Let A be the region enclosed by the parabola and the line . Then the maximum area of the rectangle inscribed in the region A is

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Solution

Q.
Let A be the region enclosed by the parabola $y^{2} = 2 x$ and the line $x = 24$ . Then the maximum area of the rectangle inscribed in the region A is _____. [2024]

Ans.
(128)

$y^{2} = 2 x and x = 24$

$∴$   Area of the rectangle, $A = (24 - x) 2 y$

$= (24 - \frac{y^{2}}{2}) 2 y$

   $= 48 y - y^{3}$

$\frac{d A}{d y} = 48 - 3 y^{2}$

$\frac{d A}{d y} = 0 \Rightarrow 48 - 3 y^{2} = 0 \Rightarrow y^{2} = 16 \Rightarrow y = \pm 4$

Now, $\frac{d^{2} A}{d y^{2}} = - 6 y$

   $\frac{d^{2} A}{d y^{2}} |_{y = 4} = - 24 < 0 (Maximum)$

$\frac{d^{2} A}{d y^{2}} |_{y = - 4} = 24 > 0 (Minimum)$

Hence, for maximum area, $y = 4 \Rightarrow x = 8$

Hence, maximum area $= (24 - 8) \times 2 \times 4 = 128$

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