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

Let the observer stand at a distance $x$ from the wall. Then the angle $	heta$ subtended at the observer's eye by the painting is given by $	an 	heta = \frac{h}{d+x}$. We want to maximize $	heta$, so we want to maximize $	an 	heta$. This happens when $	heta$ is as big as possible, which happens when $	an 	heta$ is as big as possible.

Recall the formula $	an^2 	heta + 1 = \sec^2 	heta$. Since we want to maximize $	an 	heta$, we want to minimize $\sec 	heta$, and the minimum value of $\sec 	heta$ is $1$. Therefore, we want to minimize $	an^2 	heta$, which happens when $	heta$ is as close to $90^\circ$ as possible (since the tangent function goes to infinity as $	heta$ approaches $90^\circ$).

So, we want to maximize the denominator of the fraction $\frac{h}{d+x}$ while keeping $h$ fixed. This happens when $x$ is as small as possible, so the observer should stand as close to the wall as possible. Since the painting is hung a distance $d$ above the observer's eye, the distance from the observer to the painting is $d+x$, and we want to minimize this distance. The shortest distance between a point and a line is the perpendicular distance, so we want to stand directly in front of the painting, at a distance $x$ such that the height of the painting is equal to the distance from the observer's eye to the bottom of the painting. This distance is $d+h$, so we want to solve the equation $h = d+x$ for $x$, which gives $x = h-d$.

Therefore, the observer should stand a distance $x=h-d$ from the wall to get the best view. Option (D) gives this answer.

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