00:01
In our problem we have a rx length cone with radius r, height, h, and side lens of l.
00:08
We have two equations.
00:10
The first one is for the surface area and the second one is for the volume.
00:16
By using the equation, by using the equation of area, first we want to define that l square equal root h squared plus r squared.
00:30
Next we can use the equation of a surface area which will be equal a we take the square of it equal by squared r squared or double i h squared plus r squared from this we can get a function for the height which is equal h squared equal a squared minus by r bar four over by squared r squared.
01:13
We have another equation for the volume, which is equal volume, equal 1 over 3 by r squared, which we can substitute the height inside this equation.
01:27
We will get 1 over 3 by r squared of deploy root a squared, minus by r r 4 over over by r this r will be removed with this squared and this by we will have v equal 1 over 3 r a squared minus by x squared minus by square r bar 4 or bar half.
02:28
We want the volume to be maximum...