An open-top rectangular box is being constructed to hold a volume of 200 in3. The base of the box is made from a material costing 5 cents/in2. The front of the box must be decorated, and will cost 9 cents/in2. The remainder of the sides will cost 3 cents/in2.

Find the dimensions that will minimize the cost of constructing this box. Please show your answers to at least 4 decimal places.

Front width:    in.
Depth:    in.
Height: 

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 \( h \)**: \(\)

An open-top rectangular box is being constructed to hold a volume of 200 in3. The base of the box is made from a material costing 5 cents/in2. The front of the box must be decorated, and will cost 9 cents/in2. The remainder of the sides will cost 3 cents/in2. Find the dimensions that will minimize the cost of constructing this box. Please show your answers to at least 4 decimal places.

Learn how to minimize the cost of constructing an open-top rectangular box with a volume constraint of 200 cubic inches. This problem involves optimization using partial derivatives, covering concepts in calculus, cost functions, and practical geometry. Suitable for students in Grades 11-12 and college.

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