length of rectangle -- x
width of rectangle --- y
b) radius of semicircle is y/2
c) distance = 2 lengths + 2 halfcircles
= 2x +2(1/2) π(y/2)^2
= 2x + π(y^2)/4
2x + (1/4)πy^2 = 200
times 4
8x + πy^2 = 800
x = (800 - πy^2)/8
d) -- poorly worded question.
Since they had you solve for x in c) they should have asked for the area in terms of y , not x
check your typing.
BTW, I would have defined the radius of the semicircle as r
then the width of the rectangle would be 2r, and we can avoid some of these nasty fractions.
An indoor physical fitness room consists of a rectangular region with a semicircle on each end. The perimeter of the room is to be a 200-meter running track.
a) Draw a figure that visually represents the problem. Let x and y represent the length and width of the rectangular region.
b) Determine the radius of the semicircular ends of the track.
c) Use the result of part b to write an equation in terms of x and y, for the distance traveled in one lap around the track. Solve for x.
d) Use the result of part c to write the area A of the rectangular region as a function of x.
1 answer