Your task is to design a rectanglular industrial warehouse consisting of three separe spaces of equal size. The wall materials cos $500 per linear meter and your company allocotes $1,200,00 for the part of the project involving the walls. Which dimensions maximize the area of the warehouse? What is the area of each compare

Let's solve this problem step by step to find the dimensions that maximize the area of the warehouse and then determine the area of each of the three separate spaces.

### Step 1: Total Budget for the Walls

The budget allocated for the walls is $1,200,000, and the cost per linear meter of wall is $500. Thus, the total length of all the walls (perimeter and internal) can be calculated as:
\[
\text{Total Wall Length} = \frac{1,200,000}{500} = 2400 \text{ meters}
\]

### Step 2: Understanding the Layout

Since the warehouse is divided into three equal spaces, we will have:
- Two internal walls, each parallel to the width $ W $ of the warehouse, of length $ L $.
- Two walls of length $ L $ (the length of the warehouse).
- Two walls of width $ W $ (the width of the warehouse).

The total length of the walls can be expressed as:
\[
\text{Total Wall Length} = 4L + 2W
\]
Given that the total wall length is 2400 meters, we have:
\[
4L + 2W = 2400
\]

### Step 3: Express $ W $ in Terms of $ L $

Solving for $ W $:
\[
2W = 2400 - 4L
\]
\[
W = 1200 - 2L
\]

### Step 4: Maximize the Area of the Warehouse

The area $ A $ of the warehouse is:
\[
A = L \times W
\]
Substituting $ W = 1200 - 2L $ into the area equation:
\[
A(L) = L \times (1200 - 2L)
\]
\[
A(L) = 1200L - 2L^2
\]
This is a quadratic function of $ L $.

### Step 5: Find the Value of $ L $ that Maximizes the Area

To find the value of $ L $ that maximizes the area, take the derivative of $ A(L) $ with respect to $ L $ and set it to zero:
\[
\frac{dA}{dL} = 1200 - 4L = 0
\]
\[
4L = 1200
\]
\[
L = 300 \text{ meters}
\]
Substitute $ L = 300 $ meters back into the equation for $ W $:
\[
W = 1200 - 2(300) = 600 \text{ meters}
\]

### Step 6: Calculate the Total Area and Area of Each Space

The total area of the warehouse is:
\[
A_{\text{total}} = L \times W = 300 \times 600 = 180,000 \text{ square meters}
\]
Since the warehouse is divided into three equal spaces, the area of each space is:
\[
A_{\text{each}} = \frac{180,000}{3} = 60,000 \text{ square meters}
\]

### Conclusion

- **Dimensions of the warehouse**: Length $ L = 300 $ meters, Width $ W = 600 $ meters.
- **Total area of the warehouse**: 180,000 square meters.
- **Area of each of the three separate spaces**: 60,000 square meters.

Would you like more details or have any questions?

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Here are some related questions:
1. How would changing the cost per linear meter affect the dimensions?
2. What if the budget was only $1,000,000? How would that change the dimensions and area?
3. How would the design change if the warehouse had to be divided into four equal spaces?
4. What is the impact on area if the internal walls are more expensive than the perimeter walls?
5. How can you verify that these dimensions indeed give the maximum area?

**Tip:** In problems involving optimization, always check the feasibility of the solution by substituting the calculated values back into the original constraints.

Learn how to maximize the area of a warehouse divided into three equal spaces with a specified budget for wall materials. Discover the optimal dimensions and the area of each space. Suitable for high school students and anyone interested in optimization problems.

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