Question 18.

Let us assume the length of the rectangle is 'l' and breadth of the rectangle is 'b'.

The radius, l/2 and b in the above diagram form a right-angled triangle.

=> ${(\frac{l}{2})}^2 + b^2 = 2^2$

We know that the area of the rectangle is l*b, which can be obtained by considering 2 times the geometric mean of ${(\frac{l}{2})}^2$

Therefore, for the maximum area, the equality condition of AM-GM inequality should be satisfied

=> ${(\frac{l}{2})}^2$ = $b^2$ => l = 2b

=> l/b = 2/1