A $40$ feet high screen is put on a vertical wall $10$ feet above your eye-level. How far should you stand to maximize the angle subtended by the screen (from to bottom) at your eye?
My solution goes like this:
If the person is at a distance $x$ from the wall-screen arrangement, then we obtain a right-angled triangle of base $x$ feet and height $40+10$ feet. If the angle subtended from the screen to the person's eye level is $\theta$, then we have, $\tan\theta=\frac{50}{x}.$ Now, $\tan\theta$ is maximum as $\theta\to\frac{\pi}{2}$( since $\tan\theta\to\infty$ as $\theta\to\frac{\pi}{2}$) thus, $x=0.$
Now, I think something is wrong with my solution, but I just don't know the reason why? Is this solution valid? If not, where is it going wrong ?
Hint: Your method maximises the angle subtended by the screen and the 10 feet below. However, how much of that angle is due to the screen depends on distance.
To elaborate (see Empy's comment): If you're very close to the wall, you mostly see the ten feet of empty wall, because that's closer to your eyes, and a tiny strip of scree above that. When you're very far away, the height above your eye level doesn't matter much anymore, and the screen will be four times as large as the empty wall below; however, everything becomes smaller as you move away. Then, there is a sweet spot where the screen angle is maximal.