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A pencil cup with a capacity of 25π in.3 is to be constructed in the shape of a right circular cylinder with an open top. If th...Asked by Joooo
A pencil cup with a capacity of 18𝜋 in.^3 is to be constructed in the shape of a right circular cylinder with an open top. If the material for the side costs 3/16
of the cost of the material for the base, what dimensions should the cup have to minimize the construction cost?
of the cost of the material for the base, what dimensions should the cup have to minimize the construction cost?
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Answered by
mathhelper
let the radius be r inches , and the height be h inches
we know π r^2 h = 18π
h = 18/r^2
cost of base = π r^2 (1) if the cost is 1 unit of money
cost of side = (3/16)(2πrh)
cost of whole thing = π r^2 + (3/8)(πr)(18/r^2)
= π r^2 + (27/4)π / r
d(cost)/dr = 2πr - 27π/(4r^2) = 0 for a min of cost
2πr = 27π/(4r^2)
2r = 27/(4r^2)
8r^3 = 27
take the cube root, I see an abundance of perfect cubes, so
2r = 3
r = 3/2
the h = 18/r^2 = 18/(9/4) = 8
check my arithmetic
we know π r^2 h = 18π
h = 18/r^2
cost of base = π r^2 (1) if the cost is 1 unit of money
cost of side = (3/16)(2πrh)
cost of whole thing = π r^2 + (3/8)(πr)(18/r^2)
= π r^2 + (27/4)π / r
d(cost)/dr = 2πr - 27π/(4r^2) = 0 for a min of cost
2πr = 27π/(4r^2)
2r = 27/(4r^2)
8r^3 = 27
take the cube root, I see an abundance of perfect cubes, so
2r = 3
r = 3/2
the h = 18/r^2 = 18/(9/4) = 8
check my arithmetic