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?)

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?)

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's label the distance from the observer to the wall as $x$.

Using similar triangles, we can see that:

$$\tan(\theta) = \frac{h}{d+x}$$

To maximize $\theta$, we want to maximize $\tan(\theta)$. Recall that $\tan(\theta)$ increases as $\theta$ increases for angles between $0^\circ$ and $90^\circ$. Therefore, we want to maximize the value of $d+x$ in the denominator.

Rearranging the expression for $\tan(\theta)$, we get:

$$d+x = \frac{h}{\tan(\theta)}$$

Since we want to maximize $d+x$, we want to minimize $\tan(\theta)$. The minimum value of $\tan(\theta)$ is 0, which occurs when $\theta = 0^\circ$. However, this is not a realistic option since the observer must be able to see the painting. Therefore, we look for the positive value of $\theta$ that makes $\tan(\theta) = 0$ asymptotically.

That happens when $\theta = 90^\circ$, which is the maximum value for the angle that the viewer can observe the painting according to the figure.

Plugging this value into the expression we get:

$$d + x = \frac{h}{\tan(90^\circ)} = 0$$

Therefore, we want to choose the value of $x$ that minimizes $d+x$, which is $x = \sqrt{d(h+d)}$.

The answer is \boxed{\textbf{(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