SOLUTION: A cylinder with a top and bottom has a volume of 30 cubic inches. Find the minimum amount of material needed to create the can. (Surface area=2pir^2 + 2pirh & volume = pir^2h) I

Algebra ->  Volume -> SOLUTION: A cylinder with a top and bottom has a volume of 30 cubic inches. Find the minimum amount of material needed to create the can. (Surface area=2pir^2 + 2pirh & volume = pir^2h) I       Log On


   

Question 736368: A cylinder with a top and bottom has a volume of 30 cubic inches. Find the minimum amount of material needed to create the can. (Surface area=2pir^2 + 2pirh & volume = pir^2h)
I tried to create two separate equations,
30 = pir^2h (and then solve for r^2)
and
2pir^2
and then I was going to substitute the r value into the other equation. This did not work.
Thanks in advance!
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
we have two unknown variables, the radius (r) and the height (h).
Volume (V) = pir^2h


30 = pir^2h
h = 30/pir2
Substitute this into the surface area equation for h, we get
A = (2pir( 30 )/pir^2) + 2pir2)

A = (60/r) + 2pir2

Take the derivative for r

A = 60r-1 + 2pir2

A' = -60r-2 + 4pir


when slope is zero

0 = -60r-2 + 4pir


60r-2 = 4pir

Multiply both sides by r^2

60 = 4pir3


4.775 = r3

1.68 = r
h = 30/pir^2
h=30/pi*(1.68)^2
h = 3.38


A = 2pirh + 2pir^2
A = 2*pi*(1.68)(3.38) + 2pi*(1.68)^2
A = 35.68 + 17.73
A = 53.41 square inches
CHECK

V= pi*1.68^2*3.38= 29.96
you can check for maxima or minima by taking the second derivate for more detailing