To find the dimensions of the box with a square base and an open top that minimizes the amount of material used, we need to follow these steps:

Step 1: Define the variables.

• Let the side length of the square base be $x$ cm.
• Let the height of the box be $h$ cm.

Step 2: Express the volume constraint.

The volume of the box is given by: $V = x^2h$ We are told that the volume must be 32,000 cm³: $x^2h = 32000$ From this, we can express the height $h$ in terms of $x$: $h = \frac{32000}{x^2}$

Step 3: Express the surface area to minimize.

The surface area $S$ of the box is the amount of material used. Since the box has an open top, the surface area is the sum of the area of the base and the four sides: $S = x^2 + 4xh$ Substituting $h = \frac{32000}{x^2}$ into this equation: $S = x^2 + 4x \left(\frac{32000}{x^2}\right) = x^2 + \frac{128000}{x}$

Step 4: Minimize the surface area.

To minimize the surface area, we take the derivative of $S$ with respect to $x$ and set it equal to zero: $\frac{dS}{dx} = 2x - \frac{128000}{x^2}$ Set $\frac{dS}{dx} = 0$ to find the critical points: $2x - \frac{128000}{x^2} = 0$ Multiply through by $x^2$ to eliminate the fraction: $2x^3 = 128000$ Solve for $x$: $x^3 = 64000 \quad \Rightarrow \quad x = \sqrt[3]{64000} = 40 \, \text{cm}$

Step 5: Find the corresponding height.

Using $x = 40$ cm, find the height $h$: $h = \frac{32000}{x^2} = \frac{32000}{40^2} = \frac{32000}{1600} = 20 \, \text{cm}$

Step 6: Conclusion

The dimensions of the box that minimize the amount of material used are:

• Base side length $x = 40$ cm
• Height $h = 20$ cm

These dimensions satisfy the volume constraint and minimize the surface area.

Would you like to see a detailed explanation of the derivative step, or do you have any other questions?

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Here are five related questions:

1. What is the minimum surface area calculated with these dimensions?
2. How would the problem change if the box had a top?
3. Can this method be applied to a box with a rectangular base instead of a square base?
4. How would the solution change if the volume were different?
5. What if the material thickness varied for different parts of the box?

Tip: Always check the units of your final answer to ensure they align with the problem's requirements.