SOLUTION: a large soup can is to be designed so that the can will hold 16π in^3 (about 28oz) of soup. find what height and radius of the cylindrical can will minimize the amount of meta

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Question 167653: a large soup can is to be designed so that the can will hold 16π in^3 (about 28oz) of soup. find what height and radius of the cylindrical can will minimize the amount of metal needed. note: you will need to use the "minimum" program on your calculator. -- I have no idea where to even start with this problem.
Found 2 solutions by ankor@dixie-net.com, Alan3354:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
a large soup can is to be designed so that the can will hold 16π in^3 (about 28oz) of soup. find what height and radius of the cylindrical can will minimize the amount of metal needed.
:
Find the relationship between the radius and height using the V = pi%2Ar%5E2%2Ah
pi%2Ar%5E2%2Ah+=+16pi
:
h = %28%2816pi%29%29%2F%28%28pi%2Ar%5E2%29%29
Cancel pi
h = 16%2F%28r%5E2%29
:
The amt of metal used in the can is determined by the surface area
S.A = %282%2Api%2Ar%5E2%29+%2B+%282%2Api%2Ar%2Ah%29
Substitute 16%2F%28r%5E2%29 for h
S.A = %282%2Api%2Ar%5E2%29+%2B+%282%2Api%2Ar%2A%2816%2Fr%5E2%29%29
Cancel r:
S.A = %282%2Api%2Ar%5E2%29+%2B+%282pi%2816%2Fr%29%29
Or
S.A = %282%2Api%2Ar%5E2%29+%2B+%28%2832%2Fr%29pi%29
:
Put this in your graphing calc
y = %282%2Api%2Ax%5E2%29+%2B+%28%2832%2Fx%29pi%29; x = radius, y = surface area
Scale it: x: -2, +4; y: -30, +120
;
you graph should look like this:
+graph%28+300%2C+200%2C+-2%2C+4%2C+-30%2C+120%2C+%286.28%2Ax%5E2%29%2B%28100.53%2Fx%29%29+
:
You can estimate that minimum surface area about r = 2 inches
Use the minimum feature on your calc: bracket the minimum and it will
give the exact radius and surface area;
I got radius (x) = 2; giving a Minimum surface area (y) = 75.398 sq/inches
;
Did this make some sense to you?

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
From the solution above:
--------------------
Start at S.A. = 2PIr^2 + 2V/r
Set the 1st derivate of dSA/dr = 0 (derivative of SA with respect to r)
4PIr - 2V/r^2 = 0
4PIr^3 - 2V = 0
2PIr^3 - 16PI = 0
r^3 = 8
r = 2
h = 4