Cylindrical Tin

Question

The volume of a cylindrical tin can with a top and bottom is to be 16π cubic inches. If a minimum amount if tin is to be used to construct the can, what must be the height, in inches, of the can?

Answer 1

r^2=16/h

Say, the tin required is T

T= 2*pi*r*h +2*pi*r^2

Substituting r^2=16/h and taking derivative dT/dh=0

h=4 inches

 

Answer 2

Volume of a cylinder is πr²h. r = radius, h = height

πr²h = 16π

h = 16/(r²)

dh/dr = -16/(r)

Find the derivative of the volume and set it equal to zero to find the minimum amount of tin it will take to make the can.

dV/dr = 2πrh + πr²(dh/dr)

dV/dr = 2πr16/(r²) + πr²(-16/(r))

0 =  32π/(r) - 16πr

 

Solve for r 

32π/(r) = 16πr

2/(r) = r

r = Sqrt(2)

 

Plug r in to h

h = 16/(r²)

h = 16/(Sqrt(2)²)

h = 8