TITLE: Find the volume of the largest right circular cone that can be inscribed in a sphere of radius r?
QUESTION [6 upvotes]: I checked this question but didn't fully understand it. 
I know that the volume of a right circular cone is $V = \frac{1}{3}\pi x^2h$
I know that I must take the first derivative and set it equal to zero, which will find the maximum. 
My problem is how to deal with the variable $h$, the height? 
How can I rewrite that in terms of $x$ or $r$.    
I have tried to implicitly differentiate this equation, but that wasn't helpful for me.

REPLY [2 votes]: With given radius r of a sphere let the inscribed cone have height h then remaining length without radius is (h-r) let R be radius of cone then there we get a right angle triangle with r as hypotanious R as adjecent and (h-r) as opposite side. Now by pythagorus theorem ((h-r)^2)+(R^2)=(r^2). Now express R in terms of h and r. We get (R^2)=(2hr-hh). Substitute in the volume of cone equation V=(pi/3)(R^2)h. Now V=(pi/3)(2hr-hh)h. Differentiate wrt h with r as constant because radius cannot be variable and substitute to zero to get maxima. We get h=(4r/3).