Question

# Find the dimensions of the rectangle of largest area that

Negative numbers and coordinate plane
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.)
$$\displaystyle{y}={6}-{x}^{{2}}$$

2021-08-15
Area of the rectangle is A=xy Substitue $$\displaystyle{y}={6}-{x}^{{2}}$$ in the formula of the area. $$\displaystyle{A}={x}{\left({6}-{x}^{{2}}\right)}$$
$$\displaystyle={6}{x}-{x}^{{3}}$$ Diferentiate area with respect to x and equate to 0. $$\displaystyle{A}'={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({6}{x}-{x}^{{3}}\right)}$$
$$\displaystyle={6}-{3}{x}^{{2}}$$
$$\displaystyle{6}-{3}{x}^{{2}}={0}$$
$$\displaystyle{3}{x}^{{2}}={6}$$
$$\displaystyle{x}=\pm\sqrt{{{2}}}$$
Substitute $$\displaystyle{x}=\pm\sqrt{{{2}}}\ \in\ {y}={6}-{x}^{{2}}$$ to get value of height, $$\displaystyle{y}={6}-{x}^{{2}}$$
$$\displaystyle={6}-{\left(\sqrt{{{2}}}\right)}^{{2}}$$
$$\displaystyle={4}$$ Hence. the dimensions of rectangle are, $$\displaystyle{2}\sqrt{{{2}}}\times{4}.$$