Calculus 1: Open-Top Box with Square Base and Volume 32 ft³, Minimize Surface Area

Exact Midterm Problem

An open-top rectangular box with a square base is designed to hold a volume of $32\text{ ft}^3$. Find the dimensions (base length $x$ and height $h$) that minimize total surface area.

Step-by-Step Mathematical Solution

1. Constraint: $V = x^2 h = 32 \implies h = \frac{32}{x^2}$
2. Objective Function: $S(x) = x^2 + 4xh = x^2 + 4x\left(\frac{32}{x^2}\right) = x^2 + \frac{128}{x}$
3. Differentiate & Set to Zero:

$$ S'(x) = 2x - \frac{128}{x^2} = 0 \implies 2x^3 = 128 \implies x^3 = 64 \implies \mathbf{x = 4\text{ ft}} $$

1. Dimensions & Minimum Surface Area:

• Height $h = \frac{32}{16} = \mathbf{2\text{ ft}}$
• Surface Area $= 4^2 + \frac{128}{4} = 16 + 32 = \mathbf{48\text{ ft}^2}$