Thread: 3rd and final optimization problem

Trying to design an aluminum can containing a volume of 2,000 cubic centimeters.

Find the dimensions of the can that will minimize the amount of aluminum used.

The three formulas I'm working with will obviously
Volume: π r² h=2000
Vertical surface area: 2π r h
Lid and base surface area: π r² (x2)

I understand that basically what I'm trying to do is isolate one variable (for example make h=2000/π r²) so that I can now substitute that equation in for the variable h in another formula.

That will result in a quadratic equation, from which I will find the derivative, find max and min, and figure out the volume at the minimum. Am I missing something? I keep attempting this, but I keep getting stuck after I substitute 2000/πr^2 for h because I don't have a quadratic equation to work with.

you don't always need a quadratic equation to solve a optimization problem,

when the volume is ,



then,


to find the maximum point you can use differentiation,



equate this to zero to find an extrema.

Once again you come to the rescue! Thank you so very much for explaining things in a way that I can easily understand them. You have already helped me more than my textbook has.

I'm glad that I could be a help