 Hi and welcome to the session. Let us discuss the following question. Question says show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area. First of all, let us understand that if function f is defined at interval i and c belongs to interval i, first of all, let m and rc exist, then dash c is equal to 0 and rc is less than 0. Then point c is called a point of truth. Let us assume A, B, C, D is a rectangle whose length is 2x and red is 2y which is inscribed in a circle. Circle is having its center o and radius r. So, we can write let A, B, C, D be a rectangle in a given circle of radius r with center o. Now, let length of the rectangle is equal to 2x, breadth of the rectangle is equal to 2y. Now, A represents the area of the rectangle which is given by length into breadth. Now, A is equal to 2x multiplied by 2y. Now, 2x multiplied by 2y is equal to 4x5. Now, let us consider right triangle ABC. In right triangle ABC, AC square is equal to BC square plus AB square by Pythagoras theorem. We know AC is the diameter of the circle and radius of the circle is r. So, we can write AC is equal to 2y. We know diameter is the double of radius. So, we can write 2r square is equal to 2x square plus 2y square. We know AC is equal to 2r. AB is equal to 2x. AC is equal to 2y. So, here we have substituted their corresponding values in this expression. Now, simplifying we get 4r square is equal to 4x square plus 4y square. This implies r square is equal to x square plus y square. Dividing both sides by 4 we get r square is equal to x square plus y square. Now, subtracting x square from both sides we get r square minus x square is equal to y square. Or we can write y square is equal to r square minus x square. Now, this implies y is equal to under root of r square minus x square. We know area is equal to 4x5. Let us name this expression as 1. And y is equal to square root of r square minus x square. Let us name this expression as 2. Now, substituting the value of y from 2 in 1 we get A is equal to 4x multiplied by square root of r square minus x square. Now, differentiating both sides with respect to x we get dA upon dx is equal to 4x multiplied by 1 upon 2 multiplied by under root r square minus x square multiplied by minus 2x plus square root of r square minus x square multiplied by 4. Here we have applied the product rule to find the derivative of this term. Now, simplifying further we get dA upon dx is equal to 4r square minus 8x square upon square root of r square minus x square. Now, let us find out the points at which dA upon dx is equal to 0. dA upon dx is equal to 0 implies 4r square minus 8x square upon square root of r square minus x square is equal to 0. Now, multiplying both sides by square root of r square minus x square we get 4r square minus 8x square is equal to 0. Now, adding 8x square on both sides we get 4r square is equal to 8x square. Now, dividing both sides by 4 we get a r square is equal to 2x square. Now, taking square root on both sides we get r is equal to root 2 multiplied by x. Here we have neglected r equal to minus root 2x. As we know radius can never be negative. Now, we know dA upon dx is equal to 4r square minus 8x square upon square root of r square minus x square. Now, differentiating both sides with respect to x again we get d square a upon dx square is equal to square root of r square minus x square multiplied by minus 16x minus 4r square minus 8x square multiplied by 1 upon twice under root r square minus x square multiplied by minus 2x upon square root of r square minus x square whole square. Here we have applied the quotient rule to find the derivative of this term. Now, this implies d square a upon dx square is equal to r square minus x square multiplied by minus 16x plus x multiplied by 4r square minus 8x square upon under root of r square minus x square multiplied by r square minus x square. Here in the numerator we have subtracted these two terms by taking their LCM and we know square of square root of r square minus x square is equal to r square minus x square only. Now, simplifying further we get d square a upon dx square is equal to x multiplied by 8x square minus 12r square upon r square minus x square raised to the power 3 upon 2. Now, we know d a upon dx is equal to 0 at r equal to root 2x. Let us now find out the value of d square upon dx square at r equal to root 2x. This is equal to x multiplied by 8x square minus 12 multiplied by root 2x square upon root 2x square minus x square whole raised to the power 3 upon 2. Now, this is equal to x multiplied by 8x square minus 12 multiplied by 2x square upon 2x square minus x square raised to the power 3 upon 2. We know square of root 2x is equal to 2x square. Now, simplifying we get x multiplied by minus 16x square upon x square raised to the power 3 upon 2. Now, this is equal to minus 16x cube upon x square multiplied by 3 upon 2. 2 and 2 will cancel each other and we will get x cube in that denominator. This is equal to minus 16x cube upon x cube. x cube and x cube will cancel each other and we get minus 16. So, d square raised upon dx square at r equal to root 2x is equal to minus 16 which is less than 0. Now, we get at r equal to root 2x dv upon dx is equal to 0 and d square v upon dx square is less than 0. This implies r is equal to root 2x is a point of local maxima. So, we can say area of rectangle is maximum at r is equal to root 2x. Now, substituting this value of r in expression 2 we get y is equal to root 2x square minus x square. Now, this implies y is equal to 2x square minus x square. This implies y is equal to square root of we know 2x square minus x square is equal to x square. So, we get y is equal to square root of x square. Now, this implies y is equal to x. Now, if we multiply both sides by 2 we get 2y is equal to 2x. Now, 2y is equal to 2x implies all the sides of the rectangle abcd are equal. This implies abcd is of square. Now, we get out of all the rectangles inscribed in a given fifth circle the square has the maximum area. So, this is our required answer hence proved this completes the session hope you understood the session. Take care and have a nice day.