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.)

$y=6-{x}^{2}$

Cabiolab
2021-08-14
Answered

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

Answered 2021-08-15
Author has **100** answers

Area of the rectangle is A=xySubstitue $y=6-{x}^{2}$ in the formula of the area.$A=x(6-{x}^{2})$

$=6x-{x}^{3}$ Diferentiate area with respect to x and equate to 0.${A}^{\prime}=\frac{d}{dx}(6x-{x}^{3})$

$=6-3{x}^{2}$

$6-3{x}^{2}=0$

$3{x}^{2}=6$

$x=\pm \sqrt{2}$

Substitute$x=\pm \sqrt{2}\text{}\in \text{}y=6-{x}^{2}$ to get value of height,$y=6-{x}^{2}$

$=6-{\left(\sqrt{2}\right)}^{2}$

$=4$ Hence. the dimensions of rectangle are,$2\sqrt{2}\times 4.$

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