Hence, of all the rectangles inscribed in the given circle, the square has the maximum area.
Equations of the lines are y = 2x + 1, y = 3x + 1 and x + 4
Let ${\mathit{y}}_1$ = 2x + 1, ${\mathit{y}}_2$ = 3x + 1
Now area of the triangle bounded by the given lines,
= $\underset{0}{\overset{4}{}}\left({\mathit{y}}_2-{\mathit{y}}_1\right)$ dx
= $\underset{0}{\overset{4}{}}\left[\left(3\mathit{x}+1\right)-\left(2\mathit{x}+1\right)\right]$ dx
= $\underset{0}{\overset{4}{}}\mathit{x}$ dx
= $\frac{1}{2}{\left[{\mathit{x}}^2\right]}_0^4$
= $\frac{1}{2}\left(4^2-0^2\right)$
= $\frac{1}{2}$ × 16
= 8 sq. units
Thus, the area of the required triangular region is 8 square units.