To solve this problem, let's break it down step-by-step.

Step 1: Define the Volume and Surface Area Formulas

1. Volume: We have a box with a square base of side length $x$ and height $h$. The volume $V$ of the box is given by: $V = x^2 \cdot h$ Since we know the volume must be $42592 \, \text{cm}^3$, we have: $x^2 \cdot h = 42592$ Solving for $h$ in terms of $x$: $h = \frac{42592}{x^2}$

2. Surface Area: The box has a square base and an open top, so its surface area $A$ is given by: $A = x^2 + 4 \cdot (x \cdot h)$ Substituting $h = \frac{42592}{x^2}$ into the surface area formula: $A(x) = x^2 + 4 \cdot \left(x \cdot \frac{42592}{x^2}\right)$ Simplifying further: $A(x) = x^2 + \frac{170368}{x}$

Step 2: Find the Derivative $A'(x)$

To minimize the surface area, we need to find the critical points by setting the derivative $A'(x)$ equal to zero.

1. Differentiate $A(x) = x^2 + \frac{170368}{x}$ with respect to $x$: $A'(x) = 2x - \frac{170368}{x^2}$

Step 3: Set $A'(x) = 0$ and Solve for $x$

To find when $A'(x) = 0$: $2x - \frac{170368}{x^2} = 0$ Multiply both sides by $x^2$ to clear the fraction: $2x^3 = 170368$ Now, divide by 2: $x^3 = 85184$ Taking the cube root of both sides: $x = \sqrt[3]{85184}$ Calculating this value gives: $x \approx 44.3 \, \text{cm}$

Step 4: Confirm Minimum with the Second Derivative Test

1. Find the second derivative $A''(x)$: $A''(x) = 2 + \frac{2 \cdot 170368}{x^3}$

2. Evaluate $A''(x)$ at $x = 44.3$: Since $A''(x) > 0$, this confirms that the surface area function $A(x)$ is concave up at $x = 44.3$, indicating a local minimum.

Thus, the dimension $x \approx 44.3$ cm for the side length of the square base minimizes the material used.

Would you like more details on any step or have any questions?

Here are some related questions to explore further:

1. How would the formula change if the box had a closed top?
2. What if the volume requirement changed? How would that affect the optimal dimensions?
3. How does the second derivative test confirm a minimum for this function?
4. Could you solve for the height $h$ given $x \approx 44.3$?
5. Why is it necessary to multiply by $x^2$ when solving for $x$ in this problem?

Tip: When dealing with optimization problems, always check if you’re finding a minimum or maximum by using the second derivative or analyzing the function's behavior.