Let's solve the problem step by step, providing detailed explanations for each part as per Chegg guidelines. --- ### **a) Write the general formulas for the volume \( V \) and surface area \( S \).** Let the side lengths of the box be \( x \) (length), \( y \) (width), and \( z \) (height). - **Volume formula:** \[ V(x, y, z) = x y z \] *Explanation:* Volume of a box is the product of its three side lengths. - **Surface area formula (closed top):** \[ S(x, y, z) = xy + 2xz + 2yz \] *Explanation:* The surface area consists of the base (\( xy \)), two sides (\( 2xz \)), and two ends (\( 2yz \)). Only one base because the box has a closed top. --- ### **b) Express \( V(x, y) \) as a function of \( x \) and \( y \) using the surface area constraint.** Given: Surface area \( S = 96 \). \[ xy + 2xz + 2yz = 96 \] Solve for \( z \): \[ 2xz + 2yz = 96 - xy \\ 2z(x + y) = 96 - xy \\ z = \frac{96 - xy}{2(x + y)} \] Now substitute this into the volume formula: \[ V(x, y) = x y z = xy \left(\frac{96 - xy}{2(x + y)}\right) = \frac{xy(96 - xy)}{2(x + y)} \] *Explanation:* We eliminate \( z \) using the surface area constraint so that the volume is a function of just \( x \) and \( y \). --- ### **c) Why must \( x > 0 \) and \( y > 0 \)?** **Correct answer:** **D. Physical dimensions like length and width cannot be zero or negative if we want a positive, non-zero volume.** *Explanation:* Negative or zero values for \( x \) or \( y \) don't make sense for real physical dimensions of a box. --- ### **d) Find the critical point \((x, y)\) that maximizes \( V(x, y) \).** We need to find where the partial derivatives are zero. #### 1. **Set up the volume function:** \[ V(x, y) = \frac{xy(96 - xy)}{2(x + y)} \] #### 2. **Assume symmetry for maximum volume:** Usually, the optimal box occurs when \( x = y \) (by symmetry). Let's try \( x = y \): \[ z = \frac{96 - x^2}{2(2x)} = \frac{96 - x^2}{4x} \] \[ V(x, x) = x \cdot x \cdot z = x^2 \cdot \frac{96 - x^2}{4x} = \frac{x(96 - x^2)}{4} \] #### 3. **Maximize \( V(x) \):** Set the derivative with respect to \( x \) to zero: \[ V(x) = \frac{x(96 - x^2)}{4} \] \[ V'(x) = \frac{96 - x^2}{4} + \frac{x \cdot (-2x)}{4} = \frac{96 - x^2 - 2x^2}{4} = \frac{96 - 3x^2}{4} \] Set \( V'(x) = 0 \): \[ 96 - 3x^2 = 0 \implies x^2 = 32 \implies x = \sqrt{32} \] So, \( x = y = \sqrt{32} \). #### 4. **Find \( z \):** \[ z = \frac{96 - x^2}{4x} = \frac{96 - 32}{4\sqrt{32}} = \frac{64}{4\sqrt{32}} = \frac{16}{\sqrt{32}} = \frac{16}{4\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \] **Critical point:** \[ (\sqrt{32}, \sqrt{32}) \] --- ### **e) What are the dimensions \((x, y, z)\) of the box with maximal volume?** Dimensions: \[ (\sqrt{32}, \sqrt{32}, 2\sqrt{2}) \] --- ## **Final Answers (as per Chegg guidelines):** **(a)** \( V(x, y, z) = x y z \) \( S(x, y, z) = xy + 2xz + 2yz \) --- **(b)** \( V(x, y) = \frac{xy(96 - xy)}{2(x + y)} \) --- **(c)** **D. Physical dimensions like length and width cannot be zero or negative if we want a positive, non-zero volume.** --- **(d)** Critical point: \( (\sqrt{32}, \sqrt{32}) \) --- **(e)** Dimensions: \( (\sqrt{32}, \sqrt{32}, 2\sqrt{2}) \) --- If you want to see the answer in decimal form: - \( \sqrt{32} \approx 5.657 \) - \( 2\sqrt{2} \approx 2.828 \) So, approximate dimensions: **(5.66, 5.66, 2.83)** Let me know if you need the detailed work for derivatives or further steps!