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
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.
4 answers
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
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
A cylindrical metal container with open top is to hold 10 cubic feet of material. If there is
waste in construction, find the dimensions which require the least amount of material.
waste in construction, find the dimensions which require the least amount of material.
ahh yes