JEE MAIN - Mathematics (2024 - 8th April Evening Shift - No. 26)
Let $$\mathrm{A}$$ 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 $$\mathrm{A}$$ is ________.
Answer
128
Explanation
$$\begin{aligned} & y^2=2 x \\ & a=\left(\frac{1}{2}\right) \end{aligned}$$
_8th_April_Evening_Shift_en_26_1.png)
$$\begin{aligned} & A(t)=2 t \times\left(24-\frac{t^2}{2}\right) \\ & A=48 t-t^3 \end{aligned}$$
_8th_April_Evening_Shift_en_26_2.png)
$$\begin{aligned} & \frac{d A}{d t}=48-3 t^2 \\ & 48-3 t^2=0 \\ & 3 t^2=48 \\ & t^2=16 \\ & t= \pm 4 \end{aligned}$$
_8th_April_Evening_Shift_en_26_3.png)
$$\begin{aligned} & A(4)=48 \times 4-4^3 \\ & =192-64 \\ & A(4)=128 \end{aligned}$$
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