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Alex787 [66]
3 years ago
14

A track and field playing area is in the shape of a rectangle with semicircles at each end . The perimeter of the track is to be

1200 meters. What should the dimensions of the rectangle be so that the area of the rectangle is maximum? What is the maximum area?
You must use completing the square technique to solve this problem. Show all your work.
Mathematics
1 answer:
MAVERICK [17]3 years ago
5 0
We can let the radius of the semi circles be r and x can be the length of the rectangle.
The perimeter will be Perimeter = 2x + 2pr = 1200
The area of the rectangle is Area = x(2r)
 Once we solve the perimeter equation, you get x = 600.
Put this into the area equation and you get Area = (600 - pr)(2r)
The maximum then gives r = 95. 493 and area 57, 295.8
Dimensions

Height is twice the radius or 190.986. Find base use equation x = 600 - p(95.493) = 300.

This means dimensions are 300 by 190.986 
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The equation of a parabola in vertex form is

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Mice21 [21]
<h2>19.</h2><h3>Given</h3>
  • window width and height are in proportion to building width and height
  • window width and height are 11 in and 18 in, respectively
  • building height is 108 ft
<h3>Find</h3>
  • building width
<h3>Solution</h3>

The proportional relation can be written as

... (building width)/(building height) = (window width)/(window height)

Multiplying by (building height) gives

... (building width) = (building heigh) × (window width)/(window height)

... (building width) = 108 ft × (11 in)/(18 in)

... building width = 66 ft

<h2>21.</h2><h3>Given</h3>
  • map distance = 6.75 in
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  • actual distance
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The distances are in proportion, so

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