Question

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Answer 1

Let r be the radius of the circle and let x be the length and y be the breadth of the rectangle inscribed in the given circle of radius r.

It can be observed that the diagonal of the rectangle is the diameter of the circle which is 2r and ABC is a right angle triangle.

By Pythagoras theorem,

( AB ) 2 + ( BC ) 2 = ( AC ) 2 x 2 + y 2 = ( 2r ) 2 y 2 =4 r 2 − x 2 y= 4 r 2 − x 2

The area of rectangle ABCD is,

A=xy =x( 4 r 2 − x 2 )

Differentiate the area with respect to x,

A ′ ( x )= 4 r 2 − x 2 +x( −2x 2 4 r 2 − x 2 ) = 4 r 2 − x 2 − x 2 4 r 2 − x 2 = 4 r 2 − x 2 − x 2 4 r 2 − x 2 = 4 r 2 −2 x 2 4 r 2 − x 2 (1)

Put A ′ ( x )=0,

4 r 2 −2 x 2 4 r 2 − x 2 =0 4 r 2 −2 x 2 =0 4 r 2 =2 x 2 x= 2 r

Then,

y= 4 r 2 − x 2 = 4 r 2 −2 r 2 = 2 r 2 = 2 r

Since, x=y= 2 r, so the rectangle ABCD is a square.

Differentiate equation (1) with respect to x,

A ″ ( x )= 4 r 2 − x 2 ( −4x )−( 4 r 2 −2 r 2 )( ( −2x ) 2 4 r 2 − x 2 ) ( 4 r 2 − x 2 ) 2 = ( −4x )( 4 r 2 − x 2 )−( 4 r 2 x+2 x 3 ) ( 4 r 2 − x 2 ) 3 2 = −16 r 2 x+4 x 3 +4 r 2 x−2 x 3 ( 4 r 2 − x 2 ) 3 2 = −12 r 2 x+2 x 3 ( 4 r 2 − x 2 ) 3 2

Simplify further,

A ″ ( x )= −2x( 6 r 2 − x 2 ) ( 4 r 2 − x 2 ) 3 2 A ″ ( 2 r )= −2 2 r( 6 r 2 − ( 2 r ) 2 ) ( 4 r 2 − ( 2 r ) 2 ) 3 2 = −8 2 r 3 2 2 r 3 =−4

This shows that the function is negative, so the area of rectangle is maximum when x= 2 r.

Hence, it is proved that all the rectangles inscribed in a given circle are the squares having maximum area.