최댓값·최솟값과 최적화

(a) For $s(t) = 3\sin(2(t-\frac{\pi}{6})) + 5$, $s'(t) = 3\cos(2(t-\frac{\pi}{6})) \cdot 2 = 6\cos(2(t-\frac{\pi}{6}))$. It follows that $s'(t) = 0$ if and only if $\cos(2(t-\frac{\pi}{6})) = 0$, and thus $2(t-\frac{\pi}{6}) = \frac{\pi}{2} \pm k\pi$ for some integer value of $k$. Solving this most recent equation for $t$, we have $t-\frac{\pi}{6} = \frac{\pi}{4} \pm \frac{k\pi}{2}$, so $$t = \frac{\pi}{6} + \frac{\pi}{4} \pm \frac{k\pi}{2} = \frac{5\pi}{12} \pm \frac{k\pi}{2}\,.$$ Now we consider the given interval $[\frac{\pi}{6}, \frac{7\pi}{6}]$, which we can think of as $[\frac{2\pi}{12}, \frac{14\pi}{12}]$. The value $t = \frac{5\pi}{12}$ lies in this interval, as does $t = \frac{5\pi}{12} + \frac{\pi}{2} = \frac{11\pi}{12}$. Thus, we have two critical numbers in the interval at which to evaluate $s$, as well as at the endpoints. Doing so, we find:

• $s(\frac{5\pi}{12}) = 8$;
• $s(\frac{11\pi}{12}) = 2$;
• $s(\frac{\pi}{6}) = 5$; and
• $s(\frac{7\pi}{6}) = 5$. The absolute max of $s$ on $[\frac{\pi}{6}, \frac{7\pi}{6}]$ is thus $s(\frac{5\pi}{12}) = 8$, while the absolute min is $s(\frac{11\pi}{12}) = 2$. (b) From our work in (a), we know that $s$ has critical numbers at $t = \frac{5\pi}{12} \pm \frac{k\pi}{2}$. On the interval $[0, \frac{\pi}{2}]$, only $t = \frac{5\pi}{12}$ lies in the interval. Thus we have three values at which to evaluate $s$. Doing so:
• $s(\frac{5\pi}{12}) = 8$;
• $s(0) = 3\sin(2(0-\frac{\pi}{6})) + 5 = 3\sin(-\frac{\pi}{3}) + 5 = 5 - \frac{3\sqrt{3}}{2} \approx 2.402$; and
• $s(\frac{\pi}{2}) = 3\sin(2(\frac{\pi}{2}-\frac{\pi}{6})) + 5 = 3\sin(\frac{2\pi}{3}) + 5 = 5 + \frac{3\sqrt{3}}{2} \approx 7.598$. The absolute max of $s$ on $[0, \frac{\pi}{2}]$ is thus $s(\frac{5\pi}{12}) = 8$, while the absolute min is $s(0) = 5 - \frac{3\sqrt{3}}{2} \approx 2.402$. (c) On the interval from $t = 0$ to $t = 2\pi$, $s$ completes nearly two full periods, and thus achieves each of its maximum and minimum values. The absolute maximum is $8$ as determined in (a), and the absolute minimum is $2$. These occur, for example, at $s(\frac{5\pi}{12}) = 8$, while the absolute min is $s(\frac{11\pi}{12}) = 2$. There are other points at which the function achieves these values on the given interval. (d) On the interval $[\frac{\pi}{3}, \frac{5\pi}{6}] = [\frac{4\pi}{12}, \frac{10\pi}{12}]$, only the critical number $t = \frac{5\pi}{12}$ lies in the interval. Thus we have three values at which to evaluate $s$. Doing so:
• $s(\frac{5\pi}{12}) = 8$;
• $s(\frac{\pi}{3}) = 3\sin(2(\frac{\pi}{3}-\frac{\pi}{6})) + 5 = 3\sin(\frac{\pi}{3}) + 5 = \frac{3\sqrt{3}}{2} + 5 \approx 7.598$; and
• $s(\frac{5\pi}{6}) = 3\sin(2(\frac{5\pi}{6}-\frac{\pi}{6})) + 5 = 3\sin(\frac{4\pi}{3}) + 5 = 5 - \frac{3\sqrt{3}}{2} \approx 2.402$. The absolute max of $s$ on $[\frac{\pi}{3}, \frac{5\pi}{6}]$ is thus $s(\frac{5\pi}{12}) = 8$, while the absolute min is $s(\frac{5\pi}{6}) \approx 2.402$.