§3.5 Global optimization (continued)

Moving toward applications

Summary

A $20$ cm wire is cut so one piece forms a square of side $\frac{x}{4}$ and the other an equilateral triangle of side $\frac{20-x}{3}$. With total area $A(x) = \frac{x^2}{16} + \frac{\sqrt{3}}{36}(20-x)^2$, which conclusion follows?

A twice differentiable function $f$ satisfies $f''(x) \lt 0$ for every $x$. What can be concluded?

A function $g$ with critical numbers $a \lt b$ has $g' \lt 0$ on $x \lt a$, $g' \lt 0$ between them, and $g' \gt 0$ after $b$. Then $g$ has

For $p(x) = x^3 - a^2x$ on $[0, a]$ with $a \gt 0$, the critical numbers are $\pm\frac{a}{\sqrt{3}}$, and only the positive one lies inside. The absolute extrema are

A continuous function $f$ on $[a, b]$ satisfies $f'(x) \le 0$ throughout. Its absolute extrema are located

For $s(t) = 3\sin(2(t-\frac{\pi}{6})) + 5$ on $[\frac{\pi}{6}, \frac{7\pi}{6}]$, the critical numbers inside are $\frac{5\pi}{12}$ and $\frac{11\pi}{12}$. Compute the absolute maximum value.

From a $10$ by $15$ inch cardboard, squares of side $x$ are cut from each corner and the flaps folded up. The box volume as a function of $x$ is

The problem restricts the cut size to $1 \le x \le 3$, and the critical numbers of $V$ are about $1.962$ and $6.371$. The correct candidate list for the closed interval method is

Evaluating the candidates gives $V(1) = 104$, $V(1.962) \approx 132.0$, and $V(3) = 108$. The maximum and minimum volumes are

For the box $V(x) = x(10-2x)(15-2x)$, compute $V(1)$.

For the box built from a 10 by 15 sheet with corner squares cut, compute $V(3)$, the right endpoint candidate of the restricted domain.

In the box problem the critical number $x \approx 6.371$ can be ignored because it lies outside the physical domain $1 \le x \le 3$.