How to find maximum volume

I would begin as you did, by using similarity:

[MATH]\frac{h}{R}=\frac{h-r}{r}[/MATH]
Solve for \(r\):

[MATH]r=\frac{Rh}{h+R}[/MATH]
Now, the volume of the cone is:

[MATH]V=\frac{1}{3}\pi R^2h\implies h=\frac{3V}{\pi R^2}[/MATH]
Hence:

[MATH]r=\frac{R\cdot\dfrac{3V}{\pi R^2}}{\dfrac{3V}{\pi R^2}+R}=\frac{3RV}{\pi R^3+3V}[/MATH]
Maximizing the volume of the sphere is equivalent to maximizing the radius of the sphere, and so we find:

[MATH]r'=\frac{3V(3V-2\pi R^3)}{(\pi R^3+3V)^2}=0[/MATH]
From this, we obtain the critical value:

[MATH]R=\sqrt[3]{\frac{3V}{2\pi}}[/MATH]
Now, the first derivative test easily allows us to state:

[MATH]r_{\max}=r\left(\sqrt[3]{\frac{3V}{2\pi}}\right)=\sqrt[3]{\frac{4V}{9\pi}}[/MATH]
And so the maximum volume of the sphere is:

[MATH]V_{S_{\max}}=\frac{4}{3}\pi r_{\max}^3=\frac{16V}{27}[/MATH]
With [MATH]V=10\text{ m}^3[/MATH], we have:

[MATH]V_{S_{\max}}=\frac{160}{27}\text{ m}^3[/MATH]
 
You appear to be trying to optimize the volume of the cone, instead of the volume of the inscribed sphere.
 
How to get the above expression?

Looking back now, it appears I bungled the use of similarity. I should have begun with:

[MATH]\frac{\sqrt{h^2+R^2}}{R}=\frac{h-r}{r}[/MATH]
Which implies:

[MATH]r=\frac{h}{\sqrt{\left(\dfrac{h}{R}\right)^2+1}+1}[/MATH]
 
I dont know to to change to [MATH]\displaystyle \displaystyle r=\frac{Rh}{h+R}[/MATH]. Please guide me. Thanks

That was a mistake I made. Your image shows an expression for \(r\) equivalent to the latest one I posted.
 
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