We are tasked with finding the dimensions of an open-top rectangular box that minimizes the cost, given a volume constraint of $200 \text{ in}^3$ and different costs for the base, front, and other sides.

Step 1: Define the variables

• Let the width of the front of the box be $w$ (inches).
• Let the depth (side length perpendicular to the front) be $d$ (inches).
• Let the height of the box be $h$ (inches).

The volume constraint is: $w \cdot d \cdot h = 200$ Thus, $h = \frac{200}{w \cdot d}$.

Step 2: Cost function

The total cost of the box includes the base, the front side, and the other sides.

• The base area is $w \cdot d$, and it costs 5 cents per square inch. So, the cost of the base is: $\text{Base cost} = 5 \cdot w \cdot d$

• The front face area is $w \cdot h$, and it costs 9 cents per square inch. So, the cost of the front is: $\text{Front cost} = 9 \cdot w \cdot h$

• The remaining sides include:

  • Two sides, each with area $h \cdot d$, and they cost 3 cents per square inch. The total cost of these two sides is: $\text{Side cost} = 3 \cdot 2 \cdot h \cdot d = 6 \cdot h \cdot d$

• The back face is the same as the front but costs only 3 cents per square inch. So, the cost of the back face is: $\text{Back cost} = 3 \cdot w \cdot h$

Total cost function:

The total cost $C$ is the sum of these components: $C = 5 \cdot w \cdot d + 9 \cdot w \cdot h + 6 \cdot h \cdot d + 3 \cdot w \cdot h$ This simplifies to: $C = 5 \cdot w \cdot d + 12 \cdot w \cdot h + 6 \cdot h \cdot d$

Step 3: Substitute for $h$

Using the volume constraint $h = \frac{200}{w \cdot d}$, substitute this into the cost function: $C = 5 \cdot w \cdot d + 12 \cdot w \cdot \frac{200}{w \cdot d} + 6 \cdot \frac{200}{w \cdot d} \cdot d$ Simplify: $C = 5 \cdot w \cdot d + \frac{2400}{d} + \frac{1200}{w}$

Step 4: Minimize the cost

We now minimize $C$ with respect to $w$ and $d$. We do this by taking partial derivatives and setting them equal to zero.

1. Partial derivative with respect to $w$: $\frac{\partial C}{\partial w} = 5 \cdot d - \frac{1200}{w^2}$ Set $\frac{\partial C}{\partial w} = 0$: $5 \cdot d = \frac{1200}{w^2}$ $w^2 = \frac{1200}{5 \cdot d}$ $w = \sqrt{\frac{240}{d}}$

2. Partial derivative with respect to $d$: $\frac{\partial C}{\partial d} = 5 \cdot w - \frac{2400}{d^2}$ Set $\frac{\partial C}{\partial d} = 0$: $5 \cdot w = \frac{2400}{d^2}$ $d^2 = \frac{2400}{5 \cdot w}$ $d = \sqrt{\frac{480}{w}}$

Step 5: Solve the system of equations

We now have two equations: $w = \sqrt{\frac{240}{d}} \quad \text{and} \quad d = \sqrt{\frac{480}{w}}$

Substitute one into the other and solve for $w$ and $d$.

This requires solving the system of equations numerically or symbolically to find the optimal values of $w$, $d$, and $h$.

Conclusion:

• Front width $w$: $$$$
• Depth $d$: $$$$
• Height $h$: $$$$