Single Variable Calculus, Chapter 7, 7.6, Section 7.6, Problem 48

Suppose that a pointing in and out gallery has heigh $h$ and is hung so that its lower edge is a distance $d$ above the eye of an observer. How far from the wall should the observer stand to get the best view? In other words, where should the observer stand so as t maximize the angle $\theta$ subtended at her eyes by the pointing?




From the figure, we know that...

$\begin{equation} \begin{aligned} \alpha &= \beta + \beta\\ \\ \theta &= \alpha - \beta \qquad \Longleftarrow \text{(Equation 1)} \end{aligned} \end{equation}$


By using tangent function,
$\displaystyle \tan \alpha = \frac{h+d}{x}$ and $\displaystyle \tan \beta = \frac{d}{x}$
Thus,
$\displaystyle \alpha = \tan^{-1} \left[ \frac{h+d}{x} \right] \text{ and } \beta = \tan^{-1} \left[ \frac{d}{x} \right]$

From Equation 1,

$\begin{equation} \begin{aligned} \theta &= \alpha - \beta\\ \\ \theta &= \tan^{-1} \left[ \frac{h+d}{x} \right] - \tan^{-1} \left[ \frac{d}{x} \right] \end{aligned} \end{equation}$


If we need to maximize the angle, we set $\displaystyle \frac{d \theta}{dx} = 0$

$\begin{equation} \begin{aligned} \frac{d \theta}{dx} = 0 &= \frac{\frac{d}{dx}\left( \frac{h+d}{x} \right)}{1 + \left( \frac{h+d}{x} \right)^2} - \frac{\frac{d}{dx}\left( \frac{d}{x}\right)}{1 + \left( \frac{d}{x} \right)^2}\\ \\ 0 &= \frac{-\frac{(h+d)}{x^2}}{1 + \frac{(h+d)^2}{x^2}} - \frac{\left( - \frac{d}{x^2} \right)}{1 + \left( \frac{d^2}{x^2} \right)}\\ \\ 0 &= \frac{-\frac{(h+d)}{x^2}}{\frac{x^2+(+d)^2}{x^2}} - \frac{\frac{\left(- \frac{d}{x} \right)}{x^2 + d^2}}{x^2}\\ \\ 0 &= \frac{-(h+d)}{x^2 + (h + d)^2} + \frac{d}{x^2 + d^2}\\ \\ \frac{h+d}{x^2 + (h + d)^2} &= \frac{d}{x^2 + d^2}\\ \\ (h+d)\left( x^2 + d^2 \right) &= d \left[ x^2 + (h + d)^2 \right]\\ \\ hx^2 + hd^2 + dx^2 + d^3 &= d\left[ x^2 + h^2 + 2hd + d^2 \right]\\ \\ hx^2 + hd^2 + dx^2 + d^3 &= dx^2 + dh^2 + 2hd^2 + d^3\\ \\ hx^2 &= dh^2 + dh^2 \\ \\ hx^2 &= h \left( dh + d^2 \right)\\ \\ x^2 &= dh + d^2\\ \\ x &= \sqrt{dh + d^2} \end{aligned} \end{equation}$


It shows that the observer will get the best view when his distance from the base of the pointing is $x = \sqrt{dh + d^2}$