SOLUTION: A high school is planining to build a new playing field surrounded by a running track. The track coach wants 2 laps around the track to be 1000m. The football coach wants the recta

Question 818741: A high school is planining to build a new playing field surrounded by a running track. The track coach wants 2 laps around the track to be 1000m. The football coach wants the rectangular infield to be as large as possible. Can both coaches be satisfied? State the dimensions.
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
2(2L + 2w) = 1000
4L + 4w = 1000
4w = 1000 - 4L
w = 250 - L
a = Lw
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a = Lw
a = L(250 - L)
a = 250L - LL
a(L) = -L^2 + 250L
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the above quadratic equation is in standard form, with a=-1, b=250, and c=0
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to solve the quadratic equation, by using the quadratic formula, plug this:
-1 250 0
into this: https://sooeet.com/math/quadratic-equation-solver.php
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the quadratic vertex is a maximum at: ( L= 125, a(L)= 15625 )
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the maximum area of the infield is: a(L) = a(125) = 15625 sq.m (that's what football coach wants)
L = 125 m
w = a(125)/125= 15625 /125 = 125 m
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check track perimeter:
p = 2L + 2w
p = 2(125) + 2(125)
p = 500
two laps around track = 2p = 1000 m (correct, that's what track coach wants)
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Solve quadratic equations, quadratic formula:
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