 Hi and welcome to the session. Let us discuss the following question. Question says, show that semi-vertical angle of the cone of the maximum volume and of given slant height is tan inverse 2. First of all, let us understand that if we have given a function f defined on interval i and c belongs to interval i such that f double dash c exist, then if f dash c is equal to 0 and f double dash c is less than 0, then c is a point of local maxima. This is the key idea to solve the given question. Let us start with the solution now. Let h be the height and r be the radius of the cone. Now let v represents the volume of the cone. Volume of the cone is given by 1 upon 3 pi r square h. Now let the slant height of the cone be l. Now we know l square is equal to h square plus r square. Now we can find value of r square from this equation. Now this implies r square is equal to l square minus h square. Now let us name this expression as 1 and this expression as 2. Now substituting value of r square from 2 in 1 we get v is equal to 1 upon 3 pi multiplied by l square minus h square multiplied by h. Now this implies v is equal to 1 upon 3 pi l square h minus 1 upon 3 pi h cube. We have multiplied 1 upon 3 pi h with this bracket. Now differentiating both sides with respect to h we get dv upon dh is equal to 1 upon 3 pi l square multiplied by 1 minus 1 upon 3 pi multiplied by 3 h square. 3 and 3 will cancel each other. So we get dv upon dh is equal to 1 upon 3 pi l square minus pi h square. Now we will find all the points at which dv upon dh is equal to 0. So we will put dv upon dh is equal to 0. This implies 1 upon 3 pi l square minus pi h square is equal to 0. This implies pi multiplied by 1 upon 3 l square minus h square is equal to 0. Here we have taken pi common from both the terms. Now dividing both sides by pi we get 1 upon 3 l square minus h square is equal to 0. Now adding h square on both the sides we get 1 upon 3 l square is equal to h square. Now multiplying both sides by 3 we get l square is equal to 3x square. Now square of slant height is equal to h square plus r square. So we can write h square plus r square is equal to 3x square. Now this implies r square is equal to 2x square subtracting h square from both the sides we get r square is equal to x square. Now dividing both sides by h square we get r square upon h square is equal to 2. Now this implies square of r upon h is equal to 2. Now taking square root on both the sides we get r upon h is equal to square root of 2. Now let us assume that alpha is semi vertical angle of the cone. Now r upon h is equal to tan alpha if we see this triangle. Now clearly we can see r is the perpendicular and h is the base with respect to alpha. So r upon h is equal to tan alpha. So we can write tan alpha is equal to root 2. This further implies value of alpha is equal to tan inverse root 2. Now we will show volume is maximum when alpha is equal to tan inverse root 2 or we can say semi vertical angle of the cone is equal to tan inverse root 2. Now we know dv upon dh is equal to 1 upon 3 pi a square minus pi a square. This we have already proved above. Now we will again differentiate both the sides with respect to h to find second derivative of v. So we get d square v upon dh square is equal to 0 minus 2 pi h. Derivative of this term is 0 and derivative of this term is 2 pi h. So we get d square v upon dh square is equal to minus 2 pi h which is less than 0. We know height of the cone can never be negative. So d square v upon dh square is less than 0 for any value of h. Now this implies for alpha is equal to tan inverse root 2 volume is maximum. So we get semi vertical angle of the cone of maximum volume and of given stront height is tan inverse 2. Hence proved this completes the session. Hope you understood the session. Take care and have a nice day.