**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 AC^{2} = AB^{2} + BC^{2} is given by 4a^{2} = x^{2}+y^{2} (as AC = 2a)

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

**Consider the diagonal,**

4a^{2} = x^{2} + y^{2}

y^{2} = 4a^{2} – x^{2}

y = \(\sqrt{4a^2-x^2}\)---- (1)

Now, perimeter of the rectangle, P

P = 2x + 2y

**Substituting (1) in the perimeter of the rectangle.**

P = 2x + 2\(\sqrt{4a^2-x^2}\)------ (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:**

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

**[squaring on both sides]**

**[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:**

**and if u and v are two functions of x, then**

**Now, consider the value of**

so the function P is maximum at x = a√2.

**Now substituting x = a√2 in equation (1):**

**As x = y = a√2 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.**