I'm lost on where to begin with the problem... Someone help explain how the problem should be done.
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 single-lane running track.
a.) determine the radius of the semicircular ends of the room. Determine the distance, in terms of y, around the inside edge of the two semicircular parts of the track.
b.) use the result of part (a) to write an equation in terms of x & y ,for the distance traveled in one lap around the track. solve for y.
c.) use the result of part (b) to write the area A of the rectangular region as a function of x. What dimensions will produce a maximum area of the rectange?
as usual, draw a diagram. The track consists of two parallel stretches (the length of the rectangle, x) and a complete circle (the two semi-circular ends, diameter y).
2x + πw = 200
Now, unless you can provide some more info, you can't determine x and y. The track might be looooong and thin (big x), or fat and tall (big y).
To solve this problem, you can break it down into smaller steps. Let's start with part (a) and determine the radius of the semicircular ends of the room.
1. Let's assume the rectangular region has a length of L and a width of W.
2. The perimeter of the room is given as a 200-meter single-lane running track. Since the track is made of a straight portion and two semicircles, we can set up the equation: 2L + πr + πr = 200, where r is the radius of the semicircles.
3. Simplify the equation to 2L + 2πr = 200.
4. Since the straight portion has length L, the semicircular portion on each end will have a length of πr.
5. Combine the lengths of the straight and semicircular portions to get the distance around the inside edge of the two semicircular parts of the track: 2L + 2πr = Distance.
6. This distance is equal to 200 meters, so you can write the equation as: 2L + 2πr = 200.
Moving on to part (b), we can use the result from part (a) to write an equation in terms of x and y, and then solve for y.
1. In this problem, x represents the length of the rectangular region, and y represents the width.
2. Since the rectangular region has two semicircular ends, the equation from part (a) becomes: 2x + 2πr = Distance.
3. Substituting the value of r from part (a) into this equation, we get: 2x + 2π(200 - 2x - 2L)/2π = Distance.
4. Simplify the equation to get: 2x + 200 - 2x - 2L = Distance.
5. The goal is to solve for y, so substitute L with y in the equation to get: 2x + 200 - 2x - 2y = Distance.
6. Simplify further to get: 200 - 2y = Distance - 2x.
7. Since the distance is fixed at 200 meters, you can write the equation as: 200 - 2y = 200 - 2x.
Lastly, in part (c), we can use the result from part (b) to write the area A of the rectangular region as a function of x, and then determine the dimensions that will produce a maximum area of the rectangle.
1. The area A of a rectangle is equal to its length multiplied by its width.
2. In this problem, x represents the length of the rectangle, and y represents the width.
3. So, the area A can be written as: A = x * y.
4. From part (b), we have the equation: 200 - 2y = 200 - 2x.
5. Rearrange the equation to get: 2x = 2y.
6. Divide both sides of the equation by 2 to get: x = y.
7. Substitute this value of x into the area equation A = x * y to get A = y * y, which simplifies to A = y^2.
8. Therefore, the area A of the rectangular region is a function of y: A(y) = y^2.
9. To find the dimensions that will produce a maximum area, you can take the derivative of A(y) with respect to y, set it equal to zero, and solve for y.