1. **Problem statement:** We have a 12-inch by 16-inch piece of cardboard. We cut out squares 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 and then find the dimensions of the box. 2. **Formula for volume:** The volume $V$ of the box is given by the area of the base times the height. After cutting out squares of side $x$, the base dimensions become $(16 - 2x)$ by $(12 - 2x)$, and the height is $x$. So, $$V = x(16 - 2x)(12 - 2x)$$ 3. **Simplify the volume expression:** $$V = x(192 - 32x - 24x + 4x^2) = x(192 - 56x + 4x^2) = 192x - 56x^2 + 4x^3$$ 4. **Find critical points:** To maximize volume, find $V'$ and set it to zero. $$V' = 192 - 112x + 12x^2$$ Set $V' = 0$: $$12x^2 - 112x + 192 = 0$$ Divide by 4: $$3x^2 - 28x + 48 = 0$$ 5. **Solve quadratic:** $$x = \frac{28 \pm \sqrt{28^2 - 4 \cdot 3 \cdot 48}}{2 \cdot 3} = \frac{28 \pm \sqrt{784 - 576}}{6} = \frac{28 \pm \sqrt{208}}{6}$$ $$\sqrt{208} = 4\sqrt{13} \approx 14.42$$ So, $$x_1 = \frac{28 + 14.42}{6} \approx 7.07, \quad x_2 = \frac{28 - 14.42}{6} \approx 2.60$$ 6. **Check domain:** Since $x$ must be less than half the smaller side (6 inches), $x=7.07$ is invalid. So $x \approx 2.60$ inches. 7. **Calculate dimensions:** Bottom length = $16 - 2x = 16 - 2(2.60) = 16 - 5.2 = 10.8$ inches Bottom width = $12 - 2x = 12 - 5.2 = 6.8$ inches Height = $x = 2.60$ inches **Final answer:** Dimensions of the bottom: $10.8$ inches by $6.8$ inches Height of the box: $2.6$ inches