Find the dimensions of the rectangle of largest area that

Clifland 2021-08-13 Answered
Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius .

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Expert Answer

Layton
Answered 2021-08-14 Author has 17662 answers
We apply Pythagoras Theorem.
The diagonal of this rectangle is r+r=2r
\(\displaystyle{x}^{{2}}+{y}^{{2}}={\left({2}{r}\right)}^{{2}}\)
\(\displaystyle{x}^{{2}}+{y}^{{2}}={4}{r}^{{2}}\)
\(\displaystyle{y}^{{2}}={4}{r}^{{2}}-{x}^{{2}}\)
\(\displaystyle{y}=\sqrt{{{4}{r}^{{2}}-{x}^{{2}}}}\)
Let it be A-area of rectangle.
\(\displaystyle{A}={x}{y}\)
\(\displaystyle{A}={x}\sqrt{{{4}{r}^{{2}}-{x}^{{2}}}}\)
Now, we differentiate A(x)
\(\displaystyle{A}'{\left({x}\right)}={\left({x}\sqrt{{{4}{r}^{{2}}-{x}^{{2}}}}'\right.}\)
\(\displaystyle={x}'\sqrt{{{4}{r}^{{2}}-{x}^{{2}}}}+{\left(\sqrt{{{4}{r}^{{2}}-{x}^{{2}}}}\right)}'{x}\)
\(\displaystyle={1}\cdot\sqrt{{{4}{r}^{{2}}-{x}^{{2}}}}+{\left(-{\frac{{{x}}}{{\sqrt{{{4}{r}^{{2}}-{x}^{{2}}}}}}}\right)}\)
\(\displaystyle=\sqrt{{{4}{r}^{{2}}-{x}^{{2}}}}-{\frac{{{x}^{{2}}}}{{\sqrt{{{4}{r}^{{2}}-{x}^{{2}}}}}}}\)
\(\displaystyle={\frac{{{4}{r}^{{2}}-{x}^{{2}}-{x}^{{2}}}}{{\sqrt{{{4}{r}^{{2}}-{x}^{{2}}}}}}}\)
\(\displaystyle={\frac{{{4}{r}^{{2}}-{2}{x}^{{2}}}}{{\sqrt{{{4}{r}^{{2}}-{x}^{{2}}}}}}}\)
Solve A'(x)=0
\(\displaystyle{\frac{{{4}{r}^{{2}}-{2}{x}^{{2}}}}{{\sqrt{{{4}{r}^{{2}}-{x}^{{2}}}}}}}={0}\)
\(\displaystyle{4}{r}^{{2}}-{2}{x}^{{2}}={0}\)
\(\displaystyle{x}=\sqrt{{{2}}}{r},{x}=-\sqrt{{{2}}}{r}\)
x must be positive, so \(\displaystyle{x}=\sqrt{{{2}}}{r}\)
Largest area is: (subtitute \(\displaystyle{x}=\sqrt{{{2}}}{r}\) in equation of area)
\(\displaystyle{A}=\sqrt{{{2}}}{r}\sqrt{{{4}{r}^{{2}}-{\left(\sqrt{{{2}}}{r}\right)}^{{2}}}}\)
\(\displaystyle{A}=\sqrt{{{2}}}{r}\cdot\sqrt{{{4}{r}^{{2}}-{2}{r}^{{2}}}}\)
\(\displaystyle{A}=\sqrt{{{2}}}{r}\cdot\sqrt{{{2}{r}^{{2}}}}\)
\(\displaystyle{A}=\sqrt{{{2}}}{r}\cdot\sqrt{{{2}}}{r}\)
\(\displaystyle{A}={2}{r}^{{2}}\)
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