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A farmer wants to fence off a pasture in the shape of a rectangle. He wants to divide the pasture into two equal parts by putting a fence connecting the two longer sides at their midpoints. He has 400 meters of fencing. What should the dimensions of the pasture be in order to maximize the area of the pasture?

Let l be the length and w be the width of the pasture. The area is wl . There are 2l + 3w meters of fencing used , so 400 = 2l + 3w .Then l= 200 - (3/2)w

So the area is wl = (w)(200 - (3/2)w) =

200w - (3/2)w^2

Take the derivative and set it equal to 0

200 - 3w = 0

w = 200/3

l = 200 - (3/2)w = 200 - (3/2)(200/3) = 100

So the dimensions are l = 100 , w = 200/3

Calculus

Answers by Expert:

I can answer questions from the standard four semester Calculus sequence. I am not prepared for questions on Tensor Calculus. Everything else is welcome. Derivatives, partial derivatives, ordinary differential equations, single and multiple integrals, change of variable, vector integration (Green`s Theorem, Stokes, and Gauss) and applications.

Ph.D. in Mathematics and many years teaching Calculus at state universities.**Education/Credentials**

B.S. , M.S. , Ph.D.