Answer :

Given,


Rectangle is of maximum perimeter.


The rectangle is inscribed inside a circle.


The radius of the circle is ‘a’.



Let us consider,


‘x’ and ‘y’ be the length and breadth of the given rectangle.


Diagonal AC2 = AB2 + BC2 is given by 4a2 = x2+y2 (as AC = 2a)


Perimeter of the rectangle, P = 2(x+y)


Consider the diagonal,


4a2 = x2 + y2


y2 = 4a2 – x2


---- (1)


Now, perimeter of the rectangle, P


P = 2x + 2y


Substituting (1) in the perimeter of the rectangle.


------ (2)


For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.


Differentiating the equation (2) with respect to x:





[Since ]


------ (3)


To find the critical point, we need to equate equation (3) to zero.





[squaring on both sides]


4a2 – x2 = x2


2x2 = 4a2


x2 = 2a2




[as x cannot be negative]


Now to check if this critical point will determine the maximum diagonal, we need to check with second differential which needs to be negative.


Consider differentiating the equation (3) with x:





[Since and and if u and v are two functions of x, then ]






----- (4)


Now, consider the value of




As , so the function P is maximum at .


Now substituting in equation (1):





As the sides of the taken rectangle are equal, we can clearly say that a rectangle with maximum perimeter which is inscribed inside a circle of radius ‘a’ is a square.


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