Maximum AreaCalculusEdit
posted by Lisa .
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 + (L100) = 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 :)