the lagrange multiplier equation
posted by sunny on .
A cylindrical oilstorage tank is to be constructed for which the following costs apply:
cost per square meter
metal for sides $30.00 (cost per square meter)
combined costs of concrete base and metal bottom $37.50(cost per square meter)
top 7.50 (cost per square meter)
The tank is to be constructed with dimensions such that the cost is minimum for whetever capacity is selected.
a) one possible approch to slecting the capacity is to build the tank large enough for an additional cubic meter of capacity to cost $8.(note that this does not mean $8 per cubic meter average for the entire tank.) what is the optimal diameter and optimal height of the tank?
b) instead of the approach used in part(a), the tank is to be of such a size that the cost will be $9 per cubic meter average for the entire storage capacity of the tank. set up the lagrange multiplier equations and verify that they are satisfied by an optimal diameter of 20 m and optimal height of 15m.

b.
The cost function is given by:
C(r,h)=30(2πrh)+(37.5+7.5)πr²
=15π(4rh+3r²)
The volume, is given by
V(r,h)=πr²h
The "Lagrangian" is therefore obtained by including the constraint for a particular volume Vo,
Vo = πr²h
C(x,y) = 15π(4rh+3r²) + λ(πr&^sup2;hVo)
The first order conditions can be obtained by partially differentiating with respect to each of the variables, r and h, and equate to zero:
15π(4h+6r)+λ(2πrh) = 0 ...(1)
15π(4r) + λ(πr²) = 0 ...(2)
I will leave it to you to verify the optimal diameter and height.
Hint: λ should equal 6 in this case.
Also, verify the global unit cost per volume to be $9/m³.
Numerous references on the subject are available, for example:
http://www.slimy.com/~steuard/teaching/tutorials/Lagrange.html
http://www.economics.utoronto.ca/osborne/MathTutorial/ILMF.HTM 
answer no molum

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