# Math Help - Fitting a cylinder inside a cone

1. ## Fitting a cylinder inside a cone

So I think this is a related rates problem though i am not sure. I am in calculus AB.

A right circular cone has base radius 5 and altitude 12. A cylinder is to be inscribed in the cone so that the axis the cylinder coincides with the axis of the cone. given hat the radius of the cylinder must be between 2 and 4 inclusive, find the the value of that radius for which the laterals surface area of the cylinder is minimum. Justify you answer. (Note: The lateral surface of a cylinder does not include the bases)

2. Originally Posted by OnMyWayToBeAMathProffesor
So I think this is a related rates problem though i am not sure. I am in calculus AB.

A right circular cone has base radius 5 and altitude 12. A cylinder is to be inscribed in the cone so that the axis the cylinder coincides with the axis of the cone. given hat the radius of the cylinder must be between 2 and 4 inclusive, find the the value of that radius for which the laterals surface area of the cylinder is minimum(?). Justify you answer. (Note: The lateral surface of a cylinder does not include the bases)

Let R denote the radius of the cone and H the height of the cone. Let r denote the radius of the cylinder and h it's height.

Then you have the proportion:

$\dfrac{12-h}r=\dfrac{12}5~\implies~h=12-\dfrac{12}5 r$

The lateral surface area of a cylinder is calculated by:

$a_s = 2\pi \cdot r \cdot h$ Plug in the term for h:

$a_s(r) = 2\pi \cdot r \cdot \left( 12-\dfrac{12}5 r \right) = 24\pi r-\dfrac{24}5 \pi r^2$

Differentiate a(r) wrt r and set a'(r) = 0. Solve for r. I've got $r = \dfrac52$

$a\left(\dfrac52\right) = 30 \pi$

Now calculate

$a(4) = 19.2 \pi$

$a(2)= 28.8 \pi$

and you easily can see for which value of r the surface area is a minimum.

3. Originally Posted by OnMyWayToBeAMathProffesor
A right circular cone has base radius 5 and altitude 12....