Let's solve the problem step by step: ### 1. **Define Variables** - Let \( x \) be the length of a side of the square base (in feet). - Let \( h \) be the height of the box (in feet). ### 2. **Volume Constraint** The volume \( V \) of the box is: \[ V = x^2 h = 50 \] So, \[ h = \frac{50}{x^2} \] ### 3. **Cost Function** - **Top and Bottom:** Both are squares of area \( x^2 \) each. Total area: \( 2x^2 \). Cost per square foot: $3. - **Sides:** There are 4 sides, each of area \( x \times h \). Total area: \( 4xh \). Cost per square foot: $4. **Total cost \( C \):** \[ C = 3 \cdot (2x^2) + 4 \cdot (4xh) \] Wait, that's incorrect for the sides: The cost per square foot for sides is $4, so: \[ C = 3 \cdot (2x^2) + 4 \cdot (4xh) \] But the total area of the sides is \( 4xh \), not 4 times \( 4xh \): \[ C = 3 \cdot (2x^2) + 4 \cdot (4xh) \] No, it should be: \[ C = 3 \cdot (2x^2) + 4 \cdot (4xh) \] This is still wrong. Let's clarify: - Top and Bottom: \( 2x^2 \) at $3 per sq ft \(\to 3 \cdot 2x^2 = 6x^2 \) - Sides: \( 4xh \) at $4 per sq ft \(\to 4 \cdot 4xh = 16xh \) Wait, that's too much. It should be: - Sides: Each side \( x \cdot h \), 4 sides: \( 4xh \), cost per square foot $4, so total side cost: \( 4 \cdot 4xh = 16xh \) But that's not right. It should be just \( 4xh \cdot 4 = 16xh \). But that's not the usual way. Usually, it's: - Top and bottom: \( 2x^2 \), cost per area: 3 → \( 3 \cdot 2x^2 = 6x^2 \) - Sides: \( 4xh \), cost per area: 4 → \( 4 \cdot 4xh = 16xh \) But that's multiplying by 4 twice. So **sides cost** is: area \( 4xh \), cost per area $4, so \( 4xh \cdot 4 = 16xh \). But that's still odd. Wait! Each side is \( xh \), four sides is \( 4xh \), at $4 per square foot, so total side cost is \( 4 \times xh \times 4 = 16xh \). I think that's correct. So the cost function is: \[ C = 6x^2 + 16xh \] ### 4. **Substitute the Volume Constraint** Recall: \( h = \frac{50}{x^2} \) Plug into cost function: \[ C(x) = 6x^2 + 16x \left(\frac{50}{x^2}\right) = 6x^2 + 16 \cdot 50 \cdot \frac{1}{x} = 6x^2 + \frac{800}{x} \] ### 5. **Minimize the Cost Function** Take the derivative with respect to \( x \), set to zero: \[ C'(x) = 12x - \frac{800}{x^2} \] Set to zero: \[ 12x - \frac{800}{x^2} = 0 \\ 12x = \frac{800}{x^2} \\ 12x^3 = 800 \\ x^3 = \frac{800}{12} = \frac{200}{3} \\ x = \left(\frac{200}{3}\right)^{1/3} \] ### 6. **Find \( h \)** Recall: \[ h = \frac{50}{x^2} \] ### 7. **Final Answers** #### a) **Length of Base:** \[ \boxed{x = \left(\frac{200}{3}\right)^{1/3} \approx 3.62 \text{ feet}} \] #### b) **Height:** \[ h = \frac{50}{x^2} = \frac{50}{\left(\left(\frac{200}{3}\right)^{1/3}\right)^2} \] \[ x^2 = \left(\frac{200}{3}\right)^{2/3} \] So, \[ h = \frac{50}{\left(\frac{200}{3}\right)^{2/3}} \approx 3.80 \text{ feet} \] --- ## **Final Dimensions (rounded to two decimal places):** \[ \boxed{ \begin{aligned} \text{Length of base } (x) & = 3.62\ \text{ft} \\ \text{Height } (h) & = 3.80\ \text{ft} \end{aligned} } \]