Maximizing volume of cone inscribed within cone

[nina94]

A right circular cone is inscribed inside a larger right circular cone with a volume of 150 cm3. The axes of the cones coincide and the vertex of the inner cones touches the center of the base of the outer cone. Find the ratio of the heights of the cones that maximizes the volume of the inner cone.

So I know the figure looks like this, but I'm just not sure where to go from there. Help will be gladly appreciated.
View attachment 5511

 

[MarkFL]

Re: Calc word problem.

My approach here would be to observe that we need two things to compute the volume of the inner (smaller) cone...its base radius $r_S$ and height $h_S$. Now, we can let the height of the inner cone vary anywhere from $0$ to the height $h_L$ of the outer (larger) cone...and we can see that the radius of the inner cone will decrease linearly as its height increases. We know then if we write $r_S$ as a function of $h_S$, we will have the following point on the line:

\(\displaystyle (0,r_L),\,(h_L,0)\)

We now have two points on the line, so we can find the equation of the line...can you now express the radius of the inner cone as a function of the height of the inner cone?

\(\displaystyle r_S(h_S)=?\)

 

[MarkFL]

After 24 hours, let's wrap this up...we may write:

\(\displaystyle r_S(h_S)=-\frac{r_L}{h_L}h_S+r_L=r_L\left(1-\frac{h_S}{h_L}\right)\)

Now, since we are asked for the ratio of the heights of the two cones where the volume of the inner cone is maximized, let's call this ratio \(\displaystyle r=\frac{h_S}{h_L}\), and express the volume of the inner cone as:

\(\displaystyle V_S=\frac{\pi h_Lr_L^2}{3}r(1-r)^2\)

We need only maximize the non-constant portion of the volume, so let's write:

\(\displaystyle f(r)=r(1-r)^2\)

And we then find by equating the derivative to zero:

\(\displaystyle f'(r)=r(2(1-r)(-1))+1(1-r)^2=(1-r)(1-r-2r)=(1-r)(1-3r)\)

Thus, our critical values are:

\(\displaystyle r=\frac{1}{3},\,1\)

We see that on the interval \(\displaystyle \left(0,\frac{1}{3}\right)\), $f'$ is positive and on the interval \(\displaystyle \left(\frac{1}{3},1\right)\), $f'$ is negative, so by the first derivative test, we know:

\(\displaystyle f_{\max}=f\left(\frac{1}{3}\right)\)

From this we may conclude that the inner cone's volume is maximized when its height is one-third that of the outer cone.