Tricky Optimization: rt circ. cone, r=5, h=12, inscribed in

[alakaboom1]

would really appreciate some help with this problem...

A right circular cone has a base of radius 5 and an altitude of12. A cylinder is to be inscribed in the cone so that the axis of the cylinder coincides with the axis of the come. Given that the radius of the cylinder must be between 2 and 4 inclusive, find the value of that radius for which the lateral surface area of the cylinder is minimum.

Well, the only thing i managed to firgure out is that the lateral surface area would be : SA= pi r[sup:2kwi2cwy]2[/sup:2kwi2cwy] h since it doesnt include the bases. But im not sure how to do the actual problem. Thanks a lot if anyone can help me out!!

 

[soroban]

Re: Tricky Optimization

Hello, alakaboom1!

A right circular cone has a base of radius 5 and an altitude of 12.
A cylinder is to be inscribed in the cone so that the axis of the cylinder
coincides with the axis of the come.
Given that the radius of the cylinder must be between 2 and 4 inclusive,
find the value of that radius for which the lateral surface area of the cylinder is minimum.

i managed to firgure out that the lateral surface area would be: \(\displaystyle S \:=\:\pi r^2h\) .[1]

You already have the area function . . . but it has two variables, \(\displaystyle r\) and \(\displaystyle h.\)

We must find a relationship between \(\displaystyle r\) and \(\displaystyle h.\)

    -           *           -
    :          /|\          :
    :         / |:\       12-h
    :        /  |::\        :
    :       *---+---*       -
   12      /|   |   |\      :
    :     / |   |   |:\     :
    :    /  |   |   |::\    h
    :   /   |   |   |:::\   :
    :  /    |   |   |::::\  :
    - *-----*---*---*-----* -
      :    5    : r : 5-r :

\(\displaystyle r\) is the radius of the cylinder.
\(\displaystyle h\) is the height of the cylinder.

The two shaded triangles are similar.

\(\displaystyle \text{So we have: }\:\frac{12-h}{r} \:=\:\frac{h}{5-r} \quad\Rightarrow\quad h \:=\:12 - \tfrac{12}{5}r\)


Substitute into [1]:

. . \(\displaystyle S \;=\;\pi r^2\left(12 - \tfrac{12}{5}r\right) \quad\Rightarrow\quad S \;=\;\pi\left(12r^2 - \tfrac{12}{5}r^3\right)\)

And that is the function we must minimize . . .