Question

Find the dimensions of the rectangle of largest area that

Negative numbers and coordinate plane
ANSWERED
asked 2021-08-14
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.)
\(\displaystyle{y}={6}-{x}^{{2}}\)

Answers (1)

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, \(\displaystyle{2}\sqrt{{{2}}}\times{4}.\)
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