 Hello and welcome to the session my name is Mansi and I am going to help you with the following question. The question says show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius r is 2r divided by root 3 also find the maximum value. So let this be the cylinder that is inscribed in the circle of radius r. Let h be the height of the cylinder and let x be the diameter of the base of this cylinder. So let us start with the solution to this question. As we have seen that h is the height and x is the diameter to the base of the inscribed cylinder then we can say that h square plus x square is equal to 2r the whole square. Now this happens because this acts as a right angle triangle where h is the altitude x is the base and 2r that is r plus r that is the hypotenuse. So from Pythagoras square up we get h square plus x square equals to 2r the whole square. This implies h square plus x square is equal to 4r square and we call this 1. Now volume of the cylinder is equal to pi into square of radius into height that is equal to pi into now we say that the diameter is x so radius will be pi x by 2 the whole square and height is h that is equal to 1 by 4 pi x square h. Now from this equation we see that x square we can write as 4r square minus x square. So volume now becomes 1 by 4 pi into h 4r square minus h square this implies that volume is equal to 1 by 4 into pi h into 4r square is pi h r square minus 1 by 4 pi h cube. Therefore derivative of v with respect to h is equal to pi r square minus 3 by 4 pi h square this is equal to now we take pi common from these two we get pi into r square minus 3 by 4 h square now we know that for maxima or minima the first thing that we should have is dv by dh should be equal to 0. Now when we put dv by dh equal to 0 we get r square is equal to 3 by 4 h square and this implies that h is equal to 2r by root 3 also d2v by dh square that is the second derivative of volume with respect to height is equal to minus 3 by 4 into 2 pi h and that is equal to minus 3 by 2 pi h. Now this happens because we see that dv by dh is this now if we have to find another derivative of this with respect to h then r acts as a constant and we will just have minus 3 by 4 into twice of h into pi. So this is what we have minus 3 by 2 into pi h now at h equal to 2r by root 3 d2v by dh2 is equal to minus 3 by 2 into pi into 2r by root 3 and we see that this is a negative value. So this implies that v is maximum at h equal to 2r by root 3. Now we find out the maximum value of the volume so maximum value is equal to value of volume at h that is 2r by root 3 and that is equal to 1 by 4 into pi into 2r by root 3 into 4r square minus 4r square by 3 and this we get from the earlier step where we have proved that v is equal to 1 by 4 pi h into 4r square minus h square. So this can be further simplified to pi r by 2 root 3 into 8r square by 3 and this is equal to 4 pi rq divided by 3 root 3 cubic units. So our answer to this question would be 4 pi rq by 3 root 3 cubic units. So I hope that you understood the question and enjoyed the session. Have a good day.