Question

A poster of height 36 in. is mounted on a wall so that its lower edge is 12 in. above the eye level of an observer. How far from the wall should the observer stand so that the viewing angle subtended at his eye by the poster is as large as possible?
use optimization to find the length. @Algebra

Answer 1

From the diagram you can use tan ratios to set up a relationship between x and theta
\[\tan a = \frac{12}{x}\]
\[\tan (a+\theta) = \frac{48}{x}\]
using the angle sum identity for tangent
\[\tan (a+\theta) =\frac{ \tan a + \tan \theta}{1-\tan a \tan \theta} = \frac{48}{x}\]
substitute in value for tan(a) and solve for tan(theta) in terms of x
i'll leave the algebra for you to work through
\[\tan \theta = \frac{36x}{x^{2}+576}\]
\[\theta = \tan^{-1} (\frac{36x}{x^{2}+576})\]
Now to maximize theta, differentiate and set equal to 0
\[\frac{d}{dx} \tan^{-1} u = \frac{1}{u^{2}+1}*\frac{du}{dx}\]
\[\frac{d \theta}{dx} = \frac{(x^{2}+576)^{2}}{(36x)^{2}+(x^{2}+576)^{2}}*\frac{(36)(576)-36x^{2}}{(x^{2}+576)^{2}} = 0\]
after simplifying, this can be reduced to
\[36x^{2} = (36)(576)\]
\[x = \sqrt{576} = 24\]
so the viewing angle is maximized when viewer is 24 in. away