1. **State the problem:** We need to find the dimensions of a box with a square base and an open top that has a volume of 4 ft³ and requires the least material, meaning the surface area is minimized. 2. **Define variables:** Let the side length of the square base be $x$ ft and the height of the box be $h$ ft. 3. **Write the volume formula:** The volume $V$ of the box is given by $$V = x^2 h$$ Given $V = 4$, we have $$x^2 h = 4$$ 4. **Express height in terms of $x$:** $$h = \frac{4}{x^2}$$ 5. **Write the surface area formula:** Since the box has an open top, the surface area $S$ consists of the base and the four sides: $$S = x^2 + 4xh$$ 6. **Substitute $h$ into $S$:** $$S = x^2 + 4x \cdot \frac{4}{x^2} = x^2 + \frac{16}{x}$$ 7. **Minimize the surface area:** To find the minimum, take the derivative of $S$ with respect to $x$ and set it to zero: $$\frac{dS}{dx} = 2x - \frac{16}{x^2} = 0$$ 8. **Solve for $x$:** Multiply both sides by $x^2$ to clear the denominator: $$x^2 \cdot 2x - x^2 \cdot \frac{16}{x^2} = 0 \Rightarrow 2x^3 - 16 = 0$$ 9. **Simplify:** $$2x^3 = 16$$ $$x^3 = 8$$ 10. **Find $x$:** $$x = \sqrt[3]{8} = 2$$ 11. **Find $h$ using $x=2$:** $$h = \frac{4}{2^2} = \frac{4}{4} = 1$$ 12. **Verify minimum:** The second derivative is $$\frac{d^2S}{dx^2} = 2 + \frac{32}{x^3}$$ At $x=2$, $$2 + \frac{32}{8} = 2 + 4 = 6 > 0$$ which confirms a minimum. **Final answer:** The dimensions that minimize the material are a base side length of $2$ ft and a height of $1$ ft.