Chapter 4.6, Problem 57E

To determine

To calculate: The point $P$ on the line segment $AB$ so as to maximize the angle $\theta$ .

Answer to Problem 57E

The point $P$ be chosen on the line segment from the point $B$ to $x=4.50$ unit.

Explanation of Solution

Given information:

A point $P$ on the line segment $AB$

Formula used:

Trigonometric ratio:

  

  $\tan \theta =\frac{p}{b}$

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

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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 of a point $P$ on the line segment $AB$

  

Form the given figure the triangle $△BPC$ makes the angle $\alpha$ and the triangle $△APD$ makes the angle $\beta$

Here, it is observed that

  $\theta =\pi -\alpha -\beta$

Recall that,

Trigonometric ratio: 5

  $\tan \theta =\frac{p}{b}$

  $\begin{array}{l}\tan \alpha =\frac{2}{x}\text{and}\tan \beta =\frac{5}{3-x}\\ \Rightarrow \alpha ={\tan }^{-1}\left(\frac{2}{x}\right)\text{and}\beta ={\tan }^{-1}\left(\frac{5}{3-x}\right)\end{array}$

  $\theta =\pi -{\tan }^{-1}\left(\frac{2}{x}\right)-{\tan }^{-1}\left(\frac{5}{3-x}\right)$

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 with respect to $x$

  $\begin{array}{l}{\theta }^{\prime }\left(x\right)=\pi -{\tan }^{-1}\left(\frac{2}{x}\right)-{\tan }^{-1}\left(\frac{5}{3-x}\right)\\ =-\left(\frac{1}{1+\frac{4}{x^2}}\times -\frac{2}{x^2}\right)-\left(\frac{1}{1+\frac{25}{{\left(3-x\right)}^2}}\times -\frac{5}{{\left(3-x\right)}^2}\right)\end{array}$

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

  $\begin{array}{l}-\left(\frac{1}{1+\frac{4}{x^2}}\times -\frac{2}{x^2}\right)-\left(\frac{1}{1+\frac{25}{{\left(3-x\right)}^2}}\times -\frac{5}{{\left(3-x\right)}^2}\right)=0\\ \Rightarrow \frac{2}{x^2+4}+\frac{5}{x^2-6x+34}=0\\ \Rightarrow \frac{5}{x^2-6x+34}=-\frac{2}{x^2+4}\\ \Rightarrow -2x^2+12x-68=5x^2+20\\ \Rightarrow 7x^2-12x-88=0\end{array}$

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

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

  $\begin{array}{l}x=\frac{+12\pm \sqrt{{\left(-12\right)}^2-4\times 7\times -88}}{2\times 7}\\ =\frac{12\pm \sqrt{144+2464}}{14}\\ =\frac{12\pm \sqrt{2608}}{14}\\ =\frac{12\pm 51.06}{14}\end{array}$

Solve for different sign’s

  $\begin{array}{l}x=\frac{12+51.06}{14}\text{and }x=\frac{12-51.06}{14}\\ =4.50\text{and }=-2.79\end{array}$

Because length can’t be negative,

Therefore,

  $x=4.50$

Conclusion:

Thus the point $P$ be chosen on the line segment from the point $B$ to $x=4.50$ unit.