Regiomontanus' angle maximization problem

The 15th-century German mathematician Johannes Müller, known as Regiomontanus, posed the following optimization problem:

The two dots at eye level are possible locations of the viewer's eye.

A painting hangs from a wall. Given the heights of the top and bottom of the painting above the viewer's eye level, how far from the wall should the viewer stand in order to maximize the angle subtended by the painting and whose vertex is at the viewer's eye?

If the viewer stands too close to the wall or too far from the wall, the angle is small; somewhere in between it is as large as possible. The same problem has also been applied to finding the optimal place from which to kick a ball in rugby.

Regiomontanus' solution by elementary geometry

There is a unique circle passing through the top and bottom of the painting and tangent to the eye-level line. By elementary geometry, if the viewer's position were to move along the circle, the angle subtended by the painting would remain constant. All positions on the eye-level line except the point of tangency are outside of the circle, and therefore the angle subtended by the painting from those points is smaller.

Solution by calculus

In the present day, this problem is widely known only because it appears as an exercise in many first-year calculus textbooks. For example: 

Let

a = the height of the bottom of the painting above eye level;

b = the height of the top of the painting above eye level;

x = the viewer's distance from the wall;

α = the angle of elevation of the bottom of the painting, seen from the viewer's position;

β = the angle of elevation of the top of the painting, seen from the viewer's position.

The angle we seek to maximize is β − α. The tangent of the angle is an increasing function of the angle; therefore it suffices to maximize

$\tan(\beta -\alpha )={\frac {\tan \beta -\tan \alpha }{1+\tan \beta \tan \alpha }}={\frac {{\frac {b}{x}}-{\frac {a}{x}}}{1+{\frac {b}{x}}\cdot {\frac {a}{x}}}}=(b-a){\frac {x}{x^{2}+ab}}.$

Since b − a is a positive constant, we only need to maximize the fraction that follows it. Differentiating, we get

${d \over dx}\,{\frac {x}{x^{2}+ab}}={\frac {ab-x^{2}}{(x^{2}+ab)^{2}}}\qquad {\begin{cases}{}>0&{\text{if }}0\leq x<{\sqrt {ab\,{}}},\\{}=0&{\text{if }}x={\sqrt {ab\,{}}},\\{}<0&{\text{if }}x>{\sqrt {ab\,{}}}.\end{cases}}$

Therefore the angle increases as x goes from 0 to √(ab) and decreases as x increases from √(ab). The angle is therefore as large as possible precisely when x = √(ab), the geometric mean of a and b.

Solution by algebra

We have seen that it suffices to maximize

${\frac {x}{x^{2}+ab}}.$

This is equivalent to minimizing the reciprocal:

${\frac {x^{2}+ab}{x}}=x+{\frac {ab}{x}}.$

Observe that this last quantity is equal to

$\left({\sqrt {x}}-{\sqrt {\frac {ab}{x}}}{}\right)^{2}+2{\sqrt {ab\,{}}}.$

(Click "show" at right to see the algebraic details or "hide" to hide them.)

Recall that u² − 2uv + v². Thus when we have u² + v², we can add the middle term −2uv to get a perfect square. We have

$x+{\frac {ab}{x}}.$

If we regard x as u² and ab/x as v², then u = √x and v = √(x/ab), and so

$2uv=2{\sqrt {x}}{\sqrt {\frac {ab}{x}}}=2{\sqrt {ab\,{}}}.$

Thus we have

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} x + \frac{ab}{x} & = u^2 + v^2 = \underbrace{\left(u^2 - 2uv + v^2)}_\text{a perfect square} + 2uv \\ & = (u - v)^2 + 2uv = \left( \sqrt{x} - \sqrt\frac{ab}{x} \right)^2 + 2\sqrt{ab\,{}}. \end{align} }

This is as small as possible precisely when the square is 0, and that happens when x = √(ab).