A painting in an art gallery has height h and is hung so its 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?) (Your answer may depend on h and d.)

Question

A painting in an art gallery has height h and is hung so its 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?) (Your answer may depend on h and d.)

Answer 1

To find the distance from the wall where the observer should stand to get the best view, we need to maximize the angle θ subtended at the observer's eye by the painting.

Let's start by visualizing the situation. We have a painting hanging on a wall, with its lower edge d above the observer's eye. The height of the painting is h.

To maximize the angle θ, we want to find the position where the observer is farthest away from the wall while still being able to see the entire painting.

Let's consider a right triangle formed by the observer, the lower edge of the painting, and a point on the wall directly behind the observer. The distance from this point on the wall to the observer is the unknown distance we are looking for.

Using trigonometry, we can use the tangent function to relate the angle θ to the side lengths of the triangle:

tan(θ) = h / (d + x)

where x is the distance from the wall to the observer (which we are trying to find).

To maximize the angle θ, we need to find the value of x that makes tan(θ) as large as possible.

Note that θ must be acute (less than 90 degrees) for the observer to see the entire painting. Therefore, we are interested in the largest possible positive value of tan(θ).

Since tan(θ) is a increasing function in the range (-π/2, π/2), we can maximize θ by maximizing tan(θ).

To find the maximum value of tan(θ), we can take the derivative of tan(θ) with respect to x and set it to zero:

d/dx (tan(θ)) = 0

Simplifying this, we get:

(h / (d + x)^2) = 0

Solving this equation, we find that x = -d.

However, since x represents a distance, we cannot have a negative value. Therefore, we discard the solution x = -d.

So, the observer should stand a distance of d away from the wall to get the best view of the painting.