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 respectively
b) Determine the radius of the semicircular ends of the track. Determine the distance, in terms of y, around the inside edge of each semicircular part 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. What dimensions will produce a rectangle of maximum area?
a) To visually represent the problem, we can draw a rectangle with semicircles on each end. The rectangle represents the main area of the room where exercise can be done, and the semicircles represent the running track. The perimeter of the room is a 200-meter track, so we need to make sure that the entire perimeter is covered by the rectangle and the semicircles.
Let's label the length of the rectangle as x and the width as y.
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b) The perimeter of the room is a 200-meter track. The rectangular region contributes to two sides of the perimeter, and the semicircles contribute to the other two sides.
The length of the rectangular region contributes x meters to the track, and the width contributes y meters to the track.
Each semicircle contributes half the circumference, which is πr, where r is the radius of the semicircle. Since we have two semicircles, the contribution from both semicircles is 2 * (π * r).
Therefore, the total perimeter of the room is: x + y + 2 * (π * r).
Since the perimeter is 200 meters, we can set up the equation:
x + y + 2 * (π * r) = 200.
To determine the distance around the inside edge of each semicircular part of the track, we need to find the radius, r.
c) To solve for the radius, we need to rearrange the equation from part b to isolate r.
2 * (π * r) = 200 - x - y.
Divide both sides by 2π to solve for r:
r = (200 - x - y) / (2π).
The distance around the inside edge of each semicircular part of the track is the circumference of the semicircle, which is πr.
So, the distance is: π * [(200 - x - y) / (2π)] = (200 - x - y) / 2.
d) The distance traveled in one lap around the track is the sum of the distances for the rectangular region and the two semicircles.
The distance for the rectangular region is 2x (since we are going around the length twice).
So, the total distance for one lap is: 2x + 2 * [(200 - x - y) / 2] = 2x + 200 - x - y = x + 200 - y.
The area of the rectangular region, A, is given by length times width, which is x * y.
So, A = x * y.
To find the dimensions that will produce a rectangle of maximum area, we need to maximize A with respect to x.
Differentiating A with respect to x and setting it to zero gives us the critical point for maximum area. We can solve for x to find the dimensions that will produce maximum area.