Ex. 120 m of fencing is to be used to form three sides of a rectangular enclosure , the fourth side being an existing wall . Find the maximum possible area of the enclosure

L + 2 W = 120 so L = 120 - 2 W

A = L W
A = (120-2W)W = 120 W - 2 W^2

so
2 W^2 -120 W = - A

W^2 - 60 W = -A/2

W^2 - 60 W + 900 = -A/2 + 900

(W-30)^2 = -(1/2)(A-1800)

vertax of parabola at (30,1800)

so
W = 30
L = 120 - 60 = 60
and A = 1800 m^2

Hi Damon,

Where did you get the value 900 from?

Ah I see, that's part of completing the square.

To find the maximum possible area of the enclosure with a fixed perimeter, we can use the concept of optimization. In this case, we will use the derivative to determine the maximum area.

Let's assume the length of the rectangular enclosure is L and the width is W. We can form the three sides using L and W and the existing wall will contribute to the fourth side, making the total perimeter:

P = L + W + L = 2L + W

However, we are given that the total length of the four sides (perimeter) is fixed at 120 m. So we can write:

2L + W = 120

To express W in terms of L, we can rearrange the equation:

W = 120 - 2L

Now, we can express the area A in terms of L and W:

A = L * W

Substituting the value of W:

A = L * (120 - 2L) = 120L - 2L^2

To find the maximum area, we can take the derivative of A with respect to L and set it equal to zero:

dA/dL = 120 - 4L = 0

Solving this equation, we get:

4L = 120
L = 30

To ensure this is indeed the maximum, we can take the second derivative:

d^2A/dL^2 = -4 < 0

Since the second derivative is negative, it confirms that L = 30 is the value for which A is maximized.

Now, we can calculate the corresponding width W:

W = 120 - 2L = 120 - 2(30) = 120 - 60 = 60

So the maximum possible area of the enclosure is:

A = L * W = 30 * 60 = 1800 square meters.