Let  $A(1,-1),B(4,-2)$ and  $C(9,3)$ be the vertices of the triangle ABC. A parallelogram AFDE is drawn with vertices D,E and F on the line segments BC, CA and AB respectively. Find the maximum area of the parallelogram AFDE.

$\text{ Let }AF=x=DE\text{ and }AE=y=DF$

$\text{ As }\mathrm{△}CAB\text{ is similar to}\mathrm{△}CED$

$\text{ So, }\frac{CE}{CA}=\frac{DE}{AB}\Rightarrow \frac{b-y}{b}=\frac{x}{c}\Rightarrow y=b\left(1-\frac{x}{c}\right)$

$\text{ (Here }BC=a,AC=b\text{ and }AB=c\text{ ) }$

$\text{ Now, area of parallelogram AFDE }$

$=S=\left(AF\right)\left(EM\right)=xy\sin A\Rightarrow S=x\cdot b\left(1-\frac{x}{c}\right)\sin A$

$\text{ Note: }\sin A\text{ is fixed }$

$\text{ Now, differentiating both sides of equation (i) with respect to x we get }$

$\frac{dS}{dx}=\frac{b}{c}(c-2x)\sin A=0\Rightarrow x=\frac{c}{2}$

$\text{ Also, }{\left.\frac{d^{2}S}{d{x}^2}\right]}_{x-\frac{c}{2}}=\frac{-2b}{c}<0$

$\text{ So, }S\text{ is maximum when }x=\frac{\mathrm{c}}{2}\text{. }$

$\text{ Now, }{\mathrm{S}}_{max}=\frac{1}{4}\mathrm{bcsin}\mathrm{A}$

$\left.=\frac{1}{2}\left(\frac{1}{2}bc\sin A\right)=\frac{1}{2}[\mathrm{area}(\mathrm{△}ABC\left)\right]=\frac{1}{4}\left|\right|\begin{array}{ccc}1& -1& 1\\ 4& -2& 1\\ 9& 3& 1\end{array}\right|=\frac{20}{4}=5\text{ sq units}$