Determining maximum area (Calculus problem)?

Question

The question is:
A rancher is planning to build a rectangle corral adjacent to an existing barn that uses all of the 400 feet of fencing material that he has available. What are the dimensions of the corral that would maximize the area of the corral if he plans to use a side of the barn as one side of the corral?

I'd greatly appreceate it if someone explained how to do this answer in plain english and gave the correct answer! =]

Answer 1

The 400 feet of fencing forms the perimeter P of the rectangular area. The perimeter of a rectangle is P = 2*W + 2*L. The area of a rectangle is A = L*W. Take either the W or L from the perimeter equation and solve for it in terms of the other. For example. 2*W = P - 2*L, or W = (P - 2*L)/2. Put this expression for W into the area formula

A = L*(P - 2*L)/2

A = P*L/2 - L^2

To maximize area, take the derivative of A with respect to L to get

A' = P/2 - 2*L and set it equal to zero

P/2 - 2*L = 0, or L = P/4; This gives you the length of the area. You can get the width from the perimiter formula W = (P-2*L)/2;
Putting in L = P/4 gives W = (P - P/2)/2 = P*(1-.5)/2 = P/4.

Notice that L = W = P/4; in other words the maximum area rectangle is a square.

Answer 2

you're able to desire to place in writing a quadratic equation representing the component of the rectangle. shall we make the width W and the dimensions L and the section A: LW = A through fact the cord is one hundred cm, we additionally will say: 2L + 2W = one hundred so L + W = 50 now you utilize substitution: W = 50 - L L(50-L) = A 50L - L^2 = A once you finished the sq., you hit upon that the vertex is at (25,625), meaning the max section occurs while L = 25. If L = 25, then W additionally has to = 25, so there is your answer.

Answer 3

Area= length * width
assign one side as a variable, say x
=((400-2x)/2) * x
=(200x-x^2)