Optimization: Maximizing Enclosed Area

A rectangular paddock is fenced on all four sides using 60 meters of fence. Write the area function A(x) in terms of its width x.

A(x) = 30x - x^2, with the practical domain 0 < x < 30.

Using your function from item 1, find the width x that maximizes the area, and state the maximum area.

Differentiating yields A'(x) = 30 - 2x = 0, so x = 15 meters. The maximum area is 225 square meters.

A farmer encloses a rectangular field along a straight riverbank. No fencing is needed along the river. If 80 meters of fence are available, write the area function in terms of the width x perpendicular to the water.

The fence covers two widths and one length: 2x + y = 80, so y = 80 - 2x. The area function is A(x) = 80x - 2x^2.

Find the dimensions and maximum area for the riverbank field in item 3.

A'(x) = 80 - 4x = 0, so x = 20 meters. The length y is 80 - 2(20) = 40 meters. The maximum area is 800 square meters.

A gardener uses 120 feet of fencing to create a rectangular garden divided into two equal side-by-side pens sharing an interior fence line. If x is the width of the shared divider, express the total area in terms of x.

The fence uses three widths and two lengths: 3x + 2y = 120, so y = 60 - 1.5x. Total area A(x) = x(60 - 1.5x) = 60x - 1.5x^2.

Determine the dimensions that maximize the two-pen garden in item 5.

A'(x) = 60 - 3x = 0, so x = 20 feet. The total length is y = 60 - 1.5(20) = 30 feet. Maximum area is 600 square feet.