JEE Main 2025 — Application Of Derivatives Question with Solution

The maximum area of such a parallelogram $AFDE$, with one vertex fixed at $A$ and the other three points lying on the sides of triangle $ABC$, is half the area of triangle $ABC$.

Using the determinant formula for area of triangle with vertices $A\left(x_1,y_1\right),B\left(x_2,y_2\right),C\left(x_3,y_3\right)$ :

Area $△ABC$

$$=\frac{1}{2}\left|x_1\left(y_2-y_3^2\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$$

Substitute the coordinates:

$$\begin{array}{rl}& =\frac{1}{2}|4(1-(-3))+1((-3)-(-2))+9((-2)-1)|\\ & =\frac{1}{2}|4(4)+1(-1)+9(-3)|\\ & =\frac{1}{2}|16-1-27|=\frac{1}{2}|-12|=\frac{12}{2}=6\end{array}$$

Maximum area of parallelogram $AFDE$ $=\frac{1}{2}\times$ area of triangle $=\frac{1}{2}\times 6=3$