Maximum area of a reactangle which can be inscribed in a circle of a given radius R is

Options

A. `pi r^(2)`
B. `r^(2)`
C. `pir^(2)//4`
D. `2r^(2)`

Answer

D

Explanation

Let length of rectangel is be `2x`.
Ara of rectangle `A=2x.2sqrt(r^(2)-x^(2))=5xsqrt(r^(2)-x^(2))`
`implies(dA)/(dx)=(4(r^(2)-2x^(2)))/(sqrt(r^(2)-x^(2)))`
For maximum or minimum put
`(dA)/(dx)=0 impliesx=r/(sqrt(2))`
It can be easily checked that `(d^(2)A)/(dx^(2))lt0`
for `x=r/(sqrt(2))`

The maximum value of A is given by
`A=4 r/(sqrt(2))sqrt(r^(2)-(r^(2))/2)=2r^(2)`