Show that a rectangle of maximum perimeter which can be inscribed in a circle of radius a is a square of side .

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.