Question 169775
The can is a cylinder. 
The amount of metal is the top, bottom, and circular tube portion.
A=A(top)+A(bottom)+A(tube)
{{{A[m]=pi*r^2+pi*r^2+2*pi*r*h}}}
{{{A[m]=2*pi*r^2+2*pi*r*h}}}
The amount of paper just covers the circular tube portion.
{{{A[p]=2*pi*r*h}}}
The total cost equation is then,
Cost=Area Cost of Metal*(Metal Area)+Area Cost of Paper*(Paper Area)
{{{C=0.50*(A[m])+0.10*(A[p])}}}
{{{C=0.50*(2*pi*r^2+2*pi*r*h)+0.10*(2*pi*r*h)}}}
{{{C=pi*r^2+pi*r*h+0.2*pi*r*h}}}
{{{C=pi*r^2+1.2*pi*r*h}}}
The total equation has both r and h as variables.
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The internal volume of the can is fixed. 
{{{V=pi*r^2*h=54}}}
{{{h=54/(pi*r^2)}}}
We can substitute this expression into the total cost equation and reduce it to a one variable equation.
{{{C=pi*r^2+1.2*pi*r*h}}}
{{{C=pi*r^2+(1.2*pi*r*54)/(pi*r^2)}}}
{{{C=pi*r^2+64.8/r}}}
We can graph the function and look for a minimum. 

{{{ graph( 300, 300, -2, 5, -5, 60, 3.14*x^2+64.8/x) }}} 
Looks like we have a minimum around r=2.
We can find the exact value by taking the derivative of C with respect to r and setting it equal to zero.
{{{dC/dr=2*pi*r-64.8/r^2=0}}}
{{{2*pi*r=64.8/r^2}}}
{{{r^3=64.8/(2*pi)}}}
{{{r^3=10.313}}}
{{{highlight(r=2.177)}}}
Then from above,
{{{h=54/(pi*r^2)}}}
{{{highlight(h=3.628)}}}