1. **Problem Statement:** We have a square cardboard of side length 40 inches. Four squares of side length $x$ and two rectangular regions are cut out and discarded. The remaining shaded part folds into a rectangular box with a lid. We want to determine which statement about the volume $V$ of the box is true. 2. **Understanding the box dimensions:** - The side length of the original square is 40 inches. - Four squares of side $x$ are cut from the corners. - The height of the box after folding is $x$ (the side length of the cut squares). - The length and width of the box base are each $40 - 2x$ because the cuts reduce the length on both sides. 3. **Volume formula:** The volume $V$ of the box is given by: $$V = \text{length} \times \text{width} \times \text{height} = (40 - 2x)(40 - 2x)(x) = x(40 - 2x)^2$$ 4. **Simplify the volume expression:** $$V = x(40 - 2x)^2 = x(1600 - 160x + 4x^2) = 1600x - 160x^2 + 4x^3$$ 5. **Find critical points to determine max/min volume:** Take the derivative of $V$ with respect to $x$: $$\frac{dV}{dx} = 1600 - 320x + 12x^2$$ Set derivative equal to zero to find critical points: $$1600 - 320x + 12x^2 = 0$$ Divide entire equation by 4: $$400 - 80x + 3x^2 = 0$$ Rewrite: $$3x^2 - 80x + 400 = 0$$ 6. **Solve quadratic equation:** Using quadratic formula: $$x = \frac{80 \pm \sqrt{(-80)^2 - 4 \times 3 \times 400}}{2 \times 3} = \frac{80 \pm \sqrt{6400 - 4800}}{6} = \frac{80 \pm \sqrt{1600}}{6} = \frac{80 \pm 40}{6}$$ Two solutions: - $$x = \frac{80 + 40}{6} = \frac{120}{6} = 20$$ - $$x = \frac{80 - 40}{6} = \frac{40}{6} = \frac{20}{3} \approx 6.67$$ 7. **Determine nature of critical points:** Second derivative: $$\frac{d^2V}{dx^2} = -320 + 24x$$ Evaluate at $x=20$: $$-320 + 24 \times 20 = -320 + 480 = 160 > 0$$ So $x=20$ is a local minimum. Evaluate at $x=\frac{20}{3}$: $$-320 + 24 \times \frac{20}{3} = -320 + 160 = -160 < 0$$ So $x=\frac{20}{3}$ is a local maximum. 8. **Conclusion:** - At $x=20$, the volume is minimum. - At $x=\frac{20}{3}$, the volume is maximum. Therefore, the correct statement is: **(D) When $x = \frac{20}{3}$ inches, the box has a maximum possible volume.**