Post a New Question

Maximum Area-Calculus-Edit

posted by on .

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

  • Maximum Area-Calculus-Edit - ,

    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? :)

  • Maximum Area-Calculus-Edit - ,

    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

  • Maximum Area-Calculus-Edit - ,

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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question