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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
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