MinMax04 Fencing three sides of an area.

A park area of 5000 square meters is to be built in the shape of a rectangle along a river. Fencing will be on three sides. What is the minimum length of fencing for the desired area?

Fencing is required only on three sides of the rectangle as shown. The defining equation is for the perimeter, the variable we want to minimize: P = 2a + b . The constraint equation is from the area requirement. Stated in the form of an equation: ab = 5000 . In order to write P in terms of one variable, solve the area equation for b and substitute.

b = 5000/a so P = 2a + 5000/a = 2a + 5000 a^-1

Take the derivative of P: P' = 2 + 5000(-a^-2) and set P' = 0:

2 = 5000 / a² and a² = 2500 or a = 50 .

The second derivative of P: P'' = -5000(-2a^-3) = 5000/a³ is positive for all positive values of a, so a = 50 is a minimum.

Putting a = 50 back into the constraint equation: 50b = 5000 yields b = 100 .

The dimensions a = 50, b = 100 provide the minimum fencing requirement.

The graph of P vs. a is helpful in understanding this problem. The form P = 2a + 5000/a is most convenient for graphing. Only positive a has meaning. The first step in graphing is to look for dominant terms. There are two here. The 2a term dominates for large a and the 5000/a term dominates for small a.

In mathematical terms; as a Ž 0, P Ž +Ľ ; and as a Ž Ľ, P ť 2a .

The value of the function at a = 50, the point where the slope equals zero, is

P(50) = 2(50) + 5000/50 = 200.

With this information the curve can be sketched.