To find the dimensions of the open-top box with the smallest amount of material, we need to minimize the surface area of the box while keeping its volume constant at 4 cubic feet.

Let's denote:

• $x$ as the side length of the square base,
• $h$ as the height of the box.

Given: $x^2h = 4$

The surface area $S$ of the box (with an open top) is given by: $S = x^2 + 4xh$

First, express $h$ in terms of $x$ using the volume constraint: $h = \frac{4}{x^2}$

Substitute this into the surface area formula: $S = x^2 + 4x \left( \frac{4}{x^2} \right)$ $S = x^2 + \frac{16}{x}$

To minimize $S$, we take the derivative of $S$ with respect to $x$ and set it to zero: $\frac{dS}{dx} = 2x - \frac{16}{x^2}$

Set the derivative equal to zero: $2x - \frac{16}{x^2} = 0$

Solve for $x$: $2x = \frac{16}{x^2}$ $2x^3 = 16$ $x^3 = 8$ $x = 2$

Now, find the corresponding height $h$: $h = \frac{4}{x^2} = \frac{4}{2^2} = 1$

Thus, the dimensions of the box that use the smallest amount of material are:

• Side length of the square base: $x = 2$ feet
• Height of the box: $h = 1$ foot

Would you like more details or have any questions about the solution?

Here are 5 related questions you can ask:

1. How do you find the critical points for minimizing or maximizing a function?
2. Can you explain why the second derivative test is useful in this context?
3. How would the problem change if the box had a closed top?
4. What are the applications of optimization in real-life scenarios?
5. Can you provide another example of an optimization problem?

Tip: When solving optimization problems, always check the second derivative to ensure that you have found a minimum (or maximum) point.