To solve this problem, we need to:

1. Define the dimensions of the open-top box.
2. Set up an equation for the volume constraint.
3. Set up a cost function based on the surface area and given costs.
4. Use calculus to minimize the cost function under the volume constraint.

Let's proceed step by step.

Step 1: Define Dimensions

Let:

• $w$ be the width (front width) of the box.
• $d$ be the depth of the box.
• $h$ be the height of the box.

Step 2: Volume Constraint

The box must hold a volume of 200 in³, so we have: $w \cdot d \cdot h = 200$ This relationship will help us express one of the variables in terms of the others.

Step 3: Cost Function

The costs per square inch are as follows:

• Base: $5$ cents/in²
• Front side: $10$ cents/in²
• Other sides: $4$ cents/in²

The surface areas and costs for each part of the box are:

1. Base (bottom): Area = $w \cdot d$, Cost = $5 \cdot (w \cdot d)$.
2. Front Side: Area = $w \cdot h$, Cost = $10 \cdot (w \cdot h)$.
3. Back Side: Area = $w \cdot h$, Cost = $4 \cdot (w \cdot h)$.
4. Two Side Panels: Each has area $d \cdot h$, so total area is $2 \cdot d \cdot h$, and total cost is $4 \cdot (2 \cdot d \cdot h) = 8 \cdot d \cdot h$.

The total cost function $C$ in terms of $w$, $d$, and $h$ is: $C = 5wd + 10wh + 4wh + 8dh = 5wd + 14wh + 8dh$

Step 4: Substitute Using Volume Constraint

From $w \cdot d \cdot h = 200$, we can solve for $h$ as: $h = \frac{200}{wd}$

Substitute $h = \frac{200}{wd}$ into the cost function: $C = 5wd + 14w\left(\frac{200}{wd}\right) + 8d\left(\frac{200}{wd}\right)$ Simplify each term:

1. $5wd$ remains as is.
2. $14w \cdot \frac{200}{wd} = \frac{2800}{d}$.
3. $8d \cdot \frac{200}{wd} = \frac{1600}{w}$.

Thus, the cost function $C$ in terms of $w$ and $d$ becomes: $C(w, d) = 5wd + \frac{2800}{d} + \frac{1600}{w}$

Step 5: Minimize the Cost Function

To find the minimum cost, we can take partial derivatives of $C(w, d)$ with respect to $w$ and $d$, set them to zero, and solve the resulting equations. Let me calculate this.The real, positive solution for the dimensions (ignoring complex solutions) is:

$w = \frac{4 \cdot 980^{1/3}}{7} \approx 5.8480 \, \text{inches}$ $d = 980^{1/3} \approx 7.3014 \, \text{inches}$

Using the volume constraint $w \cdot d \cdot h = 200$, we can solve for $h$:

$h = \frac{200}{w \cdot d}$

Let me substitute these values to find $h$.The dimensions that minimize the cost of constructing the box are:

• Front width (w): 5.6759 inches
• Depth (d): 9.9329 inches
• Height (h): 3.5475 inches

These dimensions satisfy the volume constraint and minimize the material cost based on the specified prices.

Would you like more details on any of the steps or further clarification?

Here are some related questions to explore:

1. How would the cost function change if the box had a top?
2. What if the decorative front side also affected the depth cost?
3. How would a different volume constraint impact the dimensions?
4. How does changing the cost per square inch on each side affect the outcome?
5. Can these calculations be applied to optimize other types of containers?

Tip: For optimization problems with constraints, converting one variable using the constraint often simplifies the problem, making it easier to solve.