 Hello and welcome to the session. In this session we discuss the following question which says find the volume of the largest cylinder that can be inscribed in a sphere of radius r. Let's move on to the solution now. We are given a sphere of radius r. We need to find the volume of the largest cylinder that can be inscribed in the sphere of radius r. Consider this cylinder inscribed in the sphere having radius r. We consider let h be the height of the cylinder inscribed in the sphere. So this is h and we take let this capital R be the radius of the cylinder inscribed in the sphere. So this is capital R. So in this figure we have a is equal to small r, a b is equal to capital R. Now since height of the cylinder is h, so this od would be equal to h by 2. Now let's find out the volume of the cylinder. Let it be v and this is equal to pi capital R square into h. Let this be equation 1. Now next we consider the right triangle oab. In this we have oa square is equal to ob square plus ab square. By the Pythagoras theorem, now putting the values for oa, ob and ab, we get r square is equal to h upon 2 whole square plus capital R square. That is further we get r square is equal to h square upon 4 plus capital R square. And from here we get capital R square is equal to small r square minus h square upon 4. Let this be equation 2. Now next substituting value of capital R square in equation 1 we get that is in this equation we get v is equal to pi into small r square minus h square upon 4 into h. That is v is equal to pi into small r square into h minus h cube upon 4. Now we differentiate this equation with respect to h. So differentiating both sides with respect to h we get dv by dh is equal to pi into r square minus 3x square upon 4. Next we find d2v by dh2 by differentiating this equation again with respect to h. So this is equal to pi into 0 minus 6h upon 4. That is we have d2v by dh2 is equal to minus 6 by 4 pih or 2, 2 times is 4 and 2, 3 times is 6. So this is equal to minus 3 by 2 pih. Now for maxima or minima we have dv by dh equal to 0 that is pi into r square minus 3x square upon 4 is equal to 0. From here we find the value for h. So this further gives us r square is equal to 3x square upon 4 or h square is equal to 4r square upon 3. So this gives us h is equal to 2r upon root 3. So we have got the value for h that is the height of the cylinder. Now next we will find d2v by dh2 at h equal to 2r upon root 3. So this is equal to minus 3 upon 2 pi into the value for h that is 2r upon root 3. This is equal to, now this 2 cancels with this 2 and we have minus root 3 pi r that is we have d2v by dh2 at h equal to 2r by root 3 is minus root 3 into pi r and this is less than 0. Now since we have d2v by dh2 is less than 0 this implies that the volume of the cylinder is maximum and it is maximum at h equal to 2r by root 3. Since we have the volume of the cylinder that is v is equal to pi into r square h minus h cube pi 4 that is we have the volume v is equal to pi into r square h minus h cube upon 4. Now let this be equation 3. Now volume of the largest cylinder inscribed in the sphere of radius r is equal to the volume v at h equal to 2r by root 3 that is in equation 3 we put h as 2r upon root 3. So this is equal to pi into r square into 2r by root 3 minus 2r by root 3 whole cube into 1 upon 4 that is further we get pi into 2r cube by root 3 minus 8r cube by 4 into 3 root 3. Now 4 2 times is 8 so we get this is equal to pi into 2r cube by root 3 minus 2r cube by 3 root 3. So this is equal to pi into 6r cube minus 2r cube this whole upon 3 root 3 that is we have pi into 4r cube upon 3 root 3. Thus we have the volume of the largest cylinder inscribed in the sphere of radius r is equal to 4 pi r cube upon 3 root 3. Thus the required volume is equal to 4 pi r cube upon 3 root 3 cubic units. So this is our final answer. This completes the session. Hope you have understood the solution of this question.