AP EAMCET202021 Sep 2020Morning ShiftMathematicsStraight LinesActual
Maximum area of the rectangle that can be formed with the fixed perimeter ' (p ) ' ( cm )
Options
- A( p^2 8 ~cm ^2 )
- B( p^2 16 ~cm ^2 )
- C( p^2 64 ~cm ^2 )
- D( p^2 32 ~cm ^2 )
Correct answer
B. ( p^2 16 ~cm ^2 )
Step-by-step solution
Let length of adjacent sides of rectangle is (x ~cm ) and (y ~cm ) so perimeter of rectangle is (p=2(x+y) cm ) and the area (A=x y ~cm ^2 ) ( A=x ( p 2 -x ) ) for maxima ( d A d x =0 p 2 -2 x=0 x= p 4 ~cm ) and (y= p 4 ~cm ) ( ) For given perimeter of rectangle (p ), the maximum possible area (A= p 4 p 4 = p^2 16 ~cm ^2 ) Hence, option (b) is correct.