Question 117465
For a cylindrical room, h is height, r is the radius.
{{{V = (pi)r^2h}}} Volume of the cylinder
{{{S = 2(pi)r^2+2(pi)rh}}} Surface area of top and bottom plus unrolled side
{{{V=20000=(pi)r^2h}}}
You can rearrange the equation to solve for h as a function of r. 
{{{h=20000/((pi)r^2)}}}

Then substitute that back into the surface area equation,
{{{S = 2(pi)r^2+2(pi)rh}}}
{{{S = 2(pi)r^2+2(pi)r(20000/((pi)r^2))}}}
{{{S = 2(pi)r^2+40000/r}}}
Plot S using various values of r and look for a minimum.
8.1,	5350.
9.1,	4915.
10.1,	4601.
11.1,	4377.
12.1,	4225.
13.1,	4131.
14.1,	4085.
15.1,	4080.
16.1,	4112.
17.1,	4175.
18.1,	4267.
19.1,	4385.
20.1,	4527.
Looks like a minimum at r=15.1.
Now that you have r, go back to your equation for h and solve. 
{{{h=20000/((pi)r^2)}}}
{{{h=20000/((pi)15.1^2)}}}
{{{h=27.9)}}}
Make sure to check your answer. 
{{{V = (pi)r^2h}}} 
{{{V = (pi)15.1^2*27.9}}}
{{{V = 19985}}}
Close enough. 
r=15.1 m, h=27.9 m