More Optimization — Question 7

Question 7 - Solution

Let: - $x$ = side length of the square base (in cm) - $h$ = height of the box (in cm)

Volume constraint: $$V = x^2 h = 500 \Rightarrow h = \frac{500}{x^2}$$

Surface area to minimize: - One bottom square: $x^2$ - Four sides: $4(xh)$

$$A(x) = x^2 + 4xh = x^2 + 4x \cdot \frac{500}{x^2} = x^2 + \frac{2000}{x}$$

Take derivative: $$A'(x) = 2x - \frac{2000}{x^2}$$

Set derivative to zero: $$2x = \frac{2000}{x^2} \Rightarrow 2x^3 = 2000 \Rightarrow x^3 = 1000 \Rightarrow x = 10$$

Find $h$: $$h = \frac{500}{x^2} = \frac{500}{100} = 5$$

Answer: $$\boxed{ x = 10\ \text{cm}, \quad h = 5\ \text{cm} }$$

These dimensions minimize the surface area.