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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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I need help on this question so if u actually can answer it correctly and show how you got it that would be helpful thx
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\rm \looparrowright -  \frac{1}{4} (8 - 4 + 20x)

Subtract the numbers

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Factor out 4 from the expression

\rm \looparrowright   - \frac{1}{ \cancel4}  \times  \cancel4(1 + 5x)

\rm \looparrowright   - (1 + 5x)

When there is a - in front of an expression in parentheses, change the sign of each term in the expression to -

\rm \looparrowright   - 1  -  5x

  • <em>Thus, Options A and F are correct!!~</em>
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