To solve the problem, we need to minimize the surface area of the closed rectangular box while maintaining a given volume of $145 \, \text{in}^3$.

Variables:

• $x$: The side length of the square base.
• $y$: The height of the box.
• Volume constraint: $x^2y = 145$.
• Surface area to minimize: $A = 2x^2 + 4xy$.

Steps:

1. Express $y$ in terms of $x$: From the volume constraint $x^2y = 145$: $y = \frac{145}{x^2}.$

2. Substitute $y$ into the surface area equation: $A = 2x^2 + 4x \cdot \frac{145}{x^2}.$ Simplify: $A = 2x^2 + \frac{580}{x}.$

3. Minimize $A$: Differentiate $A$ with respect to $x$: $\frac{dA}{dx} = 4x - \frac{580}{x^2}.$ Set $\frac{dA}{dx} = 0$ to find critical points: $4x = \frac{580}{x^2}.$ Multiply through by $x^2$: $4x^3 = 580.$ Solve for $x$: $x^3 = 145, \quad x = \sqrt[3]{145}.$

4. Find $y$: Substitute $x = \sqrt[3]{145}$ into the volume equation: $y = \frac{145}{x^2} = \frac{145}{\left(\sqrt[3]{145}\right)^2}.$

5. Numerical Approximation: Compute $x$ and $y$ to two decimal places: $x \approx \sqrt[3]{145} \approx 5.28, \quad y \approx \frac{145}{5.28^2} \approx 5.19.$

Final Dimensions:

• $x \approx 5.28 \, \text{in}$,
• $y \approx 5.19 \, \text{in}$.

Let me know if you'd like further clarification or a breakdown of any step.

1. How is surface area related to minimization problems?
2. Why do we need to use derivatives for this problem?
3. What is the role of constraints in optimization problems?
4. Can this method be generalized for other shapes?
5. How does this method ensure the box uses the least material?

Tip: Always ensure your derivative equations are simplified before solving. It makes computation easier.