Optimization - Find the surface area of a domed cylinder?

#r = 12sqrt(3), h = 5/6-8sqrt(3)#

Explanation:

The surface area #S# for the tank would be comprised of the surface area of the body of the cylinder + area of the circle base + area of the hemisphere on the top. Recall that:

#SA = underbrace(2pirh)_ "cyl" + underbrace(pir^2)_ "base" + underbrace(2pir^2)_ "hemi" = 2pirh + 3pir^2#

We know the volume is fixed at #360pi#, which means we can use the volume of the tank to isolate a variable and remove it from the SA formula:

#V = underbrace(pir^2h)_ "cyl" + underbrace(2/3pir^3)_ "hemi" = pir^2(h+2/3r)#

#pir^2(h+2/3r) = 360pi#

#r^2(h+2/3r) = 360#

#h + 2/3r = 360/r^2#

#h = 360/r^2 - 2/3r#

This can be substituted into our formula for the SAO to eliminate #h# from the formula:

#SA = 2pih+3pir^2 = 2pir(360/r^2-2/3r)+3pir^2#

# = (720pi)/r-(4pi)/3r^2+3pir^2 = (720pi)/r+(5pi)/3r#

Finally, we can use the first derivative to find the critical value(s) of #r#, and verify that they represent a minimum as requested.

#SA' = 0#

#-(720pi)/r^2+(5pi)/3 = 0#

#(5pi)/3 = (720pi)/r^2#

#r^2 = (720pi*3)/(5pi) = 432#

#r = sqrt(432) = 12sqrt(3) #

We ignore the negative root since a negative radius would be meaningless in the context of the problem.

There are a few ways to verify this result is a minimum. I'll use the Second Derivative Test:

#SA'' = (1440pi)/r^3 > 0 " for " r = 12sqrt(3)#

Since #SA''>0#, the value we found represents a minimum.

The final part of the answer is #h#:

#h = 360/r^2 - 2/3r = 360/(sqrt(432))^2 - 2/3(12sqrt(3)) #

#h = 360/432 - 8sqrt(3) = 5/6 - 8sqrt(3)#