Find the dimensions of the isosceles triangle of largest area that can be inscri

IMLOG10ct 2021-11-20 Answered
Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r.

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Expert Answer

Wasither1957
Answered 2021-11-21 Author has 1402 answers

Step 1
From the figure, by Pythagoras theorem we can write
\(\displaystyle{\frac{{{b}}}{{{2}}}}=\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\)
\(\displaystyle{b}={2}\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\rightarrow{\left({1}\right)}\)
image

Step 2
Area of Triangle \(\displaystyle={\frac{{{1}}}{{{2}}}}\) (Base)(Height)
\(\displaystyle{A}={\frac{{{1}}}{{{2}}}}{\left({b}\right)}{\left({r}+{x}\right)}\)
Use Eqn(1) to substitute the value of b
\(\displaystyle{A}={\frac{{{1}}}{{{2}}}}{\left({2}\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\right)}{\left({r}+{x}\right)}\)
\(\displaystyle{A}{\left({x}\right)}={\left({r}+{x}\right)}\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\)
Step 3
Differentiate \(\displaystyle{A}{\left({x}\right)}\)
\(\displaystyle{A}'{\left({x}\right)}={\frac{{{d}{\left[{\left({r}+{x}\right)}\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\right]}}}{{{\left.{d}{x}\right.}}}}\)
Use Product Rule
\(\displaystyle{A}'{\left({x}\right)}=\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\cdot{\frac{{{d}{\left[{r}+{x}\right]}}}{{{\left.{d}{x}\right.}}}}+{\left({r}+{x}\right)}\cdot{\frac{{{d}{\left[\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\right]}}}{{{\left.{d}{x}\right.}}}}\)
\(\displaystyle{A}'{\left({x}\right)}=\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\cdot{1}+{\left({r}+{x}\right)}\cdot{\frac{{{d}{\left[\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\right]}}}{{{d}{\left({r}^{{{2}}}-{x}^{{{2}}}\right)}}}}\times{\frac{{{d}{\left({r}^{{{2}}}-{x}^{{{2}}}\right)}}}{{{\left.{d}{x}\right.}}}}\)
\(\displaystyle{A}'{\left({x}\right)}=\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}+{\left({r}+{x}\right)}\cdot{\frac{{{1}}}{{{2}\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}}}}\times{\left(-{2}{x}\right)}\)
\(\displaystyle{A}'{\left({x}\right)}=\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}-{\frac{{{r}{x}+{x}^{{{2}}}}}{{\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}}}}\)
Step 4
Solve for \(\displaystyle{A}'{\left({x}\right)}={0}\)
\(\displaystyle\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}-{\frac{{{r}{x}+{x}^{{{2}}}}}{{\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}}}}={0}\)
\(\displaystyle\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}={\frac{{{r}{x}+{x}^{{{2}}}}}{{\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}}}}\)
Multiply both sides by \(\displaystyle\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\)
\(\displaystyle{r}^{{{2}}}-{x}^{{{2}}}={r}{x}+{x}^{{{2}}}\)
\(\displaystyle{2}{x}^{{{2}}}+{r}{x}-{r}^{{{2}}}={0}\)
Using the quadratic formula, we can write
\(\displaystyle{x}={\frac{{-{r}\pm\sqrt{{{r}^{{{2}}}-{4}{\left({2}\right)}{\left(-{4}^{{{2}}}\right)}}}}}{{{2}{\left({2}\right)}}}}\)
\(\displaystyle{x}={\frac{{-{r}\pm\sqrt{{{9}{r}^{{{2}}}}}}}{{{4}}}}\)
\(\displaystyle{x}={\frac{{-{r}\pm{3}{r}}}{{{4}}}}\)
\(\displaystyle{x}={\frac{{-{r}-{3}{r}}}{{{4}}}}\) or \(\displaystyle{x}={\frac{{-{r}+{3}{r}}}{{{4}}}}\)
\(\displaystyle{x}={\frac{{-{4}{r}}}{{{4}}}}\) or \(\displaystyle{x}={\frac{{{2}{r}}}{{{4}}}}\)
\(\displaystyle{x}=-{r}\) or \(\displaystyle{x}={\frac{{{r}}}{{{2}}}}\)
Step 5
But x cannot be negative so it must be \(\displaystyle{\frac{{{r}}}{{{2}}}}\)
Substitute \(\displaystyle{x}={\frac{{{r}}}{{{2}}}}\) in Eqn(1). To get
\(\displaystyle{b}={2}\sqrt{{{r}^{{{2}}}-{\frac{{{r}^{{{2}}}}}{{{4}}}}}}={r}\sqrt{{{3}}}\)
Also the height is \(\displaystyle{x}+{r}={\frac{{{r}}}{{{2}}}}+{r}={\frac{{{3}}}{{{2}}}}{r}\)
For Maximum Area: Base must be \(\displaystyle{r}\sqrt{{{3}}}\) and the height must be \(\displaystyle{\frac{{{3}}}{{{2}}}}{r}\)
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Elizabeth Witte
Answered 2021-11-22 Author has 642 answers

Step 1
The equation of circle will be
\(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={r}^{{{2}}}\)
image

Step 2
Now, area of triangle
\(\displaystyle{A}={\frac{{{1}}}{{{2}}}}{\left({2}{x}\right)}{\left({r}+{y}\right)}\)
\(\displaystyle={x}{\left({r}+{y}\right)}\)
\(\displaystyle{A}{\left({y}\right)}=\sqrt{{{r}^{{{2}}}-{y}^{{{2}}}}}{\left({r}+{y}\right)}\)
triangle derivative
\(\displaystyle{A}'{\left({y}\right)}=\sqrt{{{r}^{{{2}}}-{y}^{{{2}}}}}+{\left({r}+{y}\right)}\cdot{\frac{{-{2}{y}}}{{{2}\sqrt{{{r}^{{{2}}}-{y}^{{{2}}}}}}}}\)
\(\displaystyle={\frac{{{r}^{{{2}}}-{r}{y}-{2}{y}^{{{2}}}}}{{\sqrt{{{r}^{{{2}}}-{y}^{{{2}}}}}}}}\)
equating to zero
\(\displaystyle{\frac{{{r}^{{{2}}}-{r}{y}-{2}{y}^{{{2}}}}}{{\sqrt{{{r}^{{{2}}}-{y}^{{{2}}}}}}}}={0}\)
\(\displaystyle{r}^{{{2}}}-{r}{y}-{2}{y}^{{{2}}}={0}\)
\(\displaystyle{y}={\frac{{{r}\pm\sqrt{{{9}{r}^{{{2}}}}}}}{{-{4}}}}\Rightarrow{y}=-{r}\ {\quad\text{or}\quad}\ {y}={\frac{{{r}}}{{{2}}}}\)
rejecting, \(\displaystyle{y}=-{r}\)
Therefore, we check to see \(\displaystyle{y}={\frac{{{r}}}{{{2}}}}\) is maximum
\(\displaystyle{A}'{\left({y}\right)}{>}{0}\) for \(\displaystyle{y}{<}{\frac{{{r}}}{{{2}}}}\) and \(\displaystyle{A}'{\left({y}\right)}{<}{0}\) for \(\displaystyle{y}{>}{\frac{{{r}}}{{{2}}}}\)
\(\displaystyle\therefore{y}={\frac{{{r}}}{{{2}}}}\) is maximum
\(\displaystyle{x}^{{{2}}}+{\left({\frac{{{r}}}{{{2}}}}\right)}^{{{2}}}={r}^{{{2}}}\Rightarrow{x}={\frac{{\sqrt{{{3}}}}}{{{2}}}}{r}\)
Dimension of triangle with maximum area is base
\(\displaystyle={2}{x}={2}\cdot{\frac{{\sqrt{{{3}}}}}{{{2}}}}{r}=\sqrt{{{3}}}{r}\)
\(\displaystyle{h}={r}+{y}={r}+{\frac{{{r}}}{{{2}}}}={\frac{{{3}}}{{{2}}}}{r}\)
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