Answer:
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
