 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question says, show that the right circular cone of least curved surface and given volume has an altitude equal to root 2 times the radius of the base. Let us now start with the solution. Let R be the radius of the base, H be the height of the cone and L be the slant height of the cone. Now clearly we can see this figure represents the cone having radius of the base as R height of the cone as H and slant height of the cone is L. Now volume of the cone that is V is equal to 1 upon 3 pi R square H. Now this further implies H is equal to 3V upon pi R square. Now let us assume that C represents the curved surface area of this cone. So we can write let C be the curved surface area of the cone that is C is equal to pi RL. Now squaring both the sides of this equation we get C square is equal to pi square multiplied by R square multiplied by L square. Now we know L square is equal to H square plus R square. We will substitute this value of L square in this equation and we get C square is equal to pi square multiplied by R square multiplied by H square plus R square. Now let us name this equation as 1 and this equation as 2. Now we will substitute this value of H from equation 1 in equation 2. Now substituting this value of H in equation 2 we get C square is equal to pi square multiplied by R square multiplied by square of 3V upon pi R square plus R square. Now this further implies C square is equal to pi square multiplied by R square multiplied by 9V square upon pi square multiplied by R raised to the power 4 plus R square. We know square of 3 is 9, square of V is V square, square of pi is pi square and square of R square is R raised to the power 4. Now multiplying this bracket with pi square R square we get 9V square upon R square plus pi square multiplied by R raised to the power 4. Multiplying these two terms we get this term and multiplying these two terms we get this term. Now let us assume that C square is equal to constant K. So we get K is equal to 9V square upon R square plus pi square multiplied by R raised to the power 4. Now we have K is equal to 9V square upon R square plus pi square multiplied by R raised to the power 4. Now differentiating both the sides with respect to R we get dK upon dr is equal to minus 18V square upon R cube plus pi square multiplied by 4R cube. Now we are given in the question that this cone has least curved surface. Now we will find the value of H in terms of radius R at which this cone has least curved surface area. Now we know 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. Also F C is the local minimum value of F. Now to find the value of C we will first put dK upon dr is equal to 0. Now we can write for maxima or minima dK upon dr must be equal to 0. Now clearly we can see dK upon dr is equal to sum of these two terms. So we will substitute these two terms for dK upon dr in this equation and we get minus 18V square upon R cube plus 4 pi square R cube is equal to 0. Now adding this term on both the sides of this equation we get 4 pi square R cube is equal to 18V square upon R cube. Now multiplying both the sides of this equation by R cube we get 4 pi square R raised to the power 6 is equal to 18V square. Now we know V is equal to 1 upon 3 pi R square H so we will substitute this value of V in this equation and we get 4 pi square R raised to the power 6 is equal to 18V square of 1 upon 3 pi R square H. This further implies 4 pi square R raised to the power 6 is equal to 2 pi square R raised to the power 4 by multiplied by H square. Clearly we can see square of 1 upon 3 is 1 upon 9 and we will cancel common factor 9 from numerator and denominator both and we are left with 2. Square of pi is pi square, square of R square is R raised to the power 4 and square of H is X square. Now dividing both the sides of this equation by 2 pi square R raised to the power 4 we get 2 R square is equal to H square. Now taking square root on both the sides we get root 2 R is equal to H or we can simply write H is equal to root 2 R. Now let us check if the curved surface area is minimum for this value of H. Now we know dK upon dr is equal to 4 pi square R cube minus 18V square upon R cube. Now differentiating both the sides with respect to R again we get d square K upon dr square is equal to 12 pi square R square plus 54V square. Upon R raised to the power 4, now clearly we can see both of these terms are positive and sum of positive terms is always greater than 0. So we can say d square K upon dr square is greater than 0 for any value of V and R. Now clearly we can see dK upon dr is equal to 0 for this value of H and also d square K upon dr square is greater than 0 for this value of H. So this implies that curved surface area is minimum at H is equal to root 2 R. So this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.