 Hi and welcome to the session. Let us discuss the following question. Question says show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin inverse 1 upon 3. First of all, let us understand that if we are given a function f defined on interval i and c belongs to interval i such that f double dash c exists then c is a point of local maxima if f dash c is equal to 0 and f double dash c is less than 0. This will work as key idea for solving the given question. Now let us start with the solution. Let h be the height and r be the radius of the cone. Now let s be the surface area of the cone and l be the slant height. Now given surface area of the cone that is s is equal to pi rn where r is the radius of the cone and l is the slant height. Now this implies l is equal to s upon pi r dividing both sides by pi r we get l is equal to s upon pi r. Now clearly we can see l square is equal to x square plus r square square of slant height is equal to square of height plus square of radius. Now from this equation we can find out the value of x square. x square is equal to l square minus r square on we can write h is equal to l square minus r square taking square root on both the sides we get h is equal to under root of l square minus r square. Now let us name this expression as 1 this expression as 2. Now we know volume of the cone is equal to 1 upon 3 pi r square h. Now let us name this expression as 3. Now from 2 we know h is equal to square root of l square minus r square. Now substituting this value of h in 3 we get v is equal to 1 upon 3 pi r square under root of l square minus r square. Now from 1 we know l is equal to s upon pi r. So we will substitute this value of l in this expression. Now we get v is equal to 1 upon 3 pi r square under root of s upon pi r square minus r square. Now simplifying we get v is equal to 1 upon 3 pi r square square root of s square minus pi square r raised to the power 4 upon pi r. Pi and pi will cancel each other and r will get cancelled with r. So we get v is equal to 1 upon 3 r square root of s square minus pi square r raised to the power 4. Now squaring both the sides we get v square is equal to 1 upon 9 r square multiplied by s square minus pi square r raised to the power 4. Now this can be written as 1 upon 9 r square s square minus 1 upon 9 pi square r raised to the power 6. So we get v square is equal to 1 upon 9 r square s square minus 1 upon 9 pi square r raised to the power 6. Now differentiating both sides with respect to r we get 2 v multiplied by dv upon dr is equal to 1 upon 9 s square multiplied by 2 r minus 1 upon 9 pi square multiplied by 6 r raised to the power 5. Now this implies dv upon dr is equal to 1 upon 9 multiplied by 2 s square r minus 6 pi square r raised to the power 5 upon 2 v. Now we will find all the points at which dv upon dr is equal to 0. So we will put dv upon dr equal to 0. Now this implies 1 upon 9 multiplied by 2 s square r minus 6 pi square r raised to the power 5 upon 2 v is equal to 0. Now multiplying both sides by 2 v we get 1 upon 9 multiplied by 2 s square r minus 6 pi square r raised to the power 5 is equal to 0. Multiplying both sides by 9 we get 2 s square r minus 6 pi square r raised to the power 5 is equal to 0. Now if we take 2 r common from both of these terms we get 2 r multiplied by s square minus 3 pi square r raised to the power 4 is equal to 0. Now this implies dividing both sides by 2 r we get s square minus 3 pi square r raised to the power 4 is equal to 0. Now this implies s square is equal to 3 pi square r raised to the power 4. Now dividing both sides by 3 pi square we get s square upon 3 pi square is equal to r raised to the power 4 or r raised to the power 4 is equal to s square upon 3 pi square. Now taking square root on both the sides we get r square is equal to s upon root 3 pi. Now let us check if volume is maximum at r square is equal to s upon root 3 pi we know 2 v multiplied by dv upon dr is equal to 2 upon 9 s square r minus 6 upon 9 pi square r raised to the power 5. This we have already proved above. Now again we will differentiate both the sides with respect to r to get second derivative of v. So we can write 2 v multiplied by d square v upon dr square plus 2 multiplied by square of dv upon dr. Here we have applied the product rule to find the derivative of this term is equal to 2 upon 9 s square minus 6 upon 9 pi square multiplied by 5 r raised to the power 4. Now we get the value of d square v upon dr square as equal to 2 upon 9 s square minus 30 upon 9 pi square r raised to the power 4 minus 2 multiplied by square of dv upon dr upon 2 v. Now we will find out value of d square v upon dr square at r square equal to s upon root 3 pi. So we can write at r square equal to s upon root 3 pi value of d square v upon dr square is equal to 2 upon 9 s square minus 30 upon 9 pi square multiplied by s square upon 3 pi square minus 2 multiplied by 0 square upon 2 v. We know at r square is equal to s upon root 3 pi dv upon dr is equal to 0. So we can write d square v upon dr square is equal to 2 upon 9 s square minus we will cancel pi square and pi square. Now we know we will cancel common factor 3. Now this is equal to minus 10 upon 9 s square. Now this term will be equal to 0. We get 2 upon 9 s square minus 10 upon 9 s square upon 2 v. Now this implies d square v upon dr square is equal to minus 8 upon 9 s square upon 2 v. We know 2 upon 9 s square minus 10 upon 9 s square is equal to minus 8 upon 9 s square. Now simplifying this we get d square v upon dr square is equal to minus 4 s square upon 9 v which is less than 0. Clearly we can see for any value of volume and surface area d square v upon dr square will remain negative. We also know that volume and surface area can never have negative value. Now we can write at r square equal to s upon root 3 pi dv upon dr is equal to 0 and d square v upon dr square is less than 0. Now by using key idea we get r square is equal to s upon root 3 pi is a point of local maxima or we can say maximum value of cone occurs at r square is equal to s upon root 3 pi. Now we will prove the required condition that is semi vertical angle of the cone is equal to sign inverse of 1 upon root 3. So we can write let semi vertical angle of the cone is equal to alpha. From expression 1 we know l is equal to s upon pi r. So we can write l is equal to s upon pi r using expression 1. Now we also know that r square is equal to s upon root 3 pi. This we have proved above we can reduce the value of s from this expression. Now we get s is equal to root 3 pi multiplied by r square. Let us name this expression as 4. Now we will substitute this value of s in this expression. Now we get l is equal to root 3 multiplied by pi r square upon pi r. Now pi and pi will get cancel and r will cancel 1 r. So we get l is equal to root 3 multiplied by r. Now this implies l upon r is equal to root 3 dividing both sides by r we get l upon r is equal to root 3. Now clearly we can see in this triangle alpha is the semi vertical angle of the cone with respect to alpha. This is the perpendicular. This is the height and this is the high continuous and we know sin alpha is equal to r upon l. So from this expression we get r upon l is equal to 1 upon root 3. Taking reciprocal of both the sides we get r upon l is equal to 1 upon root 3. Now clearly we can see r upon l is equal to sin alpha. So we can write sin alpha is equal to 1 upon root 3. This further implies alpha is equal to sin inverse of 1 upon root 3. We were required to show that semi vertical angle of right circular cone of a given surface area and maximum volume is sin inverse 1 upon root 3. So we have shown that cone is having maximum volume and semi vertical angle is equal to sin inverse 1 upon root 3. Hence proved. This completes the session. Hope you understood the session. Take care and have a nice day.