We have 900 ft of fencing and we want to construct a back yard. If we are using the building as part of the barrier for the yard what are the dimensions that allow maximum area?

I am not even sure how to start though I know I am supposed to calculate the derivative of the setup. Thank you

Here is the image. imgur F5NJVqp

perimeter = 2 W + L = 900

so
L = (900 - 2W)

A = W L
A = W (900 -2W)
A = 900 W - 2 W^2
now you can complete the square for the parabola and find the vertex or you can use calculus

dA/dW = 0 for max = 900 - 4 W
W = 225

900 = 2 W + L
900 = 450 + L
L = 450

glad you learned calculus and did not have to find the vertex of that parabola? :)

Oh, although I can not see your image I see your note on your earlier question answered by Steve.

If only 100 feet of the building is used then
2 W + L + (L-100) = 900
2 W + 2 L - 100 = 900
2 W + 2 L = 1000
then continue as before

THANK YOU SO MUCH! I got 250 for each dimension :)

To find the dimensions that allow for the maximum area, we need to set up an equation and consider a mathematical concept called optimization. In this case, the optimization problem is to maximize the area.

Let's assume the length of the back of the yard is "L" and the width on the sides is "W". We can use the given 900 ft of fencing to form the perimeter of the yard, so:

Perimeter = 2L + W = 900 ft

Solving this equation for W, we get:

W = 900 ft - 2L

Now, the area of the yard, A, is given by the product of length and width:

A = L * W = L * (900 ft - 2L)

To find the dimensions that maximize the area, we need to find the critical points of the area function. We can do this by taking the derivative of A with respect to L, setting it equal to zero, and solving for L.

Let's start by finding the derivative:

dA/dL = 900 - 4L

Setting this equal to zero and solving for L:

900 - 4L = 0
4L = 900
L = 225 ft

Now, substituting this value of L back into the equation for W, we can find the corresponding value of W:

W = 900 ft - 2L = 900 ft - 2(225 ft) = 900 ft - 450 ft = 450 ft

Therefore, the dimensions that allow for the maximum area are: length = 225 ft and width = 450 ft.

Keep in mind that this solution assumes a rectangular-shaped yard.