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

Options

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

Answer

D

Explanation

Area of rectangle `A=2x.sqrt(r^(2)-x^2)`
`=4xsqrt(r^(2)-x^(2))`
`(dA)/(dx)=(4(r^(2)-2x^(2)))/(sqrt(r^(2)-x^(2)))`
For maximum or minimum put `(dA)/(dx)=0`
`Rightarrow x=r//sqrt2`
It can be easily checked that `(d^2A)/(dx^(2))lt 0` for this value of x.
`therefore ` A is maximum for `x=(r)/(sqrt2)` and the maximum value of A is givn by
`A=4(r)/(sqrt2)sqrt(r^(2)-(r^(2))/(2))=2r^(2)`