A painting in an art gallery has height $h$ and is hung so that lower edge is a distance $d$ above the eye of an observer (as in the figure). How far from the wall should the observer stand to get the best view? (In other words, where should the observer stand so as to maximize the angle $\theta$ subtended at his eye by the painting?) Select the correct answer.

A) $x = \sqrt { h ( d + h ) }$
B) $x = h ^ { 2 } ( h + d ) ^ { 2 }$
C) $x = d ^ { 2 } ( h + d ) ^ { 2 }$
D) $x = \sqrt { d ( h + d ) }$
E) $x = h ^ { 2 } ( h - d ) ^ { 2 }$

Question

A painting in an art gallery has height $h$ and is hung so that lower edge is a distance $d$ above the eye of an observer (as in the figure). How far from the wall should the observer stand to get the best view? (In other words, where should the observer stand so as to maximize the angle $\theta$ subtended at his eye by the painting?) Select the correct answer.

A) $x = \sqrt { h ( d + h ) }$ 
B) $x = h ^ { 2 } ( h + d ) ^ { 2 }$ 
C) $x = d ^ { 2 } ( h + d ) ^ { 2 }$ 
D) $x = \sqrt { d ( h + d ) }$ 
E) $x = h ^ { 2 } ( h - d ) ^ { 2 }$

Quizplus · Accepted Answer

We can use the tangent function to relate the angle $\theta$ to the distance $x$ from the observer to the wall: $\tan\theta = \frac{h}{x-d}$. To maximize $\theta$, we want to maximize the tangent, which occurs when the angle is 45 degrees (or $\frac{\pi}{4}$ radians). Thus, we have $\tan\frac{\pi}{4}=\frac{h}{x-d}$, so $x-d=h$ and $x=d+h$. Therefore, the correct answer is D, $x=\sqrt{d(h+d)}$.

A Painting in an Art Gallery Has Height hhh And Is Hung So That Lower Edge Is a Distance

A Painting in an Art Gallery Has Height $h$ And Is Hung So That Lower Edge Is a Distance