Chapter 4.6, Problem 21E

To determine

To calculate: The dimension of the isosceles triangle of largest area that can be inscribed in a circle of radius $r$

Answer to Problem 21E

The dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius $r$ are $\sqrt{3}r\text{and }\frac{r}{2}$

Explanation of Solution

Given information:

The isosceles triangle of largest area that can be inscribed in a circle of radius $r$

Formula used:

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

Let $2x$ be the base length of the isosceles triangle and $h$ be the height.

Area of right angle triangle is $A=\frac{1}{2}\times 2x\times h$

If the two function $f\left(x\right)\text{ and }g\left(x\right)$ are differentiable then the product is differentiable and ,

  ${\left(fg\right)}^{\prime }=f{g}^{\prime }+g{f}^{\prime }$

Quadratic formula: let the quadratic equation is $a{x}^2+bx+c=0,\left(a\ne 0\right)$ Thus, if $b^2-4ac\ge 0$ then the roots of the quadratic equation is given by

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

And

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3.   $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Calculation:

As per the given problem

Draw the diagram the isosceles triangle that can be inscribed in a circle of radius $r$

  

Recall that, Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is: ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

In right angle triangle $△ADO$

  $\begin{array}{l}{x}^2+y^2=r^2\\ \Rightarrow x^2=r^2-y^2\end{array}$

Take square root on both sides

  $x=\pm \sqrt{r^2-y^2}$

But length can’t be negative,

Therefore,

  $x=\sqrt{r^2-y^2}\mathrm{......}\left(1\right)$

Recall that, let $2x$ be the base length of the isosceles triangle and $h$ be the height.

Area of right angle triangle is $A=\frac{1}{2}\times 2x\times h\mathrm{......}\left(2\right)$

Substitute $x=\sqrt{r^2-y^2}$ , $h=y+r$ and simplified

  $\begin{array}{l}A\left(y\right)=\frac{1}{2}\times 2\left(\sqrt{r^2-y^2}\right)\times \left(y+r\right)\\ =\left(\sqrt{r^2-y^2}\right)\times \left(y+r\right)\end{array}$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Differentiate on both sides,

  $A^{\prime }\left(y\right)=\frac{d}{dy}\left\{\left(\sqrt{r^2-y^2}\right)\times \left(y+r\right)\right\}$

Recall that, if the two function $f\left(x\right)\text{ and }g\left(x\right)$ are differentiable then the product is differentiable and ,

  ${\left(fg\right)}^{\prime }=f{g}^{\prime }+g{f}^{\prime }$

  $\begin{array}{l}{A}^{\prime }\left(y\right)=\frac{d}{dy}\left\{\left(\sqrt{r^2-y^2}\right)\times \left(y+r\right)\right\}\\ =\left(\sqrt{r^2-y^2}\right)\times \frac{d}{dy}\left(y+r\right)+\left(y+r\right)\frac{d}{dy}\left(\sqrt{r^2-y^2}\right)\\ =\left(\sqrt{r^2-y^2}\right)\times 1+\left(y+r\right)\left\{\frac{1}{2\sqrt{r^2-y^2}}\times -2y\right\}\\ =\left(\sqrt{r^2-y^2}\right)-\left(y+r\right)\left\{\frac{y}{2\sqrt{r^2-y^2}}\right\}\\ =\left(\sqrt{r^2-y^2}\right)-\left\{\frac{y^2+r}{\sqrt{r^2-y^2}}\right\}\end{array}$

Solve for $A^{\prime }\left(x\right)=0$ , and simplified

  $\begin{array}{l}\left(\sqrt{r^2-y^2}\right)-\left\{\frac{y^2+ry}{\sqrt{r^2-y^2}}\right\}=0\\ \Rightarrow \sqrt{r^2-y^2}=\frac{y^2+ry}{\sqrt{r^2-y^2}}\\ \Rightarrow y^2+ry=r^2-y^2\\ \Rightarrow 2y^2+ry-r^2=0\end{array}$

Recall that, Quadratic formula: let the quadratic equation is $a{x}^2+bx+c=0,\left(a\ne 0\right)$ Thus, if $b^2-4ac\ge 0$ then the roots of the quadratic equation is given by

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

  $\begin{array}{l}y=\frac{-r\pm \sqrt{r^2-4\times 2\times -r^2}}{2\times 2}\\ =\frac{-r\pm \sqrt{r^2+8r^2}}{4}\\ =\frac{-r\pm \sqrt{9r^2}}{4}\\ =\frac{-r\pm 3r}{4}\end{array}$

Solve for different sign’s

  $\begin{array}{l}y=\frac{-r+3r}{4}\text{or}y=\frac{-r-3r}{4}\\ y=\frac{2r}{4}\text{or}y=\frac{-4r}{4}\\ y=\frac{r}{2}\text{or}y=-r\end{array}$

But length can’t be negative,

Therefore,

  $y=\frac{r}{2}$

Now, the height of the isosceles triangle:

  $h=y+r$

Substitute $y=\frac{r}{2}$ ,to get

  $\begin{array}{l}h=\frac{r}{2}+r\\ =\frac{r+2r}{2}\\ =\frac{3r}{2}\end{array}$

Substitute $y=\frac{r}{2}$ in equation $\left(1\right)$ , to get

  $\begin{array}{l}x=\sqrt{r^2-y^2}\\ =\sqrt{r^2-{\left(\frac{r}{2}\right)}^2}\\ =\sqrt{r^2-\frac{r^2}{4}}\\ =\sqrt{\frac{4r^2-r^2}{4}}\\ =\sqrt{\frac{3r^2}{4}}\\ =\frac{\sqrt{3}r}{2}\end{array}$

The base length of the isosceles triangle is $2x$

Therefore,

  $x=\sqrt{3}r$

Now, Substitute $y=\frac{r}{2}$ and $x=\sqrt{3}r$ in equation $\left(2\right)$ and simplified

  $\begin{array}{l}A=2\times \sqrt{3}r\times \frac{r}{2}\\ =\sqrt{3}{r}^2\end{array}$

Conclusion:

The dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius $r$ are $\sqrt{3}r\text{and }\frac{r}{2}$