 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. The question says, show that the right circular cylinder of given surface and maximum volume is such that its height is equal to diameter of the base. 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 f dash c is equal to 0 and f double dash c is less than 0. Then c is called point of local maxima. This is the key idea to solve the given question. Let us start with the solution now. Let s be the given surface area of the cylinder whose radius is r and height is h. So we can write let s be the given surface area of a cylinder whose radius is r and height is h. Now we know s is equal to 2 pi r square plus 2 pi r h. Now we can deduce value of h in terms of s and r from this expression. We get h is equal to s minus 2 pi r square upon 2 pi r. Now let us name this expression as 1. Now we know volume of cylinder is equal to pi r square h where r is the radius of cylinder and h is the height. Now we know h in this expression 1 is s minus 2 pi r square upon 2 pi r. Now we will substitute this value of h in this expression. Let us name this expression as 2. Now we can write substituting value of h from 1 into we get v is equal to pi r square multiplied by s minus 2 pi r square upon 2 pi r. Now we will cancel common factor pi r from numerator and denominator both and get v is equal to r multiplied by s minus 2 pi r square upon 2. Now this implies v is equal to 1 upon 2 multiplied by r s minus 2 pi r cube we have multiplied r by s and 2 pi r square in this expression. Now differentiating both sides with respect to r we get dv upon dr is equal to 1 upon 2 multiplied by s minus 6 pi r square. Now we will find the points at which dv upon dr is equal to 0. Now dv upon dr is equal to 0 implies 1 upon 2 multiplied by s minus 6 pi r square is equal to 0 this implies s minus 6 pi r square is equal to 0 multiplying both sides by 2 we get s minus 6 pi r square equal to 0. Now we will add 6 pi r square on both sides we get s is equal to 6 pi r square. Now we will substitute s is equal to 6 pi r square in expression 1. Now we can write substituting value of s is equal to 6 pi r square in expression 1 we get h is equal to 6 pi r square minus 2 pi r square upon 2 pi r. Now we know 6 pi r square minus 2 pi r square is equal to 4 pi r square so we get h is equal to 4 pi r square upon 2 pi r. Now we will cancel common factor 2 pi r in numerator and denominator and we get h is equal to 2 r. Now we know 2 r is equal to diameter so we get h is equal to diameter. Now we will show volume is maximum when height of the selector is equal to diameter of the phase. Now we know dv upon dr is equal to 1 upon 2 multiplied by s minus 6 pi r square. Now differentiating both sides with respect to r again we get d square v upon dr square is equal to 1 upon 2 multiplied by 0 minus 12 pi r. Now this is equal to 1 upon 2 multiplied by minus 12 pi r. We know 0 minus 12 pi r is equal to minus 12 pi r. Now we will cancel common factor 2 from numerator and denominator and we get minus 6 pi r. Now this is less than 0. We know value of r can never be negative so for every value of r d square v upon dr square will be less than 0. Now we get that s is equal to 6 pi r square dv upon dr is equal to 0 and d square v upon dr square is less than 0. So this implies point s is equal to 6 pi r square is a point of local maxima or we can say volume is maximum at s is equal to 6 pi r square. We have already proved that at s is equal to 6 pi r square h is equal to 2 r that is height of the cylinder is equal to diameter of its space. So we can write at s is equal to 6 pi r square h is equal to 2 r. This we have already proved above. This is our required proof. This completes the session. Hope you understood the session. Take care and have a nice day.