Optimization

$f$ is a polynomial, so it is continuous on $[-1, 4]$; the EVT says a global max and min exist.

$f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$.

$f'(x) = 0$ at $x = 1$ and $x = 3$, both inside $(-1, 4)$. $f'$ is never undefined.

$\begin{array}{c|cccc} x & -1 & 1 & 3 & 4 \\ \hline f(x) & -14 & 6 & 2 & 6 \end{array}$

Largest value $6$ (at $x = 1$ and at $x = 4$); smallest value $-14$ (at $x = -1$).

$f$ is continuous wherever $x \neq 0$, so on all of $[1, 3]$; global max and min exist.

$f'(x) = 1 - \dfrac{4}{x^2}$.

$1 - \dfrac{4}{x^2} = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$. Only $x = 2$ is in $(1, 3)$; drop $x = -2$. ($f'$ is undefined only at $0$, not in the interval.)

$f(1) = 5$, $f(2) = 2 + 2 = 4$, $f(3) = 3 + \tfrac{4}{3} = \tfrac{13}{3} \approx 4.33$.

Largest: $5$ at $x = 1$. Smallest: $4$ at $x = 2$.

$x$ and $\sin x$ are continuous everywhere, so $f$ is continuous on $[0, \pi]$; global max and min exist.

$f'(x) = 1 - 2\cos x$.

$\cos x = \tfrac{1}{2}$ gives $x = \tfrac{\pi}{3}$ in $(0, \pi)$ (the other solution, $\tfrac{5\pi}{3}$, is outside).

$f(0) = 0$, $f\!\left(\tfrac{\pi}{3}\right) = \tfrac{\pi}{3} - \sqrt{3} \approx -0.68$, $f(\pi) = \pi - 0 = \pi$.

Largest: $\pi$ at $x = \pi$. Smallest: $\tfrac{\pi}{3} - \sqrt{3}$ at $x = \tfrac{\pi}{3}$.

$x^{2/3} = \left(\sqrt[3]{x}\right)^2$ is continuous everywhere, so $f$ is continuous on $[-1, 4]$; global max and min exist.

$f(x) = x^{5/3} - 5x^{2/3}$, so $f'(x) = \tfrac{5}{3}x^{2/3} - \tfrac{10}{3}x^{-1/3} = \dfrac{5(x - 2)}{3x^{1/3}}$.

$f'(x) = 0$ at $x = 2$; $f'(x)$ is undefined at $x = 0$. Both are inside $(-1, 4)$.

$f(-1) = 1 \cdot (-6) = -6$, $f(0) = 0$, $f(2) = -3\sqrt[3]{4} \approx -4.76$, $f(4) = -\sqrt[3]{16} \approx -2.52$.

Largest: $0$ at $x = 0$. Smallest: $-6$ at $x = -1$.

(a) $f(0) = 0$, $f(x) > 0$ for $x > 0$, and $\displaystyle\lim_{x\to\infty} f(x) = 0$. Since $f(1) = 0.3 > 0$, past some $R$ we have $f(x) < 0.3$; on $[0, R]$ the EVT gives a max $\ge 0.3$, and nothing past $R$ beats it. So a global max exists.

(b) $f'(x) = \dfrac{3(x^2 + 9) - 3x(2x)}{(x^2 + 9)^2} = \dfrac{3(9 - x^2)}{(x^2 + 9)^2}$, zero at $x = 3$ (and $x = -3$, outside the domain).

$f' > 0$ on $(0, 3)$ and $f' < 0$ on $(3, \infty)$.

$f$ rises up to $x = 3$ and falls after it, so $f(3) = \dfrac{9}{18} = \dfrac{1}{2}$ is the global max.

As $x \to 0^+$, $\dfrac{16}{x} \to \infty$; as $x \to \infty$, $x^2 \to \infty$. Both ends shoot up: a global min exists, a global max does not.

$f'(x) = 2x - \dfrac{16}{x^2} = \dfrac{2(x^3 - 8)}{x^2}$, zero only at $x = 2$.

$f' < 0$ on $(0, 2)$, $f' > 0$ on $(2, \infty)$.

Decreasing then increasing with one critical point, so $f(2) = 4 + 8 = 12$ is the global min. No global max, since $f \to \infty$.

As $x \to -\infty$, $e^x \to 0$ and $-3x \to \infty$; as $x \to \infty$, $e^x$ dominates $3x$, so $f \to \infty$. Both ends go up.

$f'(x) = e^{x} - 3 = 0 \Rightarrow x = \log 3$.

$f' < 0$ for $x < \log 3$ and $f' > 0$ for $x > \log 3$. (Also $f''(x) = e^x > 0$ everywhere.)

One critical point, decreasing then increasing: $f(\log 3) = 3 - 3\log 3$ is the absolute min.

Base $(24 - 2x)$ by $(9 - 2x)$, height $x$. Maximize $V = x(24 - 2x)(9 - 2x)$.

The sheet size is the constraint, already used: $V(x) = 4x^3 - 66x^2 + 216x$.

$0 \le x \le 4.5$ (the cut cannot pass half of $9$).

$V'(x) = 12x^2 - 132x + 216 = 12(x - 2)(x - 9)$; only $x = 2$ is in the domain.

Closed interval: $V(0) = 0$, $V(4.5) = 0$, $V(2) = 2 \cdot 20 \cdot 5 = 200$. So $x = 2$ gives the global max.

Cut $2$ cm squares; the box is $20 \times 5 \times 2$ cm.

Width $w$, length $2w$, height $h$. Minimize $S = 2w^2 + 2wh + 2(2w)h = 2w^2 + 6wh$ (one base, four sides).

$2w^2 h = 36 \Rightarrow h = \dfrac{18}{w^2}$, so $S(w) = 2w^2 + \dfrac{108}{w}$.

$w > 0$.

$S'(w) = 4w - \dfrac{108}{w^2} = 0 \Rightarrow w^3 = 27 \Rightarrow w = 3$.

$S''(w) = 4 + \dfrac{216}{w^3} > 0$ for every $w > 0$: concave up on the whole domain, so $w = 3$ is the global min.

Width $3$ cm, length $6$ cm, height $h = \dfrac{18}{9} = 2$ cm. Surface area $18 + 36 = 54\ \text{cm}^2$.

Radius $r$, height $h$. Minimize $S = 2\pi r^2 + 2\pi r h$ (top, bottom, side).

$\pi r^2 h = 128\pi \Rightarrow h = \dfrac{128}{r^2}$, so $S(r) = 2\pi r^2 + \dfrac{256\pi}{r}$.

$r > 0$.

$S'(r) = 4\pi r - \dfrac{256\pi}{r^2} = 0 \Rightarrow r^3 = 64 \Rightarrow r = 4$.

$S' < 0$ for $0 < r < 4$ and $S' > 0$ for $r > 4$: decreasing then increasing, so $r = 4$ is the global min.

$r = 4$ cm, $h = \dfrac{128}{16} = 8$ cm; $S = 32\pi + 64\pi = 96\pi\ \text{cm}^2$.

$C = 5(2\pi r^2) + 2(2\pi r h) = 10\pi r^2 + 4\pi r h$.

$\pi r^2 h = 40\pi \Rightarrow h = \dfrac{40}{r^2}$.

$C(r) = 10\pi r^2 + \dfrac{160\pi}{r}$, with $r > 0$.

$C'(r) = 20\pi r - \dfrac{160\pi}{r^2} = 0 \Rightarrow r^3 = 8 \Rightarrow r = 2$.

$C'' = 20\pi + \dfrac{320\pi}{r^3} > 0$ for all $r > 0$, so $r = 2$ is the global min. Then $h = \dfrac{40}{4} = 10$ and $C = 40\pi + 80\pi = 120\pi$.

Street side $x$, depth $y$. Cost $10x + 5x + 5(2y) = 15x + 10y$; maximize $A = xy$.

$15x + 10y = 600$.

$y = 60 - 1.5x$, so $A(x) = 60x - 1.5x^2$ with $0 \le x \le 40$.

$A'(x) = 60 - 3x = 0 \Rightarrow x = 20$.

$A(0) = A(40) = 0$ and $A(20) = 20 \cdot 30 = 600$, so $x = 20$ is the global max.

Under water: $\sqrt{9 + x^2}$ km; on land: $8 - x$ km. Cost (in thousands of dollars) $C = 5\sqrt{9 + x^2} + 4(8 - x)$.

The width $3$ and the distance $8$ are fixed; Pythagoras gives the underwater length.

Already one variable: $C(x) = 5\sqrt{9 + x^2} + 32 - 4x$, with $0 \le x \le 8$.

$C'(x) = \dfrac{5x}{\sqrt{9 + x^2}} - 4 = 0 \Rightarrow 25x^2 = 16(9 + x^2) \Rightarrow x^2 = 16 \Rightarrow x = 4$.

Closed domain: $C(0) = 47$, $C(4) = 25 + 16 = 41$, $C(8) = 5\sqrt{73} \approx 42.7$. Smallest is $41$, so $x = 4$ is the global min.

A point on the curve: $(x, \sqrt{x})$, $x \ge 0$.

$D^2 = (x - 4)^2 + (\sqrt{x} - 0)^2 = (x - 4)^2 + x$.

$(D^2)' = 2(x - 4) + 1 = 0 \Rightarrow x = \tfrac{7}{2}$.

$(D^2)'' = 2 > 0$ on the whole domain, so $x = \tfrac{7}{2}$ is the global min (check: $D^2(0) = 16$ is bigger).

$D^2 = \tfrac{1}{4} + \tfrac{7}{2} = \tfrac{15}{4}$, so $D = \dfrac{\sqrt{15}}{2}$ at $\left(\tfrac{7}{2}, \sqrt{\tfrac{7}{2}}\right)$.

A point on the curve: $(x, x^2)$.

$D^2 = (x - 3)^2 + x^4$.

$(D^2)' = 2(x - 3) + 4x^3 = 2(2x^3 + x - 3) = 2(x - 1)(2x^2 + 2x + 3) = 0 \Rightarrow x = 1$ (the quadratic has negative discriminant).

$2x^2 + 2x + 3 > 0$ always, so $(D^2)'$ has the sign of $x - 1$: negative then positive. One critical point, so $x = 1$ is the global min.

$D^2 = 4 + 1 = 5$, so $D = \sqrt{5}$, at $(1, 1)$.

A point on the curve: $\left(x, \tfrac{4}{x}\right)$, $x > 0$.

$D^2 = x^2 + \dfrac{16}{x^2}$.

$(D^2)' = 2x - \dfrac{32}{x^3} = 0 \Rightarrow x^4 = 16 \Rightarrow x = 2$ (positive root).

$D^2 \to \infty$ as $x \to 0^+$ and as $x \to \infty$, and $x = 2$ is the only critical point, so it is the global min.

$D^2 = 4 + 4 = 8$, $D = 2\sqrt{2}$, at $(2, 2)$.