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.
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 r^{2 }h=2000
Vertical surface area: 2π r h
Lid and base surface area: π r^{2 }(x2)
I understand that basically what I'm trying to do is isolate one variable (for example make h=2000/π r^{2}) 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.