1. **State the problem:** We have a square piece of cardboard 2 ft wide. We cut out a square of side length $x$ from each corner and fold up the sides to form an open box. We want to find the value of $x$ that maximizes the volume of the box. 2. **Define variables:** - Side length of the original square cardboard: 2 ft - Side length of the cut-out squares: $x$ ft - Height of the box after folding: $x$ ft - Length and width of the base after cutting and folding: $2 - 2x$ ft 3. **Write the volume function:** The volume $V$ of the box is given by the product of height, length, and width: $$V = x(2 - 2x)(2 - 2x) = x(2 - 2x)^2$$ 4. **Simplify the volume function:** $$V = x(4 - 8x + 4x^2) = 4x - 8x^2 + 4x^3$$ 5. **Find the critical points by differentiating:** $$\frac{dV}{dx} = 4 - 16x + 12x^2$$ 6. **Set derivative equal to zero to find critical points:** $$4 - 16x + 12x^2 = 0$$ Divide entire equation by 4: $$1 - 4x + 3x^2 = 0$$ 7. **Solve quadratic equation:** $$3x^2 - 4x + 1 = 0$$ Using quadratic formula: $$x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6}$$ 8. **Calculate roots:** - $$x = \frac{4 + 2}{6} = 1$$ - $$x = \frac{4 - 2}{6} = \frac{2}{6} = \frac{1}{3}$$ 9. **Check domain:** Since $x$ is the side length of the cut-out square, it must satisfy $0 < x < 1$ (because cutting out squares larger than 1 ft would not leave any base). 10. **Evaluate volume at critical points and endpoints:** - At $x = 0$: $$V = 0$$ - At $x = \frac{1}{3}$: $$V = \frac{1}{3} (2 - 2 \cdot \frac{1}{3})^2 = \frac{1}{3} (2 - \frac{2}{3})^2 = \frac{1}{3} \left(\frac{4}{3}\right)^2 = \frac{1}{3} \cdot \frac{16}{9} = \frac{16}{27} \approx 0.5926$$ - At $x = 1$: $$V = 1 \cdot (2 - 2 \cdot 1)^2 = 1 \cdot 0^2 = 0$$ 11. **Conclusion:** The largest volume is approximately $0.5926$ cubic feet when the cut-out square side length is $\frac{1}{3}$ ft.