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Calculus/Propane tank optimization


A company is designing propane tanks that are cylindrical with hemispherical ends. Assume that the company wants tanks that will hold 1000 cubic feet of gas, and that the ends are more expensive to make, costing $5 per square foot, while the cylindrical barrel between the ends costs $2 per square foot. Determine the minimum cost to construct such a tank

Assume the tank has radius r and height x.
The volume is then 2*pi*r^2 on each end and 2*pi*r*x on the sides.
The cost is then 5*2*2*pi*r^2 for both ends and 2*2*pi*r*x for the sides.
That is, 20*pi*r^2 + 8*pi*r*x.

The volume of the tank is (4/3)*pi*r^3 + pi(r^2)x = pi(r^2)(4r/3 + x).
Since the volume is 1000, set 1000 =  pi(r^2)(4r/3 + x) and solve for x.
(1000 - pi(r^2)(4r/3) = pi(r^2)x, so x = (1000 - pi(r^2)(4r/3)/(pi(r^2)).
That works out to be x = 1000/(pi(r^2)) - 4r/3.

Put that back into 20*pi*r^2 + 8*pi*r*x for x.
Simplify out into two terms and reduce r where necessary.

This gives the only variable as r, so differentiate with respect to r.

Set that formula to 0 and solve for r.

Using this value of r, the value of x can be found.


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