Let's solve step by step.

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1. Write an equation for $y$ in terms of $x$:

The volume of the box is $144 \, \text{in}^3$. The box has a square base with side length $x$, so its area is $x^2$. The height of the box is $y$, so: $\text{Volume} = \text{Base Area} \times \text{Height}$ $x^2 y = 144$ $y = \frac{144}{x^2}$

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2. Write a function $f(x)$ for the material needed to create the box:

The surface area of the box includes:

• The bottom (square): $x^2$
• The top (square): $x^2$
• The four rectangular sides: $4 \times (x \cdot y) = 4xy$

The total surface area is: $\text{Surface Area} = 2x^2 + 4xy$ Substitute $y = \frac{144}{x^2}$ into $4xy$: $f(x) = 2x^2 + 4x \cdot \frac{144}{x^2}$ $f(x) = 2x^2 + \frac{576}{x}$

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3. Find $f'(x)$ and $f''(x)$:

First derivative $f'(x)$:

$f'(x) = \frac{d}{dx} \left( 2x^2 + \frac{576}{x} \right)$ $f'(x) = 4x - \frac{576}{x^2}$

Second derivative $f''(x)$:

$f''(x) = \frac{d}{dx} \left( 4x - \frac{576}{x^2} \right)$ $f''(x) = 4 + \frac{1152}{x^3}$

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4. Minimize $f(x)$:

To minimize the material, find the critical points of $f(x)$. Set $f'(x) = 0$: $4x - \frac{576}{x^2} = 0$ $4x = \frac{576}{x^2}$ $4x^3 = 576$ $x^3 = 144$ $x = \sqrt[3]{144} \approx 5.24 \, \text{in}$

Find $y$:

Substitute $x$ into $y = \frac{144}{x^2}$: $y = \frac{144}{(5.24)^2} \approx \frac{144}{27.46} \approx 5.24 \, \text{in}$

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Final Dimensions:

• $x = 5.24 \, \text{in}$
• $y = 5.24 \, \text{in}$

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Would you like additional details or explanations for any step?

Here are 5 questions for further exploration:

1. What is the significance of critical points in finding the minimum material?
2. How would the results change if the box had no top?
3. Can this problem be solved using a numerical approximation method?
4. What would the dimensions be if the volume was doubled to $288 \, \text{in}^3$?
5. How does the second derivative confirm a minimum in this problem?

Tip: For optimization problems, always verify critical points using the second derivative or a test for minimum/maximum.