An outdoor track consists of a rectangular region with a semi-circle on each end. If the perimeter of the track must be 200 mete
rs, find the dimensions that will make the area of the rectangular region as large as possible.
1 answer:
Answer:
Length = 50m
Width = 31.84m
Step-by-step explanation:
Perimeter of a track = 200m
The perimeter of a track with two semi-circular end and two rectangular regions = P
Perimeter of a semi-circle = ½(2πr)
Length of a rectangle = x
Perimeter of the rectangle = x+x
P = ½(2πr) + ½(2πr)+x + x
P = 2πr + 2x
200 =2πr + 2x
2πr = 200 – 2x
2πr = 2(100 – x)
r = 2(100 – x) / 2π
r = (100-x)/π
total area of the rectangular region =A
A= x(2r)
= (x)2[(100-x)/π]
= 2x[(100-x)/π]
A = (200x – 2x^2)/π
Differentiate A with respect to x
dA/dx = (200 -4x)/π
at critical point, first derivative vanishes(dA/dx = 0)
(200 -4x)/π = 0
-4x =-200
x =-200/-4
x = 50
this means length (x) = 50m
put x=50 into r = (100-x)/π
r= (100-50)/π
r =50/π
width = 2r
width = 2(50/π)
= 2(19.62)
= 31.84m
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