SOLUTION: Hello friends. I have a problem here that I am hoping someone can help me with . A right cylinder has a volume of 2pi cubic inches . Find the can dimensions that require the least

Algebra ->  Finance -> SOLUTION: Hello friends. I have a problem here that I am hoping someone can help me with . A right cylinder has a volume of 2pi cubic inches . Find the can dimensions that require the least       Log On


   

Question 962508: Hello friends. I have a problem here that I am hoping someone can help me with . A right cylinder has a volume of 2pi cubic inches . Find the can dimensions that require the least amount of aluminum to be used in the can construction? Thank you in advance !!!
Answer by hkwu(60) About Me  (Show Source):
You can put this solution on YOUR website!
The volume is described by
pi%2Ar%5E2%2Ah=2pi
so
h=%282pi%29%2F%28pi%2Ar%5E2%29=2%2Fr%5E2
The surface area is described by
SA=2pi%2Ah%2B2%2Api%2Ar
SA=2pi%2A%282%2Fr%5E2%29%2B2%2Api%2Ar
SA=%284pi%29%2Fr%2B2%2Api%2Ar%5E2
Taking the derivative, we get
d%2F%28dr%29%2ASA=-4pi%2Fr%5E2%2B4pi%2Ar
d%2F%28dr%29%2ASA=%28-4pi%2B4pi%2Ar%5E3%29%2Fr%5E2
which is equal to zero when
-4pi%2B4pi%2Ar%5E3=0
-1%2Br%5E3=0
r%5E3=1
r=1
If you do the first derivative test, you'll find that this is indeed a minimum value. Thus
h=2%2F1%5E2=2
so the optimal dimensions are 1in radius by 2in height.