Thread: Minimizing with Lagranges' Multipliers

So this is the exact problem:

An industrial container is in the shape of a cylinder with hemispherical ends. The container must hold 1000 liters of fluid. Determine the radius r and length h that minimize the ammount of material used in the construction of the tank.

What I understand from the problem is that the surface area S(r,h)=2πr² + 2πrh (according to what the surface area of a cilinder should be) and that the volume given by 2πr²h=1000 will be the restriction. When I apply lagrange i get the following system of equations:

2πr²h=1000
λ4πrh=4πr+2πh
λ4πr²=2πr

Solving that for λ I get λ=1\2r, and when I substitute that in the other equation, I get that r=0 which is ridiculous, because that would mean that in order to minimize the amount of material used for the tank the surface area would have to be zero! Obviously there's something wrong with the way I'm tackling the problem, so is there any way any of you guys could help?

Originally Posted by monkeybone114

S(r,h)=2πr² + 2πrh

Close. Surface area of the complete sphere (hemisphere on both ends) and lateral surface area of the cyllinder.

Originally Posted by monkeybone114

volume...2πr²h=1000

Not very close. Try Volume of the sphere and volume of the cyllinder.

The calculus will do you no good if the geometry is off.

What has gone wrong should send up red flags. Obviously r = 0 will minimize the materials, but the constraint has not been applied.

Unless I have forgotten what we are doing here, you must also remember the the method of Lagrange Multipliers gives critical points. It is the student's responsibility to determine min or max.

How would I determine wether it's a min or max?

Originally Posted by monkeybone114

How would I determine wether it's a min or max?

Calculate the volume for your critical value of r. Then vary r slightly and see what the volume becomes. If the volume increases your critical point represents a local minimum, if the volume decreases then your critical point represents a local maximum.

-Dan