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

Let the radius of the circle to be R.

When a rectangle is inscribed in a circle , the diameter of the circle will act as the diagonal for the rectangle.

So assume that the sides of the rectangle to be a, b

applying pythagoras theorm

Then a² + b² = 4R²  ===> b = (4R² - a² )^1/2

The Perimeter of the rectangle is 2(a+b) = 2 (a + (4R² - a² )^1/2)

differetiating the equattion with respect to a and put it equal to 0

= 2(1- (2a)/2(4R² - a² )^1/2)=0

= 2(1- (a)/(4R² - a² )^1/2)=0

= 1= (a)/(4R² - a² )^1/2

= a=(4R² - a² )^1/2

= a²= (4R² - a² )

= 2a² = 4R²

= a = R(2)^1/2