A cylindrical oil-storage 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;h-Vo)

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

1200

a) To find the optimal diameter and height of the tank, we can set up a cost function in terms of the variables, diameter (D) and height (H).

Let's assume the tank has a capacity of V cubic meters. The volume of a cylindrical tank is given by V = π(D/2)^2H.

The cost of the metal for the sides would be given by the surface area of the sides multiplied by the cost per square meter, which is 30.00. The surface area of the sides can be calculated as follows:
Surface area of sides = 2π(D/2)H

The cost of the concrete base and metal bottom would be given by the area of the base (π(D/2)^2) plus the area of the metal bottom (π(D/2)^2), multiplied by the cost per square meter (37.50).
Cost of base and bottom = (π(D/2)^2 + π(D/2)^2) * 37.50

The cost of the top would be given by the area of the top (π(D/2)^2) multiplied by the cost per square meter (7.50).
Cost of top = π(D/2)^2 * 7.50

Therefore, the total cost function, C, would be:
C(D, H) = (2π(D/2)H * 30.00) + ((4π(D/2)^2) * 37.50) + (π(D/2)^2 * 7.50)

To find the optimal diameter and height, we need to minimize this cost function subject to the constraint V = π(D/2)^2H + 1, where 1 is the additional cubic meter of capacity that costs $8.

We can solve this problem using calculus by taking the partial derivatives of the cost function with respect to D and H, and then equating them to zero. This will give us critical points which we can evaluate to find the minimum cost.

b) To set up the Lagrange multiplier equations, we introduce a Lagrange multiplier, λ, for the constraint V = π(D/2)^2H - 1. The objective function in this case would be the total cost function, C(D,H), as described in part (a).

The Lagrange equations are:
∂C/∂D = λ * ∂(V-1)/∂D
∂C/∂H = λ * ∂(V-1)/∂H
V = π(D/2)^2H - 1

We would need to differentiate C with respect to D, H, and λ, and set them equal to the corresponding derivatives of the constraint equation. By solving these equations simultaneously, we can find the optimal diameter and height that satisfy the constraint and minimize the cost per cubic meter average.