Chapter 4.6, Problem 58E

To determine

To calculate: the distance from the wall should the observer stand to get the best view.

Answer to Problem 58E

The observer stand to get the best view is $x=\sqrt{hd+d^2}$ .

Explanation of Solution

Given information:

A painting in an art gallery has height $h$ and is hung so that its lower edge is a distance $d$ above the eye of an observer

Formula used:

Trigonometric ratio:

  

  $\tan \theta =\frac{p}{b}$

Trigonometric formula:

  $\tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

Division Rule: If the two functions $f\left(x\right)\text{and }g\left(x\right)$ are differentiable then the quotient is differentiable and,

  ${\left(\frac{f}{g}\right)}^{\prime }=\frac{g{f}^{\prime }-f{g}^{\prime }}{g^2}$

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 painting in an art gallery has height $h$ and is hung so that its lower edge is a distance $d$ above the eye of an observer

  

Let $x$ denote the distance from the observer to the well, and let $\alpha$ be the angle right below $\theta$ . Then

Recall that,

Trigonometric ratio:

  $\tan \theta =\frac{p}{b}$

  $\tan \left(\alpha +\theta \right)=\frac{h+d}{x},\tan \alpha =\frac{d}{x}$

Recall that,

Trigonometric formula:

  $\tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

  $\begin{array}{l}\tan \left(\alpha +\theta \right)=\frac{h+d}{x}\\ \Rightarrow \frac{\tan \alpha +\tan \theta }{1-\tan \alpha \tan \theta }=\frac{h+d}{x}\\ \Rightarrow \frac{\frac{d}{x}+\tan \theta }{1-\frac{d}{x}\times \tan \theta }=\frac{h+d}{x}\\ \Rightarrow \tan \theta \left(x+\frac{hd}{x}+\frac{d^2}{x}\right)=h\\ \Rightarrow \tan \theta =\frac{hx}{x^2+hd+d^2}\end{array}$

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 with respect to $x$

  $\frac{d}{dx}\tan \theta =\frac{d}{dx}\left(\frac{hx}{x^2+hd+d^2}\right)$

Recall that,

Division Rule: If the two functions $f\left(x\right)\text{and }g\left(x\right)$ are differentiable then the quotient is differentiable and,

  ${\left(\frac{f}{g}\right)}^{\prime }=\frac{g{f}^{\prime }-f{g}^{\prime }}{g^2}$

  $\begin{array}{l}=\frac{\left(x^2+hd+d^2\right)\times h-hx\times 2x}{{\left(x^2+hd+d^2\right)}^2}\\ =\frac{h^{2}d+d^{2}h-h{x}^2}{{\left(x^2+hd+d^2\right)}^2}\end{array}$

Solve for $\frac{d}{dx}\tan \theta =0$ , and simplified

  $\begin{array}{l}\frac{h^{2}d+d^{2}h-h{x}^2}{{\left(x^2+hd+d^2\right)}^2}=0\\ \Rightarrow h^{2}d+d^{2}h-h{x}^2=0\\ \Rightarrow x^2=hd+d^2\end{array}$

Take square root on both sides,

  $x=\sqrt{hd+d^2}$

Conclusion:

Thus the observer stand to get the best view is $x=\sqrt{hd+d^2}$