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(B) The curves are $y_1 = x^2-4$ and $y_2 = \frac{4-x^2}{2}$.Let the vertices of the rectangle be $C(h, y_2)$ and $D(-h, y_2)$ on the upper curve,and

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) The curves are $y_1 = x^2-4$ and $y_2 = \frac{4-x^2}{2}$.Let the vertices of the rectangle be $C(h, y_2)$ and $D(-h, y_2)$ on the upper curve,and $B(h, y_1)$ and $A(-h, y_1)$ on the lower curve,where $h > 0$.The width of the rectangle is $2h$ and the height is $y_2 - y_1 = \frac{4-x^2}{2} - (x^2-4) = 2 - \frac{x^2}{2} - x^2 + 4 = 6 - \frac{3x^2}{2}$.Substituting $h$ for $x$,the area $A(h) = 2h 	imes (6 - \frac{3h^2}{2}) = 12h - 3h^3$.To find the maximum area,differentiate with respect to $h$: $\frac{dA}{dh} = 12 - 9h^2$.Setting $\frac{dA}{dh} = 0$,we get $9h^2 = 12$,so $h^2 = \frac{4}{3}$,which means $h = \frac{2}{\sqrt{3}}$.The maximum area is $A = 12(\frac{2}{\sqrt{3}}) - 3(\frac{2}{\sqrt{3}})^3 = \frac{24}{\sqrt{3}} - 3(\frac{8}{3\sqrt{3}}) = \frac{24}{\sqrt{3}} - \frac{8}{\sqrt{3}} = \frac{16}{\sqrt{3}} = \frac{16\sqrt{3}}{3}$.Using $\sqrt{3} \approx 1.732$,$A \approx \frac{16 	imes 1.732}{3} \approx 9.237$.The integer closest to $9.237$ is $9$.

$A$ rectangle with its sides parallel to the $X$-axis and $Y$-axis is inscribed in the region bounded by the curves $y=x^2-4$ and $y=\frac{4-x^2}{2}$. The maximum possible area of such a rectangle is closest to the integer.

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