Question

a painting in an art gallery has height h and is hung so that 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 θ subtended at his eye by the painting?)

Explanation:

Step1: Define variables and use trigonometry

Let the distance from the wall be \(x\). Let \(\alpha\) be the angle from the eye - level to the lower edge of the painting and \(\beta\) be the angle from the eye - level to the upper edge of the painting. Then \(\tan\alpha=\frac{d}{x}\) and \(\tan\beta=\frac{h + d}{x}\). And \(\theta=\beta-\alpha\), so \(\tan\theta=\tan(\beta - \alpha)=\frac{\tan\beta-\tan\alpha}{1+\tan\beta\tan\alpha}\). Substituting \(\tan\alpha=\frac{d}{x}\) and \(\tan\beta=\frac{h + d}{x}\) into the formula, we get \(\tan\theta=\frac{\frac{h + d}{x}-\frac{d}{x}}{1+\frac{d(h + d)}{x^{2}}}=\frac{h}{x+\frac{d(h + d)}{x}}\).

Step2: Use the AM - GM inequality or calculus for maximization

Method 1: AM - GM inequality

For positive \(a=x\) and \(b = \frac{d(h + d)}{x}\), by the AM - GM inequality \(a + b\geq2\sqrt{ab}\). Here \(a=x\) and \(b=\frac{d(h + d)}{x}\), so \(x+\frac{d(h + d)}{x}\geq2\sqrt{d(h + d)}\). The equality holds when \(x=\frac{d(h + d)}{x}\).

Method 2: Calculus

Let \(y = \tan\theta=\frac{h}{x+\frac{d(h + d)}{x}}\), and we can also consider \(y\) as a function of \(x\). Let \(u=x+\frac{d(h + d)}{x}\), then \(y=\frac{h}{u}\). First, find the derivative of \(u\) with respect to \(x\): \(u^\prime=1-\frac{d(h + d)}{x^{2}}\). Set \(u^\prime = 0\) (since we want to minimize \(u\) to maximize \(y\) as \(y=\frac{h}{u}\)). Solving \(1-\frac{d(h + d)}{x^{2}}=0\) gives \(x^{2}=d(h + d)\).

Answer:

The observer should stand at a distance of \(\sqrt{d(d + h)}\) from the wall.