240 feet of fencing material is going to be used. What should the dimensions be to produce a maximum enclosed area.

This will be a circle with a circumference

2*pi*r=240-->r=120/pi
(max)A=pi*r^2=14400/pi square feet

To find the dimensions that produce a maximum enclosed area with 240 feet of fencing material, we can use calculus. Let's assume the rectangular enclosure has dimensions length (L) and width (W).

To maximize the area, we need to find the critical points of the area function with respect to L and W, and then determine which one of them corresponds to a maximum.

The perimeter of the rectangle is given by the equation: 2L + 2W = 240, or simplified: L + W = 120.

We can rewrite the equation to solve for one variable in terms of the other. Let's solve for W in terms of L:
W = 120 - L

The area (A) of the rectangle is given by the equation: A = L * W.

Substitute W in terms of L into the area equation:
A = L * (120 - L)
A = 120L - L^2

To find the critical points, we take the derivative of the area equation with respect to L and set it equal to zero:
dA/dL = 120 - 2L = 0

Solving for L:
120 - 2L = 0
2L = 120
L = 60

Substitute L = 60 back into the equation for the perimeter to find W:
W = 120 - L
W = 120 - 60
W = 60

So, the dimensions that produce a maximum enclosed area when 240 feet of fencing material is used are L = 60 feet and W = 60 feet.