An indoor track is made up of a rectangular region with two semi-circles at the ends. The distance around the track is 400 meters. What dimensions for the rectangular region maximize the area of the rectangular region?

This equation is a 'constraint.' It simplifies to πR + L = 400. This equation can be solved for R if we wish to find L first, or for L if we wish to find R first. Solving for L, we get L = 400 - πR.

We wish to maximize the area of the rectangular region. That area is represented by A = L·W, which is equivalent here to A = L·2R = 2RL. We are to maximize this area by finding the correct R and L values.

We have already solved the constraint equation for L: L = 400 - πR. We can substitute this 400 - πR for L in

the area formula given above: A = L·2R = 2RL = 2R)(400 - πR). This product has the form of a quadratic: A = 800R - 2πR². Because the coefficient of R² is negative, the graph of this parabola opens down. We need to find the vertex of this parabola to obtain the value of R that maximizes the area of the rectangle:

-b ± √(b² - 4ac)

Using the quadratic formula, we get R = ------------------------

2a

-800 ± √(6400 - 4(0)) -1600

or, in this particular case, R = ------------------------------------- = ---------------

2(-2π)

-800

or R = ----------- = 200/π

-4π

and so L = 400 - πR (see work done above)

These are the dimensions that result in max area of the rectangle:

width of rectangle = 2R = (200/π) = 400/π meters

length of rectangle = 400 - π(200/π) = 400 - 200 = 200 meters