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Cartesian plane

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Question

The maximum area of the convex polygon formed by joining the points A(0, 0), B(2t²,0), C(18,2), $D (\frac{8}{t^{2}}, 4)$ and E(0,2), where $t \in R - {0}$ and interior angle at vertex B is greater than or equal to 90° is _______ .

Moderate

Solution

Slope of BC is $\frac{2 - 0}{18 - 2 t^{2}} \geq 0$
$\Rightarrow 18 - 2 t^{2} \geq 0 \Rightarrow t^{2} - 9 \leq 0 \Rightarrow t \in [- 3, 3]$
Now, area of polygon ABCDE
= area of trapezium ABCE + area of triangle CDE
$\begin{array}{l} = \frac{1}{2} (2 t^{2} + 18) \times 2 + \frac{1}{2} \times 18 \times 2 \\ = 36 + 2 t^{2} \leq 54 \end{array}$
So, required maximum area is 54 sq. units.

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