SATHEE: Application of Derivatives 4 Question 53

Application of Derivatives 4 Question 53

56. Let $A (p^{2}, - p) B (q^{2}, q), C (r^{2}, - r)$ 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 $A E = y, △ A B C$ and $△ E D C$ are similar.

$∴ \frac{A B}{E D} = \frac{A C}{C E}$

$\begin{aligned} \Rightarrow & \frac{c}{x} & = \frac{b}{b - y} \\ \Rightarrow & b x & = c (b - y) \Rightarrow x & = \frac{c}{b} (b - y) \end{aligned}$

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

$\frac{d z}{d y} = 0$

$\Rightarrow \frac{c}{b} (b - 2 y) = 0 \Rightarrow y = \frac{b}{2}$

Clearly, $\frac{d^{2} z}{d y^{2}} = - \frac{2 c}{b} < 0, \forall y$

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

$z = \frac{c}{b} b - \frac{b}{2} \cdot \frac{b}{2} \cdot \sin A = \frac{1}{4} b c \sin A$

$\begin{aligned} = \frac{1}{2} area of △ A B C \\ = \frac{1}{2} \times \frac{1}{2} | \begin{array}{ccc} p^{2} & - p & 1 \\ q^{2} & q & 1 \\ r^{2} & - r & 1 \end{array} | \end{aligned}$

Applying $R_{3} \to R_{3} - R_{1}$ and $R_{2} \to R_{2} - R_{1}$

$\begin{aligned} = \frac{1}{4} | \begin{array}{ccc} p^{2} & - p & 1 \\ q^{2} - p^{2} & q + p & 0 \\ r^{2} - p^{2} & - r + p & 0 \end{array} | \\ = \frac{1}{4} (p + q) (r - p) | \begin{array}{ccc} p^{2} & - p & 1 \\ q - p & 1 & 0 \\ r + p & - 1 & 0 \end{array} | \\ = \frac{1}{4} (p + q) (r - p) (- q - r) \\ = \frac{1}{4} (p + q) (q + r) (p - r) \end{aligned}$