1. **State the problem:** We want to find the dimensions of an open rectangular box with a square base that maximizes volume, given that the surface area is 48 ft$^2$. 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 surface area formula:** Since the box is open at the top, the surface area $S$ is the area of the base plus the area of the four sides: $$S = x^2 + 4xh$$ Given $S = 48$, we have: $$x^2 + 4xh = 48$$ 4. **Express height $h$ in terms of $x$:** $$4xh = 48 - x^2$$ $$h = \frac{48 - x^2}{4x}$$ 5. **Write the volume formula:** Volume $V$ is base area times height: $$V = x^2 h$$ Substitute $h$: $$V = x^2 \times \frac{48 - x^2}{4x} = \frac{x(48 - x^2)}{4}$$ 6. **Simplify volume expression:** $$V = \frac{48x - x^3}{4} = 12x - \frac{x^3}{4}$$ 7. **Find critical points by differentiating $V$ with respect to $x$:** $$\frac{dV}{dx} = 12 - \frac{3x^2}{4}$$ 8. **Set derivative to zero to find maxima:** $$12 - \frac{3x^2}{4} = 0$$ Multiply both sides by 4: $$48 - 3x^2 = 0$$ $$3x^2 = 48$$ $$x^2 = 16$$ $$x = 4$$ (only positive root since length must be positive) 9. **Find corresponding height $h$:** $$h = \frac{48 - 4^2}{4 \times 4} = \frac{48 - 16}{16} = \frac{32}{16} = 2$$ 10. **Check second derivative to confirm maximum:** $$\frac{d^2V}{dx^2} = -\frac{3x}{2}$$ At $x=4$: $$\frac{d^2V}{dx^2} = -\frac{3 \times 4}{2} = -6 < 0$$ This confirms a local maximum. **Final answer:** The box with largest volume has base side length $4$ ft and height $2$ ft.