A track and field playing area is in the shape of a rectangle with semicircles at each end. See the figure. The inside perimeter
1 answer:
Answer:
Step-by-step explanation:
Let x be the track straights lengths
Let y be the track ends diameter and the other rectangle side lengths.
1800 = 2x + πy
y = (1800 - 2x) / π
A = xy
A = x((1800 - 2x) / π
A = (1/π)(1800x - 2x²)
dA/dx = (1/π)(1800 - 4x)
0 = (1/π)(1800 - 4x)
0 = 1800 - 4x
4x = 1800
x = 450 m
y = (1800 - 2(450)) / π
y = 900/π or approximately 286.5 m
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S + a = 300...s = 300 - a
8s + 12a = 3200
8(300 - a) + 12a = 3200
2400 - 8a + 12a = 3200
-8a + 12a = 3200 - 2400
4a = 800
a = 800/4
a = 200 <==== 200 adult tickets
s = 300 - a
s = 300 - 200
s = 100 <=== 100 student tickets
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