# A physical fitness room consists of a rectangular region with a semicircle on each end. If the perimeter of the room is to be a 200 meter running track, how do find the dimensions that will make the area of the rectangular regions as large as possible?

##### 1 Answer

# { ("length", =50 \ m), ("width", =50/pi \ m ~~ 15.92 \ m) :} #

#### Explanation:

Let us set up the following variables:

# { (x, "length of rectangular region (m)"), (r, "Radius of semi-circle (m)"), (P, "Total perimeter of the room (m"^2")"), (A, "Total Area of the rectangular region (m"^2")") :} #

The perimeter of the given room, is that of two identical semicircular ends, and the two rectangular region lengths. Thus:

# P = 1/2(2pir) + 1/2(2pir) + x + x #

# \ \ \= 2pir + 2x #

But we are given that the total perimeter is

# 2pir + 2x =200 => r = (100-x)/(pi)#

Now the total area of the rectangular region is:

# A = (x)(2r) #

# \ \ \ = (x)2((100-x)/(pi)) #

# \ \ \ = (200x-2x^2)/(pi) #

For maximum or minimum, we seek a critical point of the area function,

# (dA)/dx = (200-4x)/pi#

At a critical point, the first derivative vanishes, thus we require:

# f'(x) = 0 => 200-4x=0 #

# :. x=50 \ m #

We can visually confirm from the graph:

graph{(200x-2x^2)/(pi) [-50, 150, -400, 2000]}

With

that the solution is both consistent and corresponds to a maximum, and the dimensions of the sought room are thus:

# { ("length", =x, =50 \ m), ("width", =2r,=50/pi \ m ~~ 15.92 \ m) :} #