More Optimization — Question 10

Question 10 - Solution

Let: - $x$ = length of the base side (square base) - $h$ = height of the box - Volume: $V = x^2 h \Rightarrow h = \frac{V}{x^2}$

Surface Area (SA): The box has no top, so: $$A(x) = \text{Base} + \text{4 sides} = x^2 + 4xh$$ $$A(x) = x^2 + 4x\left( \frac{V}{x^2} \right) = x^2 + \frac{4V}{x}$$

Minimize $A(x)$:

Differentiate: $$A'(x) = 2x - \frac{4V}{x^2}$$

Set $A'(x) = 0$: $$2x - \frac{4V}{x^2} = 0 \Rightarrow 2x^3 = 4V \Rightarrow x^3 = 2V \Rightarrow x = \sqrt[3]{2V}$$

Then: $$h = \frac{V}{x^2} = \frac{V}{\left( \sqrt[3]{2V} \right)^2} = \frac{V}{\sqrt[3]{4V^2}} = \sqrt[3]{\frac{V}{4}}$$

Answer: $$\boxed{ x = \sqrt[3]{2V}, \quad h = \sqrt[3]{\frac{V}{4}} }$$

These dimensions give the minimum surface area for the box.