A figure shows a rectangle split into three equal vertical spaces. The wall materials cost $500 per linear meter and your company allocates $4,800,000 for that part of the project involving the walls. Which dimensions maximize the area of the warehouse? (Give exact answers. Use symbolic notation and fractions where needed.)

To maximize the area of the warehouse, we need to relate the cost of the wall materials to the dimensions of the rectangle and then optimize the area. Let's break the problem down step by step.

### Step 1: Define the variables
Let:
- $ L $ be the length of the warehouse (the horizontal side of the rectangle),
- $ H $ be the height of the warehouse (the vertical side of the rectangle).

The rectangle is split into three equal vertical sections, so each vertical division has a height of $ \frac{H}{3} $.

### Step 2: Calculate the total cost of the walls
The warehouse has four walls: two vertical and two horizontal. The vertical walls will each have a length $ H $, and the horizontal walls will each have a length $ L $.

- The cost of each vertical wall is proportional to its area, which is $ H \times 1 $ (since the cost is per meter of the wall). Since there are two vertical walls, the total cost for the vertical walls is:
  \[
  \text{Cost of vertical walls} = 2H \times 500 = 1000H
  \]

- The cost of each horizontal wall is proportional to its length $ L $, and since there are two horizontal walls, the total cost for the horizontal walls is:
  \[
  \text{Cost of horizontal walls} = 2L \times 500 = 1000L
  \]

Thus, the total cost of the walls is:
\[
\text{Total wall cost} = 1000H + 1000L
\]

### Step 3: Relate the cost to the budget
The total budget allocated for the walls is $4,800,000. Therefore, we have the equation:
\[
1000H + 1000L = 4,800,000
\]
Simplifying this:
\[
H + L = 4800
\]

### Step 4: Maximizing the area
The area $ A $ of the warehouse is:
\[
A = L \times H
\]
We want to maximize this area. From the previous equation, solve for $ H $ in terms of $ L $:
\[
H = 4800 - L
\]
Substitute this expression for $ H $ into the area formula:
\[
A = L \times (4800 - L) = 4800L - L^2
\]

### Step 5: Find the value of $ L $ that maximizes the area
To maximize the area, we take the derivative of $ A $ with respect to $ L $ and set it equal to zero:
\[
\frac{dA}{dL} = 4800 - 2L
\]
Set the derivative equal to zero:
\[
4800 - 2L = 0
\]
Solving for $ L $:
\[
2L = 4800 \quad \Rightarrow \quad L = 2400
\]

Now, substitute $ L = 2400 $ into the equation $ H = 4800 - L $:
\[
H = 4800 - 2400 = 2400
\]

### Step 6: Conclusion
The dimensions that maximize the area of the warehouse are:
- Length $ L = 2400 $ meters,
- Height $ H = 2400 $ meters.

The area is:
\[
A = 2400 \times 2400 = 5,760,000 \, \text{square meters}.
\]

These are the dimensions that maximize the area, given the wall material cost constraint.

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Would you like to see more details about the optimization process, or do you have any questions? Here are a few related questions:

1. How would the cost function change if the cost per meter of wall material were different?
2. What would happen if the warehouse had a fixed height instead of a fixed budget for walls?
3. How would the area change if we used more divisions along the height?
4. What is the effect of reducing the number of vertical walls on the total cost and area?
5. Can we apply the same method to other types of buildings with different shapes or materials?

Tip: When optimizing areas or volumes with cost constraints, always set up the problem in terms of a single variable (like length or height) to simplify the calculus.

Learn how to maximize the area of a warehouse given a cost constraint on wall materials. This problem involves optimizing the warehouse dimensions under a $4,800,000 budget for walls with a cost of $500 per linear meter. The optimal dimensions are found using calculus and algebra. Suitable for students in Grades 11-12.

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