TS EAMCET 03 May 2019 Shift 2 Solved Paper

Write the relation from ΔOAB.
r2=x2+y2
x2=4−y2
Area of the rectangle is 2xy
Substitute the value of x
A‌‌=2√4−y2⋅y
A2‌‌=4y2(4−y2)
Differentiate the equation
‌

dA2

dy

=4(8y−4y3)
For maximum and minimum
‌

dA2

dy

=0 8y−4y3=0 y2=2 y=√2
Therefore, the smallest side of the rectangle to have the largest area is √2