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?

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

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

Step-by-step explanation:

The distance around the track (400 m) has two parts:  one is the circumference of the circle and the other is twice the length of the rectangle.

Let L represent the length of the rectangle, and R the radius of one of the circular ends.  Then the length of the track (the distance around it) is:

Total = circumference of the circle + twice the length of the rectangle, or

         =                    2πR                    + 2L    = 400 (meters)  

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

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