You want the highest point, but you must stay on a path. Lagrange multipliers solve it with one picture and one equation.
Hook
Highest Point, Fixed Path
Find the highest point, but you must stay on a path. How?
- Maximize f
- Stay on the path
Setup
Contours of f
Draw the hill as contour lines. Higher rings mean higher ground.
- f(x, y) is height
- Rings are equal height
Constraint
The Constraint
Now add a rule. You may only walk along this curve.
- g(x, y) = 0
- Walk only on the curve
Sliding
Slide Along the Curve
Slide along it. While the curve crosses contours, you can still climb.
- Crossing contours
- f keeps rising
Tangency
Tangent at the Best Point
At the best point, the curve just touches one contour. They are tangent.
- Curve touches one contour
- No more climbing
Gradients
Two Gradients
The gradient of f points uphill, perpendicular to its contour. The gradient of g is perpendicular to the curve.
- grad f ⟂ contour
- grad g ⟂ constraint
Insight
Gradients Line Up
Tangent means the same perpendicular. So both gradients point along one line.
- Same tangent, same normal
- Parallel at the optimum
Equation
∇f = λ∇g
One equation captures it. Gradient of f equals lambda times gradient of g.
- One equation
- Lagrange condition
Payoff
Next: Lesson 37
Lambda just scales one arrow into the other. Lesson thirty-seven puts numbers to it. Follow for more.
- λ scales the arrow
- Worked numbers next