The maximum area (in sq. units) of a rectangl | vedclass

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(C) Let the vertices of the rectangle on the parabola be $(\alpha, 12 - \alpha^2)$ and $(-\alpha, 12 - \alpha^2)$,where $\alpha > 0$.The base of the r

Vedclass · Accepted Answer

$32$

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(C) Let the vertices of the rectangle on the parabola be $(\alpha, 12 - \alpha^2)$ and $(-\alpha, 12 - \alpha^2)$,where $\alpha > 0$.The base of the rectangle lies on the $x-$axis,so the length of the base is $2\alpha$ and the height is $12 - \alpha^2$.The area $A$ of the rectangle is given by $A = 	ext{length} 	imes 	ext{height} = 2\alpha(12 - \alpha^2) = 24\alpha - 2\alpha^3$.To find the maximum area,we differentiate $A$ with respect to $\alpha$:$\frac{dA}{d\alpha} = 24 - 6\alpha^2$.Setting $\frac{dA}{d\alpha} = 0$,we get $24 - 6\alpha^2 = 0$,which implies $\alpha^2 = 4$,so $\alpha = 2$ (since $\alpha > 0$).To verify this is a maximum,we check the second derivative: $\frac{d^2A}{d\alpha^2} = -12\alpha$. At $\alpha = 2$,$\frac{d^2A}{d\alpha^2} = -24 The maximum area is $A = 2(2)(12 - 2^2) = 4(12 - 4) = 4(8) = 32$ sq. units.

The maximum area (in sq. units) of a rectangle having its base on the $x-$axis and its other two vertices on the parabola $y = 12 - x^2$ such that the rectangle lies inside the parabola,is

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