Application of Derivatives 4 Question 53
####56. Let $A\left(p^{2},-p\right) B\left(q^{2}, q\right), C\left(r^{2},-r\right)$ be the vertices of the triangle $A B C$. A parallelogram $A F D E$ is drawn with vertices $D, E$ and $F$ on the line segments $B C, C A$ and $A B$, respectively. Using calculus, show that maximum area of such a parallelogram is $\frac{1}{4}(p+q)(q+r)(p-r)$.
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Answer:
Correct Answer: 56. (5)
Solution:
- Let $A F=x$ and $A E=y, \triangle A B C$ and $\triangle E D C$ are similar.
$\therefore \quad \frac{A B}{E D}=\frac{A C}{C E}$
$ \begin{aligned} \Rightarrow & & \frac{c}{x} & =\frac{b}{b-y} \\ \Rightarrow & b x & =c(b-y) \quad \Rightarrow \quad x & =\frac{c}{b}(b-y) \end{aligned} $
Let $z$ denotes the area of par allelogram $A F D E$.
Then,
$ z=x y \sin A $
$\Rightarrow$
$ z=\frac{c}{b}(b-y) y \cdot \sin A $
On differentiating w.r.t. y we get $\frac{d z}{d y}=\frac{c}{b}(b-2 y) \sin A$ and $\frac{d^{2} z}{d y^{2}}=\frac{-2 c}{b} \sin A$
For maximum or minimum values of $z$, we must have
$ \frac{d z}{d y}=0 $
$\Rightarrow \quad \frac{c}{b}(b-2 y)=0 \quad \Rightarrow \quad y=\frac{b}{2}$
Clearly, $\quad \frac{d^{2} z}{d y^{2}}=-\frac{2 c}{b}<0, \forall y$
Hence, $z$ is maximum, when $y=\frac{b}{2}$.
On putting $y=\frac{b}{2}$ in Eq. (i), we get
the maximum value of $z$ is
$ z=\frac{c}{b} \quad b-\frac{b}{2} \cdot \frac{b}{2} \cdot \sin A=\frac{1}{4} b c \sin A $
$ \begin{aligned} & =\frac{1}{2} \text { area of } \triangle A B C \\ & =\frac{1}{2} \times \frac{1}{2}\left|\begin{array}{ccc} p^{2} & -p & 1 \\ q^{2} & q & 1 \\ r^{2} & -r & 1 \end{array}\right| \end{aligned} $
Applying $R_{3} \rightarrow R_{3}-R_{1}$ and $R_{2} \rightarrow R_{2}-R_{1}$
$ \begin{aligned} & =\frac{1}{4}\left|\begin{array}{ccc} p^{2} & -p & 1 \\ q^{2}-p^{2} & q+p & 0 \\ r^{2}-p^{2} & -r+p & 0 \end{array}\right| \\ & =\frac{1}{4}(p+q)(r-p)\left|\begin{array}{ccc} p^{2} & -p & 1 \\ q-p & 1 & 0 \\ r+p & -1 & 0 \end{array}\right| \\ & =\frac{1}{4}(p+q)(r-p)(-q-r) \\ & =\frac{1}{4}(p+q)(q+r)(p-r) \end{aligned} $
