최댓값·최솟값과 최적화

(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 $s\left( \frac{5\pi}{6k} \right) = \frac{1}{2}$.