 Hi and welcome to your session. Let us work out the following question which says, the sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is v, when the side of square is double the radius of the circle. So in this question we are given combined perimeter of circle and square which is equal to k and we have to prove that combined area of circle and square is v, when the side of square is double the radius of the circle. Let us now begin with this solution. Let r be the radius of the circle side of the square. Combine perimeter, combine area and the square is equal to perimeter of the circle that is 2 pi r plus perimeter of the square that is 4 x and in the question we are given that combined perimeter of circle and square is equal to k. Now this implies x is equal to k minus 2 pi r divided by 4. a is equal to area of the circle that is pi r square plus area of the square that is x square. We have to prove that this area is least when side of the square is double the radius of the circle. Now x is equal to k minus 2 pi r divided by 4. So this is equal to pi r square plus k minus 2 pi r divided by 4 whole square. So a is equal to this. Let us name this as equation number one. Now differentiating with respect to we get b a by b r equals to 2 into k minus 2 pi r divided by 4 into 4 and this is equal to pi r minus k pi by 4 b a by d 0. Now this implies substitute the value of k at minus pi square r by 2 minus pi x plus pi square r by 2 is equal to 0. Now this and this gets cancelled out and we are left with 2 pi r minus pi x equals to 0 and this implies x is equal to 2 r. Now we will find b to a by d r 2 this is equal to so by second derivative test we can say that this minimum is equal to 2 r. This is the session bye and take care.