-1/2} \approx 0.607 \frac{a}{2b} = 0.3035 \frac{a}{b}$
• $r\left( \frac{2}{b} \right) = a \cdot \frac{2}{b} \cdot e^{-2} \approx 0.135 \frac{2a}{b} = 0.270 \frac{a}{b}$
• $r\left( \frac{1}{b} \right) = a \cdot \frac{1}{b} \cdot e^{-1} \approx 0.368 \frac{a}{b}$ The absolute max of $r$ is therefore $r\left( \frac{1}{b} \right) \approx 0.368 \frac{a}{b}$ and the absolute min is $r\left( \frac{2}{b} \right) \approx 0.270 \frac{a}{b}$. (c) For $w(x) = a(1-e^{-bx})$ on $[b, 3b]$, we first note that $w'(x) = a(be^{-bx}) = abe^{-bx}$. Since $a$ and $b$ are both positive and so is $e^{-bx}$ for every value of $x$, we see that $w'(x) > 0$ for every value of $x$. Because $w$ is always increasing (specifically on $[b,3b]$), $w$ has its absolute minimum at the left endpoint ($g(b) = a(1-e^{-b^2})$) and its absolute maximum at the right endpoint ($g(3b) = a(1-e^{-3b^2})$). (d) Finally, for $s(x) = \sin(kx)$ on $\left[\frac{\pi}{3k}, \frac{5\pi}{6k}\right]$, we know $s'(x) = k\cos(kx)$, and $\cos(kx) = 0$ when $kx = \frac{\pi}{2} \pm n\pi$ for some integer $n$, so $x = \frac{\pi}{2k} \pm n\pi = \frac{\pi \pm 2n\pi}{2k}$. These $x$-values are the critical numbers of $s$. Note that when $n = -1, 0, 1$, the corresponding critical numbers are $$x = -\frac{\pi}{2k}, \frac{\pi}{2k}, \frac{3\pi}{2k}$$ Comparing the values of $-\frac{\pi}{2}$, $\frac{\pi}{2k}$, and $\frac{3\pi}{2k}$ to $\frac{\pi}{3}$ and $\frac{5\pi}{6}$, we see that only $\frac{\pi}{2k}$ lies in the given interval $\left[\frac{\pi}{3k}, \frac{5\pi}{6k}\right]$, and thus $x = \frac{\pi}{2k}$ is the only critical number at which we need to evaluate $s$. Evaluating $s$ at the endpoints and relevant critical number, we find:
• $s\left( \frac{\pi}{3k} \right) = \sin\left( k \cdot \frac{\pi}{3k} \right) = \sin\left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}$;
• $s\left( \frac{5\pi}{6k} \right) = \sin\left( k \cdot \frac{5\pi}{6k} \right) = \sin\left( \frac{5\pi}{6} \right) = \frac{1}{2}$; and
• $s\left( \frac{\pi}{2k} \right) = \sin\left( k \cdot \frac{\pi}{2k} \right) = \sin\left( \frac{\pi}{2} \right) = 1$. Hence the absolute max of $s$ on $\left[\frac{\pi}{3k}, \frac{5\pi}{6k}\right]$ is $s\left( \frac{\pi}{2k} \right) = 1$ and the absolute min of $s$ is $s\left( \frac{5\pi}{6k} \right) = \frac{1}{2}$.