Following Problem, from Jim Fowler's Mooculus class:
A painting is mounted on a wall. The bottom of the painting is 5 feet above eye level, and the top of the painting is 14 feet above eye level. If you stand directly underneath the painting, you cannot see it at all. Similarly, if you stand very far away, you cannot see it very well. At some distance away from the wall, the angle subtended by the painting, whose vertex is your eye, is maximized. How far away must you stand from the wall to achieve this maximal viewing angle?
I draw this picture
And reason that
$$
\cos \theta = \dfrac{14}{\sqrt{5^2+x^2}}\\
\theta = \arccos{\dfrac{14}{\sqrt{5^2+x^2}}}\\
\dfrac{d\theta}{dx} = \dfrac{d}{dx}\arccos{\dfrac{14}{\sqrt{5^2+x^2}}}\\
\dfrac{d\theta}{dx} =\frac{14 x}{\sqrt{1 – \frac{196}{x^{2} + 25}} \left(x^{2} + 25\right)^{\frac{3}{2}}}
$$
I use sympy to get the last step. When I solve $\dfrac{d\theta}{dx}=0$, I only get $x=0$, which is obviously not the correct solution.
Tips?
Best Answer
Your expression for $\cos\theta$ isn't correct, since the cosine ratio holds only in right triangles.
Observe that the angle $\theta$ is given by
$$\theta=\tan^{-1}\left(\frac{19}{x}\right)-\tan^{-1}\left(\frac{5}{x}\right)=\tan^{-1}\left(\frac{14x}{x^2+95}\right)$$
Since we want to maximize, we want to find the values for which $\theta$ is maximized, or for which $$\tan\theta=\frac{14x}{x^2+95}$$ is maximized.
I am sure you can take it from here.