You are given feet of flexible fencing material to enclose a rectangular pen. Suppose that one length of the longer side is composed of a straight brick wall, and it is not necessary to use the flexible fencing material for this side length. Find the maximum area you can enclose and the dimensions of the maximum pen.

Max. area = L^{2} / 8 ftx^{2}

Dimensions of maximum pen are:

x = L/2 ft (the largest side of the rectangle

y = L/4 ft

L is the length of flexible fencing material

Step-by-step explanation:

The straight brick wall with the fencing flexible material will form a rectangle of sides "x" and "y"

So we will use two concepts:

1. Area of rectangle : A=x*y

2. The length of the flexible fencing material L

According to problem statement

L = y + x + y or L = 2y + x 2y = L - x and y = (L-x) / 2 (1)

And as Area of the pen is f (a) = x*y (2)

Replacing in equation (2) "y" by its value from equation (1) we have f (a) as function of one variable "x"

f (x) = x * (L - x) / 2

We should get the derivative of f (a) and call it f' (a)

f¨ (x) = 1 * (L - x) / 2 + (x * (-1/2)) or f¨ (x) = (L - x) / 2 - x/2

The second derivative f¨¨ (a) = - 1/2 (negative meaning a maximun for the function)

Equating f¨ (a) = 0

f¨ (x) = (L - x) / 2 - x/2 = 0

L/2 - x/2 - x/2 = 0

L/2 - x = 0

x = L/2 (the largest side f the rectangle)

To get "y" (the other side) we go to equaton (1) and replace x for its value as function of L

y = (L - x) / 2 y = (L - L/2) / 2 y = ((2L-L) / 2)) / 2 y = (2L-L) / 4

y = L/4

And maximum area would be Amax. = x * y = L/2 * L/4 Amax. = L^{2}/8 square feet