 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. The question says, prove that the volume of largest cone that can be inscribed in a sphere of radius r is 8 upon 27 of the volume of the sphere. Let us start with the solution now. Let O be equal to OA equal to r be the radius of the given sphere. Now let BD is equal to r be the radius and AD is equal to h be the height of the cone inscribed in a given sphere. Now volume of the cone V is equal to 1 upon 3 pi r square h. Let us name this expression as 1. Now clearly we can see in this figure OD is equal to AD minus OA. Now we know AD is equal to h that is the height of the cone and OA is equal to radius of the sphere that is r. So we can write OD is equal to h minus r. Now if we apply Pythagoras theorem in right triangle OBD we get OB square is equal to OD square plus BD square. Now substituting their corresponding values we get r square is equal to OD we know is equal to h minus r. So we can write h minus r whole square plus BD is equal to r so we get r square is equal to h minus r whole square plus r square. Now here we will apply the formula for A minus B whole square. We know A minus B whole square is equal to A square plus B square minus 2 AB. So we can write this bracket as h square plus r square minus 2 HR r square and r square will cancel each other. Now we get h square minus 2 HR plus r square is equal to 0. Now this implies r square is equal to 2 HR minus h square. Now let us name this expression as 2. Now substituting the value of r square from expression 2 in expression 1 we get v is equal to 1 upon 3 pi multiplied by 2 HR minus h square multiplied by h. Now this implies v is equal to 2 upon 3 pi h square r minus 1 upon 3 pi h cube. Now differentiating both sides with respect to h we get d v upon d h is equal to 2 upon 3 pi multiplied by 2 h multiplied by r minus 1 upon 3 pi multiplied by 3 h square. Now this implies d v upon d h is equal to 4 upon 3 pi h r minus 3 and 3 will cancel each other minus pi h square. Now we will find all the values of h at which d v upon d h is equal to 0. So we will put d v upon d h is equal to 0. This implies 4 upon 3 pi h r minus pi h square is equal to 0. This implies 4 upon 3 pi h r is equal to pi h square. Adding pi h square on both sides we get this expression. Now this implies 4 upon 3 pi r is equal to pi h dividing both sides by h we get this expression. Now dividing both sides by pi we get 4 upon 3 r is equal to h or we can simply write it as h is equal to 4 upon 3 r. Now let us find out if volume is maximum at h is equal to 4 upon 3 r. For that we will have to find second derivative of v with respect to h we know d v upon d h is equal to 4 upon 3 pi r h minus pi h square. Now differentiating both sides with respect to h again we get d square v upon d h square is equal to 4 upon 3 pi r minus 2 pi h. Now let us find out value of d square v upon d h square at h is equal to 4 upon 3 r. Now this is equal to 4 upon 3 pi r minus 2 pi multiplied by 4 upon 3 r. Now this is further equal to 4 upon 3 pi r minus 8 upon 3 pi r or we can simply write it as minus 4 upon 3 pi r. These are like terms we know 4 upon 3 minus 8 upon 3 is equal to minus 4 upon 3 so we get minus 4 upon 3 pi r. Now this value is less than 0. Now clearly we can see at h is equal to 4 upon 3 r d v upon d h is equal to 0 and d square v upon d h square is less than 0. This implies there exists maxima at h is equal to 4 upon 3 r. Now to find out the value of maximum volume at h is equal to 4 upon 3 r. First of all let us calculate r square r square is equal to 2 h r minus h square from expression 1. Now r square is equal to if we substitute value of h here we get 2 multiplied by 4 upon 3 r multiplied by r minus 4 upon 3 r whole square. Now this implies r square is equal to 8 upon 3 r square minus 16 upon 9 r square. We know square of 4 upon 3 is equal to 16 upon 9 and square of r is r square here 2 multiplied by 4 is equal to 8 and r multiplied by r is r square. Now this implies 24 r square minus 16 r square upon 9 is equal to r square. This implies r square is equal to 8 upon 9 r square. Now we know volume of cone is equal to 1 upon 3 pi r square h from expression 1. Now substituting this value of r square and value of h we get maximum volume. So we get 1 upon 3 multiplied by pi multiplied by 8 upon 9 r square multiplied by 4 upon 3 r. Now we can write volume is equal to 8 upon 27 multiplied by 4 upon 3 pi r cube now 4 upon 3 pi r cube is equal to volume of square. So we can write this is equal to 8 upon 27 multiplied by volume of square. Thus we get volume of cone we know v is the volume of cone is equal to 8 upon 27 multiplied by volume of square or we can say volume of the largest cone is equal to 8 upon 27 of the volume of the square. Hence proved this completes the session. Hope you understood the session. Take care and have a nice day.