# Thread: MAX Surface Area

1. ## MAX Surface Area

what is the volume of the cylinder to maximiz the surface Area of the cone?

the things we know is the Height of the cone wich is 12, and the Radius of the cone which is 4.

2. Are you sure you've phrased the problem correctly?.

Are you sure it isn't, "find the cylinder of max volume that can be inscribed in a cone of the given dimensions"?.

The surface area of the cone is ${\pi}rL={\pi}(4)(\sqrt{12^{2}+4^{2}})=16{\pi}\sqrt {10}$, where L is the slant height of the cone.

So, I don't think that's what needs maximized.

I may be misunderstanding, though.

It can be shown that the cylinder of max volume that can be inscribed in a cone is 4/9 the volume of the cone.

3. it says, find the volume of the cylinder to max the Surface area??
So im guessing I have to find the volume of the cylinder in respect to the cone then find the max SA of the cone...... I have know IDEAAAA thats how the question is...
volume of the cylinder to max the Surface area!!

4. Originally Posted by hossein
what is the volume of the cylinder to maximiz the surface Area of the cone?

the things we know is the Height of the cone wich is 12, and the Radius of the cone which is 4.
Originally Posted by hossein
it says, find the volume of the cylinder to max the Surface area??
So im guessing I have to find the volume of the cylinder in respect to the cone then find the max SA of the cone...... I have know IDEAAAA thats how the question is...
volume of the cylinder to max the Surface area!!
Easy, the answer is that the volume of the cone is infinite because a cylinder of infinite height will produce an infinite surface area for the cone.

Not to mention that you are given the height and base radius of the cone, so you can determine the surface area of the cone.

As galactus said, there's something wrong with this problem.

-Dan

5. We can go ahead and find the max cylinder.

Let H and R be the radius and height of the cone and r and h be the radius and height of the cylinder.

By similar triangles:

$\frac{H-h}{H}=\frac{r}{R}$

$\frac{12-h}{12}=\frac{r}{4}$

$h=12-3r$

Sub into cylinder volume formula:

$V={\pi}r^{2}(12-3r)$

$\frac{dV}{dr}=-3{\pi}r(3r-8)$

$-3{\pi}r(3r-8)=0$

$r=\frac{8}{3}, \;\ h=4$

Therefore the max volume of the cylinder is:

$V={\pi}(\frac{8}{3})^{2}(4)=\frac{256\pi}{9}$

The volume of the cone is $64\pi$

$\frac{\frac{256\pi}{9}}{64\pi}=\frac{4}{9}$

As stated before, the cylinder is 4/9 the volume of the cone.

The surface area is then $2\pi(\frac{8}{3})(4)+2{\pi}(\frac{8}{3})^{2}=\frac {320\pi}{9}$

I reckon this is what was wanted. The statement is rather ambiguous.