 Hello students, let's work out the following problem. It says, show that the maximum volume of the cylinder which can be inscribed in a sphere of radius 5 root 3 centimeter is 500 by centimeter cube. So, let's now move on to the solution. We have to maximize the volume of cylinder which can be inscribed in sphere of radius 5 root 3. We have to show that the maximum volume is 500 by cubic centimeter. So, let the radius of cylinder be x centimeter and the height of the cylinder is x centimeter. We are given that the radius of sphere is 5 root 3 centimeter. Now, this is the sphere in which we have a cylinder. This is the center of the sphere. Now, we have to find the height of the cylinder in terms of the radius of the sphere and the radius of the cylinder. Now, the radius of the cylinder is x centimeter. So, the diameter will be 2x centimeter and diameter will be 10 root 3 since the radius is 5 root 3 and diameter is double of the radius. So, it would be 10 root 3. So, now the height of the cylinder is square plus 2x square is equal to 2r square where r is the radius of the sphere. So, we have x square plus 4x square is equal to 4r square. So, this implies x square is equal to 4r square minus x square upon 4. Now, the volume of the cylinder is given by the formula pi r square h where r is the radius and h is the height. So, now here r is x. So, the volume of the cylinder which is noted by V and this is given by pi r square h here radius is this. So, it is x square is 4r square minus h square upon 4 into h. So, this is equal to pi by 4 into 4r square h minus h cube. Now, for maximum or minimum we need to first differentiate V with respect to h. So, dV by dh is equal to pi by 4 into 4r square minus 3h square. Now, putting dV by dh is equal to 0 we have 4r square minus 3h square is equal to 0. So, this implies 4r square is equal to 3h square. So, this implies taking square root on both sides we have 2r is equal to root 3h. So, this implies h is equal to 2r by root 3. Now, we need to find the second order derivative of V with respect to h it would be pi by 4 into minus 6h. Now, we find the value of d square V by dh square at h is equal to 2r by root 3 to check the maximum or minimum. If it is negative then we say that the volume is maximum and if it comes out to be positive volume will be minimum. So, we have pi by 4 into minus 6h is 2r by root 3. So, this is equal to minus root 3r pi which is negative. So, this implies volume is maximum for h is equal to 2r by root 3. Now, we have to find the maximum volume. Volume is pi r square r square is 4r square minus h square upon 4. So, it is 4r square minus h square upon 4 into h. Now, h is 2r by root 3. So, volume is pi by 4 into 4r square minus h square that is 2r by root 3 square into h that is 2r by root 3 as this is pi by 4 into 4r square minus 4r square by 3 into 2r by root 3. So, again pi by 4 into taking 4r square common 1 minus 1 by 3 to 2r by root 3. Now again pi by 4 into r square 4 gets cancelled with 4 1 minus 1 by 3 is 2 by 3 into 2r by root 3. Now again 2 into 2 is 4 4 gets cancelled with 4 and we have pi r cube upon 3 root 3. Now r is the radius of this sphere which is pi root 3. So, this is pi 5 root 3 cube upon 3 root 3. So, we have pi here we have made a mistake since we have already cancelled 4 here. So, here we will be having 2 into 2. So, the volume will be 4 pi r cube upon 3 root 3. So, here also we need to multiply this with 4. So, we have pi into 4 into 5 cube is 125 root 3 cube is 3 root 3 upon 3 root 3. So, 3 root 3 gets cancelled with 3 root 3 and the volume is equal to 500 pi and that is what we had to prove. We also need to write the unit of the volume which is centimeter cube. So, this completes the question and the session. Bye for now. Take care. Have a good day.