The area (in sq. units) of the largest rectan | vedclass

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(A) Let the coordinates of vertex $C$ be $(t, t^{2}-1)$ where $0 < t < 1$. Since the parabola is symmetric about the $y$-axis,the coordinates of verte

Vedclass · Accepted Answer

$\frac{4}{3 \sqrt{3}}$

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(A) Let the coordinates of vertex $C$ be $(t, t^{2}-1)$ where $0 The length of the rectangle is $2t$ and the height is $|t^{2}-1| = 1-t^{2}$ (since the rectangle is below the $x$-axis).The area $A$ of the rectangle is given by $A = (2t)(1-t^{2}) = 2t - 2t^{3}$.To find the maximum area,we differentiate $A$ with respect to $t$:$\frac{dA}{dt} = 2 - 6t^{2}$.Setting $\frac{dA}{dt} = 0$,we get $6t^{2} = 2$,which implies $t^{2} = \frac{1}{3}$,so $t = \frac{1}{\sqrt{3}}$.The maximum area is $A = 2(\frac{1}{\sqrt{3}}) - 2(\frac{1}{\sqrt{3}})^{3} = \frac{2}{\sqrt{3}} - \frac{2}{3\sqrt{3}} = \frac{6-2}{3\sqrt{3}} = \frac{4}{3\sqrt{3}}$ sq. units.

The area (in sq. units) of the largest rectangle $ABCD$ whose vertices $A$ and $B$ lie on the $x$-axis and vertices $C$ and $D$ lie on the parabola $y = x^{2}-1$ below the $x$-axis,is

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