Question

A farmer has 300 ft of fencing with which to enclose a rectangular pen next to a barn. The barn itself will be used as one of the sides of the enclosed area.
What is the maximum area that can be enclosed by the fencing?

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ft²

Answer 1

Asked and answered elsewhere.
brainly.com/question/8970296

Answer 2

Answer

Maximum area A_max = 11250 ft^2

Step-by-Step Explanation

Declaring Variables:-

The length of the rectangle = y

The width of the rectangle = x

Solution:-

- The perimeter of a rectangle can be expressed using the above two variables as follows:

                             Perimeter (P) = 2*Length + 2*Width

                                                     = 2* ( x + y )

- Since the barn is used as one of the sides (let's say y) we can subtract y

we don't need fencing for this side. The length of the fence required L is:

                            Length (L) = P - y

                                                = 2*x + y

- We are given 300 feet of fencing. So we equate the length equal to 300 and develop a linear relationship between width and length of the barn.

                              300 = 2x + y

                              y = 300 - 2x

- The area (A) of the rectangle is given by the following expression:

                              A = Length*width

                              A = x*y

- Substitute the relationship developed between x and y in the Area (A) expression above. Then we have:

                              A = x*(300 - 2x)

                              A = 300x - 2x^2

- We will take first derivative of the expression of area (A) developed with respect to x and find the critical point of the area function by setting the first derivative A'(x) = 0.

                             A(x) = 300x - 2x^2

                             A'(x) = 300 - 4x

                             0 = 300-4x

                             x = 300 / 4 = 75 ft

- The critical point of the given function lies for the width (x) of 75 ft. We will plug in the critical value x = 75 ft back into the original function of Area and find the maximum area.

                             A(x) = 300x - 2x^2

                             A(75) = 300 (75) - 2(75)^2

                             A_max = A(75) = 11250 ft^2

- The maximum area that can be enclosed by the fencing is 11250 ft²