 Hi and welcome to the session. Today we will discuss the following question which says show that the rectangle of maximum area that can be inscribed in a circle is a square. So let's start the solution. First of all have a look at this figure in this rectangle A, B, C, D is inscribed in the circle with center O and the radius of this circle is R. Now here diagonal BD of rectangle A, B, C, D is the diameter of the circle. So that means the length of BD is 2R. Now let us assume that X and Y are the sides of the rectangle A, B, C, D. So we have rectangle A, B, C, D is inscribed in the circle with center O and radius net the sides of rectangle A, B, C, D, B, X and Y. Now we need to show that the rectangle of maximum area that can be inscribed in a circle is a square. So here the rectangle A, B, C, D with maximum area will be a square. This we need to show. So for this we will show that X is equal to Y that is all the sides of A, B, C, D are equal that means it is a square. So first of all let A be the area of rectangle A, B, C, D. Now we know that area of rectangle is length into width. So A will be equal to X into Y. Now angle BAD is a right angle that is 90 degrees. So that means triangle ABD is a right angle to triangle with BD as hypotenuse. So in right triangle ABD we have BD square is equal to AB square plus AD square by Pythagoras theorem. So this implies BD square that is 2R square is equal to AB square that is X square plus AD square that is Y square. Now let us express Y in terms of R and X. So this implies Y square is equal to 4R square minus X square thus Y is equal to square root of 4R square minus X square. Now area of the rectangle A, B, C, D, A is equal to X, Y. So let us substitute the value of Y from here. So this implies A is equal to X into square root of 4R square minus X square. Now let us differentiate both sides with respect to X. So A dash of X that is first derivative of A with respect to X is equal to X into 1 upon 2 square root of 4R square minus X square into minus 2X plus square root of 4R square minus X square into 1 using product rule which will be equal to minus X square upon square root of 4R square minus X square plus square root of 4R square minus X square. Now A dash of X is equal to 0 gives minus X square upon square root of 4R square minus X square plus square root of 4R square minus X square is equal to 0 that is minus X square plus 4R square minus X square is equal to 0 that is minus 2X square plus 4R square is equal to 0 that is 2X square is equal to 4R square that is X is equal to root 2R. Now Y is equal to square root of 4R square minus X square. So let us substitute the value of X to find out the value of Y this will be equal to square root of 4R square minus 2R square that is square root of 2R square that is root 2R thus we get X is equal to Y is equal to root 2R. Now if second derivative of A with respect to root 2R will be less than 0 then area of rectangle A, B, C, T will be maximum for X is equal to root 2R. So let us find out the second derivative of A with respect to X first. Now A dash of X is this so A double dash of X will be equal to minus X square into 1 upon 4R square minus X square whole to the power 3 by 2 into minus 2X plus 1 upon square root of 4R square minus X square into minus 2X plus 1 upon 2 into square root of 4R square minus X square into minus 2X. So A double dash of X is equal to 4R square minus 2R square root of 2R that is second derivative of A with respect to root 2R will be equal to minus 2R square into minus 2 root 2R upon 4R square minus 2R square whole to the power 3 by 2 minus 2 root 2R upon square root of 4R square minus 2R square root of 4R square minus 2R square minus 2 root 2R upon 2 into square root of 4R square minus 2R square that is 4 root 2R cube upon 2 root 2R cube minus 2 root 2R upon root 2R minus 2 root 2R upon 2 root 2R and this gives 2 minus 2 minus 1 that is minus 1 which is less than 0. So second derivative of A with respect to root 2R is less than 0 that means area of rectangle A B C D is maximum for X is equal to root 2R that is area of rectangle A B C D is maximum when X is equal to Y is equal to root 2R that is all the sides of rectangle A B C D are equal to root 2R that is we are equal to each other that means area of rectangle A B C D is maximum when A B C D is a square with this we finish this question and the session as well hope you must have understood the question goodbye take care and keep smiling.