Show-that-of-all-rectangles-inscribed-in-a-given-circle-the-square-has-a-maximum-area-

Question Number 6136 by sanusihammed last updated on 15/Jun/16

S h o w t h a t o f a l l r e c t a n g l e s i n s c r i b e d i n a g i v e n c i r c l e

t h e s q u a r e h a s a m a x i m u m a r e a .

Answered by Rasheed Soomro last updated on 15/Jun/16

A l l t h e r e c t a n g l e s i n s c r i b e d i n s a m e c i r c l e

h a v e e q u a l d i a g o n a l s a n d v i c e v e r s a .

L e t t h e m e a s u r e o f d i a g o n a l i s D

I f a a n d b a r e d i m e n t i o n s o f r e c t a n g l e

D = \sqrt{a^{2} + b^{2}}

I f s i s m e a s u r e o f t h e s i d e o f s q u a r e

D = \sqrt{s^{2} + s^{2}} = \sqrt{2} s

\sqrt{a^{2} + b^{2}} = \sqrt{2} s

s = \sqrt{\frac{a^{2} + b^{2}}{2}}

A r e a o f s q u a r e = s^{2} = \frac{a^{2} + b^{2}}{2}

A r e a o f r e c t a n g l e = a b

N o w w e h a v e t o p r o v e t h a t

A r e a o f s q u a r e > A r e a o f r e c t a n g l e

\frac{a^{2} + b^{2}}{2} > a b \forall a, b > 0 \land a \neq b

T h e r e i s a p r o v e d r e l a t i o n b e t w e e n a r i t h m a t i c

m e a n, g e o m e t r i c m e a n a n d h a r m o n i c m e a n

w h i c h w i l l h e l p u s t o p r o v e t h e a b o v e r e s u l t .

I f A, G, H a r e A . M ., p o s i t i v e G . M . a n d

H . M . b e t w e e n a n y t w o p o s i t i v e a n d u n e q u a l

n u m b e r s a a n d b t h e n

A > G > H

\frac{a + b}{2} > a b > \frac{2 a b}{a + b}

\frac{a + b}{2} > \frac{2 a b}{a + b}

\frac{{(a + b)}^{2}}{2} > 2 a b

\frac{a^{2} + 2 a b + b^{2}}{2} > 2 a b

\frac{a^{2}}{2} + a b + \frac{b^{2}}{2} > 2 a b

\frac{a^{2} + b^{2}}{2} > a b

\underset{}{A r e a o f s q u a r e} > A r e a o f r e c t a n g l e

p r o v i d e d t h a t t h e y h a v e s a m e d i a g o n a l s . I - E

t h e y a r e i n s c r i b e d i n s a m e c i r c l e

Answered by FilupSmith last updated on 15/Jun/16

An easy solution using integrals

Rectangle of height y, length x

has area :

A_{r} = \int_{0}^{x} y d u = y x

Square of sides x :

A_{s} = \int_{0}^{x} x d u = x^{2}

y x < x^{2} if y < x

therefore A_{s} > A_{r} for all y < x

∴ s q u a r e h a s g r e a t e s t a r e a

Commented by FilupSmith last updated on 15/Jun/16

oh, i didnt read the part about the cirlce

sorry, but i misread the question