Asked by Barbara
An oil can is to have a volume 1000in^3 and is to be shaped like a sylinder with a flat bottom but capped by a hemisphere. Neglect the thickness of the material of the can and find the dimensions that will minimize the total amount of material needed to construct it.
Answers
Answered by
MathMate
Using Lagrange multipliers, you will need the following variables
radius, r
height, h
Volume, V=πr^2h+(2/3)πr^3=1000
Surface Area, A = πr^2+2πrh+2πr^2
λ=Lagrange multiplier
You need to minimize A subject to V=1000 using the following objective function:
Z(r,h,V,λ)
= πr^2+2πrh+2πr^2 + λ(V-1000)
find partial derivatives with respect to r, h, & λ to get 4 equations and solve for the system of equations in three unknowns r,h and λ.
radius, r
height, h
Volume, V=πr^2h+(2/3)πr^3=1000
Surface Area, A = πr^2+2πrh+2πr^2
λ=Lagrange multiplier
You need to minimize A subject to V=1000 using the following objective function:
Z(r,h,V,λ)
= πr^2+2πrh+2πr^2 + λ(V-1000)
find partial derivatives with respect to r, h, & λ to get 4 equations and solve for the system of equations in three unknowns r,h and λ.
Answered by
MathMate
Partial derivatives:
Zr=3r+h+L(r^2+rh)=0
Zh=2 %pi r + L %pi r^2 = 0
ZL=πr^2h+(2/3)πr^3-1000 =0
Eliminating L from first two equations gives r=h
Substitute h=r in third equation gives
r=(600/%pi)^(1/3)
=5.7588 approx.
Check my arithmetic
Zr=3r+h+L(r^2+rh)=0
Zh=2 %pi r + L %pi r^2 = 0
ZL=πr^2h+(2/3)πr^3-1000 =0
Eliminating L from first two equations gives r=h
Substitute h=r in third equation gives
r=(600/%pi)^(1/3)
=5.7588 approx.
Check my arithmetic
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