# Thread: Optimization - Inscribed figures

1. ## Optimization - Inscribed figures

I've tried and tried, and visualized, but I can't find the relationship between the cone's radius/height and that of the cylinder. If someone could help me with that I'll be fine for the rest.

Determine the maximum volume of a cylinder that can be inscribed in a cone whose height is 3 cm and whose base radius is 3 cm.

2. Originally Posted by Freaky-Person
I've tried and tried, and visualized, but I can't find the relationship between the cone's radius/height and that of the cylinder. If someone could help me with that I'll be fine for the rest.

Determine the maximum volume of a cylinder that can be inscribed in a cone whose height is 3 cm and whose base radius is 3 cm.
It might be easier for you to visualize a cross-section of the cone, which looks like an isosceles triangle. Now, you want to find an equation for the volume of the cylinder that uses just one variable, and the easiest choice for this variable is the radius of the cylinder. The cross-section of the cylinder will look like a rectangle. Now, as the cylinder gets wider (via an increasing radius), it also gets shorter. The height of the cylinder is dependent on the radius, as you have noted. The exact relationship is linear: $\displaystyle h = 3 - r$. Once you see that it is a linear relationship, that equation is fairly easy to derive. So the volume of the cylinder will be $\displaystyle \pi r^2(3-r)$. Can you take it from there?

3. The cylinder will have volume 4/9th that of the cone. Let's do a general case and then you sub in your values.

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

The volume of the cylinder is $\displaystyle V={\pi}r^{2}h$

Now, we use similar triangles. $\displaystyle \frac{H-h}{H}=\frac{r}{R}$

Solve for h: $\displaystyle h=\frac{H}{R}(R-r)$

Sub into V: $\displaystyle {\pi}\frac{H}{R}(R-r)r^{2}={\pi}\frac{H}{R}(Rr^{2}-r^{3})$

$\displaystyle \frac{dV}{dr}={\pi}\frac{H}{R}(2Rr-3r^{2})={\pi}\frac{H}{R}r(2R-3r)$

DV/dr=0 when $\displaystyle r=\frac{2R}{3}$

So, the max volume is $\displaystyle \frac{4{\pi}R^{2}H}{27}=\frac{4}{9}\cdot\frac{1}{3 }{\pi}R^{2}H=\frac{4}{9}\cdot\text{volume of cone}$

Use your values now.

### a cone when it inscribed into a cylinder then what is the violume rest cylinder

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