Strongest Signal on a Circular Track

A delivery robot must stay on a circular track, and it wants the spot with the best radio signal. It can't just climb the signal hill; it has to stay on the circle. The best spot is where the signal's contour line just touches the track.

Drag on the surface to move the point, or click a red marker to jump to it.

How it works

Lagrange multipliers

In one sentence

If $S$ has a local extremum on the curve $g = 0$ at a point where $\nabla g \ne \mathbf 0$ , then $\nabla S = \lambda\nabla g$ there for some number $\lambda$ .

General formula

{\htmlData{sym=0}{\nabla S}} = {\htmlData{sym=2}{\lambda}}\,{\htmlData{sym=1}{\nabla g}}, \qquad {\htmlData{sym=3}{g(x, y)}} = x^2 + y^2 - 4 = 0

Walking along the track you can't gain signal at the best point, so the gradient of $S$ has no component along the track; it must point straight across it, like $\nabla g$ .

What the symbols mean

$\nabla S$
Signal gradient
Points toward the tower, the fastest way the signal improves.
$\nabla g$
Track normal
Points straight out from the circular track.
$\lambda$
Multiplier
How strongly the track holds the robot back: the gain from a slightly bigger track.
$g(x, y)$
Constraint
Zero exactly on the track.

The track is the constraint curve $g = 0$ , a circle of radius 2; the signal $S(x, y)$ is strongest at the tower.
At the best point on the track, a contour of $S$ is tangent to the circle, so their normals line up: $\nabla S = \lambda\nabla g$ .
Together with $g = 0$ , that gives three equations for $x$ , $y$ and $\lambda$ .
$\lambda$ measures how much better the signal would get if the track grew a little.

Learn the math

Lagrange MultipliersCalc III · Lesson 29, Applications of Partial Derivatives