Question

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 parallelogram $AFDE$.

Answer 1

Let $AF = x = DE$ and $AE = y = DF$
As $\triangle CAB$ is $\triangle CED$
So, $ \frac{C E}{C A}=\frac{D E}{A B} \Rightarrow \frac{b-y}{b}=\frac{x}{c} \Rightarrow y=b\left(1-\frac{x}{c}\right)$
(Here $BC = a , AC = b$ and $AB = c$ )
Now, area of parallelogram AFDE
$= S =( AF )( EM )= xy \sin A \Rightarrow S = x \cdot b \left(1-\frac{ x }{ c }\right) \sin A $ (Note $: \sin A$ is fixed).....(1)
Now, differentiating both sides of equation (1) with respect to $x$, we get
$\frac{d S}{d x}=\frac{b}{c}(c-2 x) \sin A=0 \Rightarrow x=\frac{c}{2}$
Also, $\left.\frac{ d ^2 s }{ dx ^2}\right]_{ x =\frac{ c }{2}}=\frac{-2 b }{ c }<0$
So, $S$ is maximum when $x =\frac{ c }{2}$
Now, $S _{\max }=\frac{1}{4} bc \sin A$

$=\frac{1}{2}\left(\frac{1}{2} bc \sin A \right)=\frac{1}{2}(\operatorname{area}(\triangle ABC ))=\frac{1}{4}\begin{vmatrix}1 & -1 & 1 \\ 4 & -2 & 1 \\ 9 & 3 & 1\end{vmatrix}\mid=\frac{20}{4}=5$ (square units.)