 Hi and welcome to the session. Let us discuss the following question. Question says, show that the right circular cone of least curved surface and given volume as an altitude equal to root 2 time the radius of the base. First of all, let us understand that if function f is defined in interval i, c belongs to interval i such that f double dash c exists. Then, x is equal to c is a point of local minima if f dash c is equal to 0 and f double dash c is greater than 0. This is the key idea to solve the given question. Now let us start with the solution. Let r and h be the radius and height of the given cone and l is the slant height of the given cone. Now volume of the cone that is v is given by 1 upon 3 pi r square h. Now multiplying both sides by 3 upon pi h we get 3 v upon pi h is equal to r square or we can simply write r square is equal to 3 v upon pi h. Now curved surface area of cone that is s is given by pi rn. Now we know l is equal to square root of h square plus r square. So we can write s is equal to pi r multiplied by square root of h square plus r square. Now squaring both sides of this expression we get s square is equal to pi square multiplied by r square multiplied by h square plus r square. Now let us name this expression as 1 and this expression as 2. Now substituting value of r square from expression 1 in expression 2 we get s square is equal to pi square multiplied by 3 v upon pi h multiplied by h square plus 3 v upon pi h. Now simplifying we get s square is equal to 3 v pi h plus 9 v square upon h square. Now differentiating both sides with respect to h we get 2s multiplied by ds upon dh is equal to 3 v pi plus 9 v square multiplied by minus 2h upon h raised to the power 4. Now simplifying further we get 2s ds upon dh is equal to 3 v pi minus 18 v square upon h cube. Now dividing both sides by 2s we get ds upon dh is equal to 1 upon 2s multiplied by 3 v pi minus 18 v square upon h cube. Now we will find all the points at which ds upon dh is equal to 0. So we will put ds upon dh is equal to 0. This implies 1 upon 2s multiplied by 3 v pi minus 18 v square upon h cube is equal to 0. Now this implies 3 v pi minus 18 v square upon h cube is equal to 0. Multiplying both sides of this expression by 2s we get this expression. Now if we take 3 v common on left hand side we get 3 v multiplied by pi minus 6 v upon h cube is equal to 0. Now this implies 3 v is equal to 0 or pi minus 6 v upon h cube is equal to 0. Now dividing both sides of this equation by 3 we get v is equal to 0. But volume of the cone can never be equal to 0. So we will neglect this value. So we can write neglecting 3 v is equal to 0. We get pi minus 6 v upon h cube is equal to 0. Now adding 6 v upon h cube on both sides we get pi is equal to 6 v upon h cube. Multiplying both sides by h cube we get pi h cube is equal to 6 v. Now dividing both sides by pi we get h cube is equal to 6 v upon pi. Now we know v is equal to 1 upon 3 pi r square h. We have already shown it above. Now we can write h cube is equal to 6 upon pi multiplied by 1 upon 3 pi r square h. Now simplifying we get h cube is equal to pi will cancel pi and we know 3 multiplied by 2 is equal to 6. So we get h cube is equal to 2 r square h. Now dividing both sides by h we get h square is equal to 2 r square. Now taking square root on both the sides we get h is equal to square root of 2 multiplied by r. We were required to prove that altitude of the cone is equal to root 2 times the radius of the base. So we can see h is equal to root 2 r. Now we will prove that curve surface area of the cone is minimum at h is equal to root 2 r. To show that surface area is minimum at h is equal to root 2 r. First of all we will find second derivative of s. We know 2 multiplied by s multiplied by ds upon dh is equal to 3 v pi minus 18 v square upon h cube. This we have already shown above. Now differentiating both sides of this equation with respect to h we get 2 multiplied by s d square s upon dh square plus ds upon dh whole square is equal to 54 multiplied by v square multiplied by h square upon h raised to the power 6. We know derivative of this term is equal to 0 and we will apply quotient rule to find the derivative of this term. Now derivative of this term is equal to 54 v square h square upon h raised to the power 6. Now dividing both sides by 2 we get s multiplied by d square s upon dh square plus ds upon dh square is equal to 27 v square h square upon h raised to the power 6. Now h square will cancel 2 h from the denominator and we get h raised to the power 4 in the denominator. So we can write s d square s upon dh square plus ds upon dh square is equal to 27 v square upon h raised to the power 4. Now substituting h is equal to root 2r in this equation we get s multiplied by d square s upon dh square plus 0 square is equal to 27 v square upon root 2r raised to the power 4. We know at h is equal to root 2r ds upon dh is equal to 0. Now this implies s multiplied by d square s upon dh square is equal to 27 v square upon 4 multiplied by r raised to the power 4. Now this further implies d square s upon dh square is equal to 27 v square upon s multiplied by 4 multiplied by r raised to the power 4. Now this is greater than 0. We know volume radius and surface area can never have negative value so second derivative is greater than 0. Now we get at h is equal to root 2r ds upon dh is equal to 0 and d square s upon dh square is greater than 0. This implies h is equal to root 2r is a point of local minima or we can say surface area is least at h is equal to root 2r. Hence proved this completes the session hope you understood the solution take care and keep smiling.