 Welcome to the session. I'm Karika and I'm going to help you to solve the calling question. The question says show that semi vertical and the right circular core of given surface area and maximum volume is sin inverse 1 by 3. Let's now begin with the solution radius, p sand height, hp height of core of given surface area s. So this is the core whose height is h, sand height is l and radius is r and it has surface area s. We have to show that semi vertical angle that is this angle theta is sin inverse 1 by 3. Now we know that total surface area of a core that is s is equal to pi r square plus pi r l. Now this implies l is equal to s minus pi r square upon pi r. Now let's assume that v is volume of core. As v is volume of core therefore v is equal to 1 by 3 pi r square h. On squaring both sides we get v square equals to 1 by 9 pi square r to the power 4 h square. We know that l square is equal to h square plus r square and this implies l square minus r square is equal to h square. So by substituting the value of h square in this we get 1 by 9 pi square r to the power 4 into l square minus r square. l is equal to s minus pi r square upon pi r. So now we will substitute the value of l. By substituting value of l we find that v square is equal to 1 by 9 pi square r to the power 4 into s minus pi r square upon pi r whole square minus r square. This implies v square is equal to 1 by 9 pi square r to the power 4 into s square plus pi square r to the power 4 minus 2 s pi r square minus pi square r to the power 4 upon pi square r square. And this is equal to 1 by 9 pi square r to the power 4 into s square minus 2 s pi r square upon pi square r square. And this is equal to 1 by 9 r square into s square minus 2 s pi r square. This is equal to 1 by 9 s into s r square minus 2 pi r to the power 4. Now let z is equal to v square. Now v is maximum or minimum according to z is maximum or minimum. Therefore now z is equal to 1 by 9 into s into s r square minus 2 pi r to the power 4. Now differentiate z with respect to r so dz by dr is equal to 1 by 9 s into 2 s r minus 8 pi r cube. Now put dz by dr as 0 so this implies 2 s r minus 8 pi r cube is equal to 0 and this implies s is equal to 4 pi r square. Now define dz by dr to this is equal to 1 by 9 s into 2 s minus 24 pi r square. You know that s is equal to 4 pi r square and this implies r square is equal to s pi 4 pi. So substitute the value of r square in this. Now this is equal to 1 by 9 s into 2 s minus 24 pi into s by 4 pi and this is equal to minus 4 s square by 9 which is less than 0. So z is maximum when s is equal to 4 pi r square and hence v is maximum when s is equal to 4 pi r square. Now s is equal to pi r l plus pi r square so this implies pi r l plus pi r square is equal to 4 pi r square and this implies pi r l is equal to 4 pi r square minus pi r square and this implies pi r l is equal to 3 pi r square. This implies l is equal to 3 r. Sine theta is equal to perpendicular upon hypotenuse. Now here perpendicular is equal to r and hypotenuse is equal to l. So sine theta is equal to r pi l is equal to 3 r. Therefore sine theta is equal to 3 r and this implies sine theta is equal to 1 by 3 and this implies theta is equal to sine inverse 1 by 3. Hence we prove that semi vertical angle of right circular pole of given surface area and maximum volume is sine inverse 1 by 3. So this completes the session. Bye and take care.