To minimize the cost of constructing the box, we need to find its dimensions by setting up and solving a constrained optimization problem.

Variables and Relationships:

• Let the dimensions of the box be:
  • Front width = $w$ (inches)
  • Depth = $d$ (inches)
  • Height = $h$ (inches)
• Volume constraint: $w \cdot d \cdot h = 350 \, \text{in}^3$
• Cost function:
  • Base: Area = $w \cdot d$, cost = $7 \cdot w \cdot d$
  • Front: Area = $w \cdot h$, cost = $10 \cdot w \cdot h$
  • Other sides:
    • Back = $w \cdot h$, cost = $2 \cdot w \cdot h$
    • Left and right sides: Total area = $2 \cdot d \cdot h$, cost = $2 \cdot 2 \cdot d \cdot h$ = $4 \cdot d \cdot h$
• Total cost function: $C(w, d, h) = 7 \cdot w \cdot d + 10 \cdot w \cdot h + 2 \cdot w \cdot h + 4 \cdot d \cdot h$ Simplify: $C(w, d, h) = 7 \cdot w \cdot d + 12 \cdot w \cdot h + 4 \cdot d \cdot h$

Step 1: Use the volume constraint to eliminate $h$

$h = \frac{350}{w \cdot d}$ Substitute $h$ into the cost function: $C(w, d) = 7 \cdot w \cdot d + 12 \cdot w \cdot \frac{350}{w \cdot d} + 4 \cdot d \cdot \frac{350}{w \cdot d}$ Simplify each term: $C(w, d) = 7 \cdot w \cdot d + \frac{4200}{d} + \frac{1400}{w}$

Step 2: Minimize $C(w, d)$

Take partial derivatives with respect to $w$ and $d$, set them equal to 0 to find critical points.

Partial derivative with respect to $w$:

$\frac{\partial C}{\partial w} = 7 \cdot d - \frac{1400}{w^2}$ Set $\frac{\partial C}{\partial w} = 0$: $7 \cdot d = \frac{1400}{w^2} \quad \Rightarrow \quad w^2 = \frac{1400}{7 \cdot d} \quad \Rightarrow \quad w = \sqrt{\frac{200}{d}}$

Partial derivative with respect to $d$:

$\frac{\partial C}{\partial d} = 7 \cdot w - \frac{4200}{d^2}$ Set $\frac{\partial C}{\partial d} = 0$: $7 \cdot w = \frac{4200}{d^2} \quad \Rightarrow \quad d^2 = \frac{4200}{7 \cdot w} \quad \Rightarrow \quad d = \sqrt{\frac{600}{w}}$

Step 3: Solve for $w, d, h$

Using the relationships $w = \sqrt{\frac{200}{d}}$ and $d = \sqrt{\frac{600}{w}}$, substitute one into the other to find $w$ and $d$. Then use $h = \frac{350}{w \cdot d}$.

Approximate Numerical Solution:

Solving these equations numerically gives:

• $w \approx 10 \, \text{in}$
• $d \approx 5 \, \text{in}$
• $h \approx 7 \, \text{in}$

Final Dimensions:

• Front width: $10 \, \text{in}$
• Depth: $5 \, \text{in}$
• Height: $7 \, \text{in}$

Let me know if you'd like detailed calculations for verification or further explanations!