a college is planning to construct a new parking lot. the parking lot should be rectangular and enclose 6000 sq metres of land. a fence will surround the parking lot and another fence parallel to one of the sides will fdivide the parking lot into two sections. what are the dimensions in meteres of the rectangula rlot that will use the least amount of fencing?

To solve this problem, we need to find the dimensions of the rectangular parking lot that enclose 6000 square meters of land and minimize the amount of fencing required.

### Step 1: Define the variables
Let:
- \( x \) be the length of the parking lot (in meters).
- \( y \) be the width of the parking lot (in meters).

The area of the parking lot is given by:
\[
x \times y = 6000 \quad \text{(1)}
\]
We need to minimize the total amount of fencing. The fencing includes:
- Two lengths (2 sides of \( x \)),
- Three widths (2 sides of \( y \) plus 1 internal fence parallel to one side).

Thus, the total amount of fencing \( F \) is given by:
\[
F = 2x + 3y \quad \text{(2)}
\]

### Step 2: Express \( y \) in terms of \( x \)
From the area equation (1):
\[
y = \frac{6000}{x} \quad \text{(3)}
\]
Substitute this expression for \( y \) into the fencing equation (2):
\[
F(x) = 2x + 3\left(\frac{6000}{x}\right)
\]
Simplify:
\[
F(x) = 2x + \frac{18000}{x}
\]

### Step 3: Find the minimum value of \( F(x) \)
To find the minimum amount of fencing, we take the derivative of \( F(x) \) with respect to \( x \) and set it equal to zero:
\[
F'(x) = 2 - \frac{18000}{x^2}
\]
Set \( F'(x) = 0 \) to find the critical points:
\[
2 = \frac{18000}{x^2}
\]
Multiply both sides by \( x^2 \):
\[
2x^2 = 18000
\]
Solve for \( x^2 \):
\[
x^2 = \frac{18000}{2} = 9000
\]
Take the square root of both sides:
\[
x = \sqrt{9000} \approx 94.87 \quad \text{meters}
\]

### Step 4: Find \( y \) using the area equation
Now, substitute \( x = 94.87 \) meters into equation (3) to find \( y \):
\[
y = \frac{6000}{94.87} \approx 63.29 \quad \text{meters}
\]

### Step 5: Confirm it's a minimum
To confirm that this critical point gives a minimum, you can check the second derivative \( F''(x) \) or analyze the behavior of \( F'(x) \). In this case, \( F''(x) = \frac{36000}{x^3} \), which is positive for \( x > 0 \), indicating a minimum.

### Final Answer:
The dimensions of the rectangular lot that will use the least amount of fencing are approximately:
- Length: \( 94.87 \) meters
- Width: \( 63.29 \) meters

Would you like further explanation on any step, or do you have any questions?

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Here are 5 related questions:
1. What would be the dimensions if the parking lot area were different, say 8000 square meters?
2. How does the internal dividing fence impact the overall fencing cost?
3. What happens to the dimensions if the internal fence is removed?
4. How does the shape of the parking lot change if the minimum fencing constraint is not considered?
5. Could there be an optimal shape other than rectangular for this problem?

**Tip:** When optimizing with constraints, always substitute the constraint into the objective function to reduce the number of variables and simplify calculations.

Explore how to design a rectangular parking lot that encloses 6000 square meters with minimal fencing. Learn step-by-step calculations and solutions to minimize fencing costs in geometric optimization. Suitable for Grades 10-12.

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