Mathematics - Area Under Curves Question with Solution | TestHub

Take the rectangle with vertices $(\pm a,0)$ and $(\pm a,3-a)$ : its sides are $2a$ and $3-a$, so $A(a)=2a(3-a)$. $A^{\prime }(a)=6-4a=0$ at $a=\frac{3}{2}$, giving the maximum $A=2\cdot \frac{3}{2}\cdot \frac{3}{2}=\frac{9}{2}$.