17.59 min read

Solving with Lagrange — A Worked Example

Let us minimize $f(x,y) = xy^3 + x^2 - y$ subject to $g(x,y) = x^2 + y^2 = 5$ (a circle of radius $\sqrt{5}$ ).

The Jacobians are $Df(x,y) = [y^3 + 2x,\; 3xy^2 - 1]$ and $Dg(x,y) = [2x,\; 2y]$ . The Lagrange condition $\lambda Dg = Df$ gives two equations: $2\lambda x = y^3 + 2x$ and $2\lambda y = 3xy^2 - 1$ .

Assuming $x, y \neq 0$ : divide the first by $x$ and the second by $y$ to isolate $2\lambda$ . Equating gives $y^3/x + 2 = 3xy - 1/y$ — a curve. Intersect this with the constraint circle $x^2 + y^2 = 5$ to get candidate points.

Special cases ( $x = 0$ or $y = 0$ ) must be handled separately. For each candidate, evaluate $f$ to find the minimum. The key point: Lagrange finds a finite list of candidates; we just evaluate $f$ at each and pick the best.

Formal View

Example 17.1 — Lagrange Computation

Minimize

f(x,y) = xy^3 + x^2 - y

s.t.

g(x,y) = x^2 + y^2 = 5

. \n\nLagrange condition:

\lambda [2x, 2y] = [y^3 + 2x,\; 3xy^2 - 1]

. \n\nFor

x, y \neq 0

: divide to get

2\lambda = y^3/x + 2 = 3xy - 1/y

. \nCombined with

x^2 + y^2 = 5

, solve numerically for constrained Lagrange points. \n\nA computer can find these intersections; the minimum is the constrained Lagrange point with the smallest

f

value.

Remark 17.2 — Practical Solution Strategy

In general, simultaneously solving the Lagrange equations and constraint can be hard. Practical approaches: (1) parameterize the constraint surface and minimize

f

restricted to it, or (2) use numerical methods. The Lagrange framework tells us *where to look*, not always *how to find*.

Why This Matters

Solving constrained systems appears in physics, economics, engineering design, and statistical estimation. Lagrange is the standard first tool.

Finding eigenvalues of symmetric matrices (next section shows this)
Maximum likelihood estimation with constraints in statistics
Optimal control: trajectory optimization with dynamics constraints
Chemical equilibrium: minimize free energy subject to stoichiometric constraints

Learning Resources

Lagrange multipliers example

Khan Academy

Works through a complete Lagrange multiplier example with a circle constraint.

12 min

Lagrange multipliers introduction

3Blue1Brown

Derives and visualizes Lagrange multipliers geometrically.

13 min

Quiz

Question 1

To find constrained optima using Lagrange multipliers, you solve:

Question 2

After finding all constrained Lagrange points, how do you identify the minimum?

Question 3

Special cases like $x = 0$ or $y = 0$ must be checked separately in Lagrange problems.

Question 4

For $\min x + y$ s.t. $x^2 + y^2 = 1$ , the Lagrange condition gives:

Question 5

Minimize $f(x,y) = x + 2y$ s.t. $x^2 + y^2 = 1$ . The constrained minimum is at:

Common Mistakes

Forgetting the constraint $g(\mathbf{x}) = k$ when solving — the Lagrange equations alone are not enough.
Treating all Lagrange points as minima without evaluating and comparing $f$ values.
Dividing by zero when eliminating $\lambda$ — always check the special cases.