Find the radius of the cylinder that produces minimum surface area. Please help me?

Question

A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. The total volume of the solid is 12 cubic centimeters. Find the radius of the cylinder that produces minimum surface area.

Here's what I did:
1. drew the solid. It looks like a pill.
2. labelled the radius of the cylinder "r" and the height "h".
3. Used the volume equation to solve for h in terms of r. I got h = (12 - 4/3*pi*r^3) / pi*r^2.
4. Plugged that value of h into the equation of the solid's surface area and simplified.
5. Differentiate but the equation looks messy. I don't know how I can set that equation equal to zero and solve for r.

Or if there's a better method, please feel free to teach me that.

Answer 1

V = 4/3π r^3 + πr^2h = 12

S = 4πr^2 + 2πrh
S = 4πr^2 + 2πr[(12 - 4/3πr^3) / πr^2]
S = 4πr^2 + 24/r - 8/3πr^2
S = 4/3πr^2 + 24/r
§

S' = 8/3πr - 24/r^2 = 0
8/3πr = 24/r^2
r^3 = 9/π
r = (9/π)^(1/3)

Edit: your approach is also our approach... you just have to simplify it or make it look more compact... not messy...

Answer 2

V=4/3*pi*r^3+pi*r^2*h = 12
Minimum of 4*pi*r^2+2pir*h =f(r,h)
If you know lagrangian multipliers
F(r,h,k)= 4*pi*r^2+2*pi*r*h+k( 4/3*pi*r^3+pi*r^2*h-12)
Take partial derivate to r and h and put them to 0
Fr =8pi*r +2pi h +k( 4pir^2+2pi r h)
Fh= 2pi r +k(pir^2)
8 r +2h+k(4r^2+2r h)=0
2r+kr^2=0 so r =0 non sense and 2+kr=0 so k=-2/r
8r+2h -8r-4h=0 so h=0 ( the cylindrical part does not exist
and 4/3 pir^3 =12 and r= (9/pi)^1/3
It looks weird but given a volume the solid of least surface is the sphere