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 the minimum surface area.

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 the minimum surface area.

precal · Answer

I need to double check my set up.

precal · Answer

I used the volume of a sphere + volume of a cylinder

precal · Answer

$$V=\frac{ 4 }{ 3 }\pi r^3 + \pi r^2h$$

precal · Answer

I also, used the Surface Area formulas
$$S=4\pi r^2+2\pi r h$$

precal · Answer

I set V = 12 and solved for h, then sub h into Surface Area formula

precal · Answer

I took the derivative of S and set it equal to zero, this is where I probably messed up because I got 9 pi as the radius

ganeshie8 · Answer

setup looks good to me

precal · Answer

9 pi is not a min value

precal · Answer

should I show my work so someone can point out my mistake, I would appreciate it

ganeshie8 · Answer

ohk.. whats the correct answer ?

precal · Answer

I don't know, this one I have no key to at all

anonymous · Answer

Yes, please show your work

precal · Answer

ok so V=12 so therefore
$$12=\frac{ 4 }{ 3 }\pi r^3+\pi r^2h$$

precal · Answer

solving for h
$$\frac{ 12-\frac{ 4 }{ 3 }\pi r ^3 }{ \pi r^2 }=h$$

precal · Answer

now I am going to sub h into the Surface Area formula$$S=4 \pi r^2+2 \pi r \left( \frac{ 12-\frac{ 4 }{ 3 }\pi r^3 }{ \pi r^2 } \right)$$

precal · Answer

see any mistakes so far?

ganeshie8 · Answer

nope, everything fine so far

precal · Answer

simplifying it a bit before I take the derivative of it is:
$$S=4\pi r^2 +24r ^{-1}-\frac{ 8 }{ 3 }\pi r^2$$

precal · Answer

$$S=\frac{ 4 }{ 3 }\pi r^2+24r ^{-1}$$

precal · Answer

now I will take the derivative, see any mistakes?

precal · Answer

$$S ' =\frac{ 8 }{ 3 }\pi r - \frac{ 24 }{ r^2 }$$

precal · Answer

I set S'=0 then I solved for r

ganeshie8 · Answer

looks good,
multiply 3r^2 and get rid of fractions

ganeshie8 · Answer

you must get cuberoots in ur answer

precal · Answer

ok did not think to take that approach

precal · Answer

need a moment to do it on paper

precal · Answer

so its the cube root of (9/pi)?

ganeshie8 · Answer

yes ! wolfram says the same : http://www.wolframalpha.com/input/?i=minimize+4pi*r%5E2%2B2pi*r*%2812-4%2F3*pi*r%5E3%29%2F%28pi*r%5E2%29%2Cr%3E0

precal · Answer

now that is the min value I was looking for, thanks.  I knew I had made a mistake solving for r....

ganeshie8 · Answer

$\large r =   \sqrt[3]{9/\pi}$  minimizes the surface area of solid

precal · Answer

thank you so much....

ganeshie8 · Answer

np :)