OPTIMIZATION PROBLEM help PLEASE

[rafeeki92]

Find the maximum volume of a right-circular cone placed upside-down in a right-circular cone of radius R and height H.

http://img451.imageshack.us/i/optimizationfi3.jpg/

All I understand is that if you view the image in profile view, you have two similar triangles...I have no idea where to go from that.

[attachment=0:1xdx6m0q]triangle.jpg[/attachment:1xdx6m0q]

 

[chrisr]

You've drawn your inner cone upside-down.

It's apex should be in the middle of the external cone's base.

For that configuration, use Tan(angle) to relate r and h to R and H.

Tan(angle) = R/H = r/(H-h).
Therefore r = R(H-h)/H.

Now write the little cone volume using h and the constants and differentiate with respect to h.

V = (1/3)(pi)r[sup:10vepe1u]2[/sup:10vepe1u]h = (1/3)(pi)R[sup:10vepe1u]2[/sup:10vepe1u](H-h)[sup:10vepe1u]2[/sup:10vepe1u]h/H[sup:10vepe1u]2[/sup:10vepe1u].

Multiply it out, differentiate the terms with respect to h,
at the maximum volume dV/dh = 0,
will yield (3h-H)(h-H) = 0.
As h = H corresponds to minimum volume = 0,
3h = H gives maximum volume.

Do the differentiation and work it through.

 

[soroban]

Hello, rafeeki92!

chrisr is right . . . You didn't copy the diagram correctly.

Find the maximum volume of a right-circular cone
placed upside-down in a right-circular cone of radius \(\displaystyle R\) and height \(\displaystyle H.\)

    -         *     -
    :        /|\    :
    :       / | \  H-h
    :      /  | r\  :
    H     * - + - * -
    :    / \::|::/ \
    :   /   \:|h/   \
    :  /     \|/     \
    - * - - - * - - - *
                  R

The large cone has radius \(\displaystyle R\) and height \(\displaystyle H.\)
The inner cone has radius \(\displaystyle r\) and height \(\displaystyle h.\)

\(\displaystyle \text{The volume of the inner cone is: }\:V \:=\:\tfrac{\pi}{3}r^2h\;\;[1]\)

\(\displaystyle \text{From similar triangles, we have: }\:\frac{H-h}{r} \:=\:\frac{H}{R}\)
. . \(\displaystyle \text{Solve for }h\!:\;\;h \:=\:\frac{H}{R}(R-r)\;\;[2]\)

\(\displaystyle \text{Substitute into [1]: }\;V \;=\;\frac{\pi}{3}r^2\cdot \frac{H}{R}(R-r) \quad\Rightarrow\quad V \:=\:\frac{\pi H}{3R}(Rr^2 - r^3)\)


\(\displaystyle \text{Solve }\frac{dV}{dr} = 0\!:\quad\frac{\pi H}{3R}(2Rr - 3r^2) \:=\:0 \quad\Rightarrow\quad r(2R-3r) \:=\:0\)

\(\displaystyle \text{And we have: } r \,=\,0\:\text{ and }\:\boxed{r\,=\,\tfrac{2}{3}R}\)

\(\displaystyle \text{Substitute into [2]: }\:h \:=\:\frac{H}{R}\left(R - \tfrac{2}{3}R\right) \quad\Rightarrow\quad\boxed{ h \,=\,\tfrac{1}{3}H}\)


\(\displaystyle \text{Substitute into [1]: }\:V \;=\;\tfrac{\pi}{3}\left(\tfrac{2}{3}R\right)^2\left(\tfrac{1}{3}H\right) \quad\Rightarrow\quad\boxed{ V \:=\:\frac{4\pi}{81}R^2H}\)