Calculus
posted by Elizabeth .
A metal cylindrical container with an open top is to hold 1 cubic foot. If there is no waste in construction, find the dimensions which will require the least amount of material.

V = πr^2 h
h = 1/(πr^2)
Surface Area (SA)
= bottom + collar of cylinder
= πr^2 + 2πrh
=πr^2 + 2πr(1/(πr^2)
= πr^2 + 2/r
dSA/dr = 2πr  2/r^2 = 0 for a min of SA
2πr = 2/r^2
r^3 = 1/π
r = 1/π^(1/3) = appr .693 ft
then h = 1/(π(.693^2)) = .683 
V = π r ² h = 1ft³
h = 1/π r ²
SA = π r ² + 2 π r h
= π r ² + 2 π r * 1/π r ²
= π r ² + 2 / r
SA ' = 2π r  2 / r² = 0
2π r = 2 / r²
r³ = 1/π
r = 1/ ³√π ft
h = 1/(π r ²)
= 1/(π (1/ ³√π) ²)
= ³√π ² / π
= 1/ ³√π ft
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