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Show that a rectangle of maximum perimeter which can be inscribed in a circle of radius a is a square of side √2 a.

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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

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.

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