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
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?
1 answer