Find the dimensions of the rectangle of largest area that

Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola. (Round your answers to the nearest hundredth.)
$y=6-{x}^{2}$
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hosentak
Area of the rectangle is A=xySubstitue $y=6-{x}^{2}$in the formula of the area.$A=x\left(6-{x}^{2}\right)$
$=6x-{x}^{3}$Diferentiate area with respect to x and equate to 0.${A}^{\prime }=\frac{d}{dx}\left(6x-{x}^{3}\right)$
$=6-3{x}^{2}$
$6-3{x}^{2}=0$
$3{x}^{2}=6$
$x=±\sqrt{2}$
Substituteto get value of height,$y=6-{x}^{2}$
$=6-{\left(\sqrt{2}\right)}^{2}$
$=4$Hence. the dimensions of rectangle are,$2\sqrt{2}×4.$