You got a small error in your post-sub equation.
\(\displaystyle \L\\S=2{\pi}r(\frac{27}{{\pi}r^{2}})+{\pi}r^{2}\)
\(\displaystyle =\L\\S={\pi}r^{2}+\frac{54}{r}\)
Differentiate:
\(\displaystyle \L\\\frac{dS}{dt}=2{\pi}r-\frac{54}{r^{2}}\)
Set to 0 and solve for r:
\(\displaystyle \L\\2{\pi}r-\frac{54}{r^{2}}=0\)
Solving for r gives \(\displaystyle \L\\r=\frac{3}{{\pi}^{\frac{1}{3}}}\approx{2.05}\)
This is the radius which will minimize the surface area and thus the cost of materials.
Now, how do we know this is a minimum and not a maximum, you ask?.
S, the surface area, is a continuous function of r on \(\displaystyle (0,{+\infty})\) with
\(\displaystyle \L\\\lim_{r\to\0^{+}}({\pi}r^{2}+\frac{54}{r})={+\infty}\)
and
\(\displaystyle \L\\\lim_{r\to+\infty}({\pi}r^{2}+\frac{54}{r})={+\infty}\)
so, S has a minimum, but no maximum on \(\displaystyle (0, {+\infty})\)
The minimum must occur at a critical point, which we found by
differentiating, setting to 0 and solving for r.
Since the only critical point is in our interval \(\displaystyle (0 , {+\infty})\),
this value of r must give the minimum value of S.
Also, try the second derivative test. If \(\displaystyle f''\left(\frac{3}{{\pi}^{\frac{1}
{3}}}\right)\) is greater than 0, then you have a relative minimum.
This equals \(\displaystyle 6{\pi}\), that's certainly > 0, so you have a
minimum.
Here's a graph for ya':
