Optimization problem, PLEASE HELP!

Question

I am asked to design a fuel tank for a fighter jet that will hold 300 liters of jet fuel. The shape of the tube is a cylinder with two hemisphere attached to the end. The side wall (the cylinder part) costs $0.1 per cm^2 to make while the rounded tops (hemispheres) cost $0.3 per cm^2 to make. What is the optimal radius that will minimize the manufacturing cost?

Answer 1

Most optimization problems in calculus require making two equations. This problem is no different. The first thing to do is to please mind the units. Let's make everything in terms of different powers of centimeters. 300 liters is 300,000 cm³. When we get the radius's final answer, it needs to be in units of centimeters. Anyways, the tank is supposed to hold 300,000 cm³. Equate this to the volume formula for a cylinder.

 

V = πr²h

300,000 = πr²h

 

The second equation is based on the cost per surface area of this cylinder. A cylinder has two circles on it, and the area of a circle is calculated by the formula πr². Because there are two circles on a cylinder, part of the surface area has the term 2πr² in it. Each area unit of the circles costs $0.3. I'm assuming you mean $0.30. Let me know if that's incorrect.

 

The total cost of the circles is therefore 0.3(2πr²). The area of the wall part of the cylinder is calculated by 2πrh, and each area unit of this costs 0.1. Therefore, the total price of the wall is 0.1(2πrh). Now here's the equation for the total price of the cylinder.

 

P = 0.3(2πr²) + 0.1(2πrh)

P = 0.6πr² + 0.2πrh

 

This is awkward. There are too many variables to make solving this equation possible. The question is asking for optimal radius, so it's best to get this whole equation in terms of only r, so that means we need to get rid of h. Use the volume equation in the beginning to get h alone so we can do substitution into the price equation.

 

300,000 = πr²h

h = (300,000)/(πr²)

 

Substitute these terms into the position of h in the price equation.

 

P = 0.6πr² + 0.2πrh

P = 0.6πr² + 0.2πr[300,000/(πr²)]

P = 0.6πr² + 60,000/r

 

We're going to need to find the minimum of the price. Extrema, which include maxima and minima, can be quite efficiently found using the first derivative. Find the derivative of P.

 

P = 0.6πr² + 60,000/r

P' = 1.2πr - 60,000/r²

 

To find exact points of extrema, set the derivative equal to zero. Hopefully you can use a graphing calculator for this, but if not, I'll show you as much algebra as possible to get this done.

 

0 = 1.2πr - 60,000/r²

 

I'm going to multiply every single term by r², so that the denominator in the last term cancels out.

 

0 = 1.2πr³ - 60,000

 

1.2πr³ = 60,000

πr³ = 50,000

r³ = 50,000/π

r ≈ 25.15 cm

 

The approximate length of the radius you need is 25.15 cm. I hope this helps, and good luck with your Calculus class!