Max Volume of cylinder

[confused_07]

What is the max possible volume of a right circular cylinder w/ a total surface area 600pi in^2(including top and bottom)?

Here's what I have:

Surface= 600pi in^2
Surface top and bottom= 2*pi*r^2
Surface curved side= 2*pi*r*h

Total Surface= (2*pi*r^2)+(2*pi*r*h)= 600pi

h= 2*pi*r*h= 600pi - (2*pi*r^2)
= [600pi - (2*pi*r^2)] / (2*pi*r)
= (300/r) - r
= (1/r)(300 - r^2)

V= pi*r^2*h
= pi*r^2 [(1/r)(300 - r^2)]
= (300pi*r) - (pi*r^3)
= (pi*r)(300 - r^2)

From here, I know I need to find dV/dr.... I think. Not looking for the answer to the problem, just a nudge to the next step. Thanks

 

[galactus]

Solve the surface area formula for h and you should get:

\(\displaystyle h=\frac{300}{{\pi}r}-r\)

Sub this into the volume formula, \(\displaystyle V={\pi}r^{2}h\)

and get \(\displaystyle \L\\300r-{\pi}r^{3}\)

Differentiate and get: \(\displaystyle \L\\300-3{\pi}r^{2}\)

Set to 0 and solve for r. h will follow. Then sub the values into the volume formula to find the max volume.

 

[confused_07]

How did you get h= (300/pi*r) - r ?

When I solved for h, I got (600pi - 2pi*r^2) / 2pi*r, which equates to (300/r)-r

Isn't that the same as (600pi / 2pi*r) - (2pi*r^2 / 2pi*r) ?

The pi should cancel out, therefore leaving you with what I cam up with. Or is that wrong?

 

[galactus]

\(\displaystyle \L\\2{\pi}r^{2}+2{\pi}rh=600\)

\(\displaystyle \L\\2{\pi}rh=600-2{\pi}r^{2}\)

\(\displaystyle \L\\h=\frac{600-2{\pi}r^{2}}{2{\pi}r}\)

\(\displaystyle \L\\h=\frac{\not{6}^{\text{3}}00}{\not{2}{\pi}r}-\frac{\sout{2{\pi}}r^{\not{2}^{\text{1}}}}{\sout{2{\pi}r}}\)

\(\displaystyle \L\\h=\frac{300}{{\pi}r}-r\)

Sub into volume formula:

\(\displaystyle \L\\V={\pi}r^{2}\left(\frac{300}{{\pi}r}-r\right)\)

\(\displaystyle \L\\\sout{{\pi}}r^{\not{2}^{\text{1}}}\cdot\frac{300}{\sout{{\pi}r}}-{\pi}r^{2}\cdot{r}\)

\(\displaystyle \L\\300r-{\pi}r^{3}\)

Differentiate:

\(\displaystyle \L\\\frac{dV}{dr}=300-3{\pi}r^{2}\)

\(\displaystyle \L\\300-3{\pi}r^{2}=0\)

Solve for r.

You should note, when you find your radius and height. The height should be twice the radius. Max volume is achieved when the height and diameter are the same.

EDIT: I JUST NOTICED THE PI IN YOUR ORIGINAL PROBLEM STATEMENT.
I USED 600, NOT 600PI. SAME METHOD, THOUGH.

Sorry about that. Paying attention, you get \(\displaystyle \L\\h=\frac{300}{r}-r\)

Sub into the volume formula you get: \(\displaystyle \L\\V=300{\pi}r-{\pi}r^{3}\)

Differentiate: \(\displaystyle \L\\\frac{dV}{dr}=300{\pi}-3{\pi}r^{2}\)

Set to 0 and solve for r: \(\displaystyle \L\\300{\pi}-3{\pi}r^{2}=0\)

r=10 and h=20

Therefore, the max volume is \(\displaystyle V={\pi}(10)^{2}(20)=2000{\pi}\)