Math Problem Statement

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?

Solution

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:

  • xx be the length of the parking lot (in meters).
  • yy be the width of the parking lot (in meters).

The area of the parking lot is given by: x×y=6000(1)x \times y = 6000 \quad \text{(1)} We need to minimize the total amount of fencing. The fencing includes:

  • Two lengths (2 sides of xx),
  • Three widths (2 sides of yy plus 1 internal fence parallel to one side).

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

Step 2: Express yy in terms of xx

From the area equation (1): y=6000x(3)y = \frac{6000}{x} \quad \text{(3)} Substitute this expression for yy into the fencing equation (2): F(x)=2x+3(6000x)F(x) = 2x + 3\left(\frac{6000}{x}\right) Simplify: F(x)=2x+18000xF(x) = 2x + \frac{18000}{x}

Step 3: Find the minimum value of F(x)F(x)

To find the minimum amount of fencing, we take the derivative of F(x)F(x) with respect to xx and set it equal to zero: F(x)=218000x2F'(x) = 2 - \frac{18000}{x^2} Set F(x)=0F'(x) = 0 to find the critical points: 2=18000x22 = \frac{18000}{x^2} Multiply both sides by x2x^2: 2x2=180002x^2 = 18000 Solve for x2x^2: x2=180002=9000x^2 = \frac{18000}{2} = 9000 Take the square root of both sides: x=900094.87metersx = \sqrt{9000} \approx 94.87 \quad \text{meters}

Step 4: Find yy using the area equation

Now, substitute x=94.87x = 94.87 meters into equation (3) to find yy: y=600094.8763.29metersy = \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)F''(x) or analyze the behavior of F(x)F'(x). In this case, F(x)=36000x3F''(x) = \frac{36000}{x^3}, which is positive for x>0x > 0, indicating a minimum.

Final Answer:

The dimensions of the rectangular lot that will use the least amount of fencing are approximately:

  • Length: 94.8794.87 meters
  • Width: 63.2963.29 meters

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


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.

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Math Problem Analysis

Mathematical Concepts

Geometry
Optimization

Formulas

Area of a rectangle
Perimeter of a rectangle

Theorems

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Suitable Grade Level

Grades 10-12