A rectangular box with a square base and open top is to have a volume of 32 cubic meters. Find the dimensions that minimize the amount of material used.

Let $x$ = side length of the square base, $h$ = height.

Constraint: $V = x^2h = 32$, so $h = \frac{32}{x^2}$

Surface area (objective function): $S(x) = x^2 + 4xh = x^2 + 4x \cdot \frac{32}{x^2} = x^2 + \frac{128}{x}$

Domain: $x > 0$ (physical constraint)

$S'(x) = 2x - \frac{128}{x^2} = \frac{2x^3 - 128}{x^2}$

Setting $S'(x) = 0$: $2x^3 = 128 \implies x^3 = 64 \implies x = 4$

$S''(x) = 2 + \frac{256}{x^3}, \quad S''(4) = 2 + \frac{256}{64} = 6 > 0$

So $x = 4$ gives a minimum. Then $h = \frac{32}{16} = 2$.

Answer: Base $4 \times 4$ meters, height $2$ meters.

A company can sell $x$ items per week at a price of $p(x) = 200 - 0.5x$ dollars each. The cost of producing $x$ items is $C(x) = 1000 + 40x$ dollars. How many items should be produced to maximize profit?

Revenue: $R(x) = xp(x) = x(200 - 0.5x) = 200x - 0.5x^2$

Profit: $P(x) = R(x) - C(x) = 200x - 0.5x^2 - 1000 - 40x = -0.5x^2 + 160x - 1000$

$P'(x) = -x + 160$

Setting $P'(x) = 0$: $x = 160$

$P''(x) = -1 < 0$

So $x = 160$ gives a maximum. The company should produce 160 items per week.

Find the point on the parabola $y = x^2$ closest to the point $(0, 1)$.

Distance from $(x, x^2)$ to $(0, 1)$: $D(x) = \sqrt{x^2 + (x^2 - 1)^2}$

To simplify, minimize $D^2$ instead: $f(x) = x^2 + (x^2 - 1)^2 = x^2 + x^4 - 2x^2 + 1 = x^4 - x^2 + 1$

$f'(x) = 4x^3 - 2x = 2x(2x^2 - 1)$

Critical points: $x = 0, \pm\frac{1}{\sqrt{2}}$

$f''(x) = 12x^2 - 2$

At $x = 0$: $f''(0) = -2 < 0$ (local maximum) At $x = \pm\frac{1}{\sqrt{2}}$: $f''\left(\frac{1}{\sqrt{2}}\right) = 12 \cdot \frac{1}{2} - 2 = 4 > 0$ (local minimum)

The closest points are $\left(\pm\frac{1}{\sqrt{2}}, \frac{1}{2}\right)$, each at distance $\frac{1}{\sqrt{2}}$ from $(0, 1)$.