Recipe

Solve a constrained optimization problem with Lagrange multipliers

Use when: you need the extrema of $f(x,y)$ subject to a single equation constraint $g(x,y)=k$

Goal: every candidate point for the constrained extremum, with the true max and min picked out by comparing $f$ across all of them

Taught in Lagrange multipliers · a later section extends this recipe with more cases

Quick recipe

Set up $\nabla f=\lambda\nabla g$ as component equations
Eliminate $\lambda$ to relate $x$ and $y$ directly
Substitute back into the constraint to solve for the candidates
Evaluate $f$ at every candidate and compare

1
Compute $\nabla f=(f_x,f_y)$ and $\nabla g=(g_x,g_y)$ , then write out $f_x=\lambda g_x$ and $f_y=\lambda g_y$ alongside the constraint $g=k$ itself.

Example

Maximize $f(x,y)=x+2y$ subject to $g(x,y)=x^2+y^2=5$ . Here $\nabla f=(1,2)$ and $\nabla g=(2x,2y)$ , so the two component equations are $1=2\lambda x$ and $2=2\lambda y$ , together with $x^2+y^2=5$ .
- Not yet: a second, independent constraint is added at the same time — covered in Lagrange multipliers with two constraints
Why: a constrained extremum can only occur where the constraint curve is tangent to a level curve of $f$ , and tangent curves have parallel gradients, which is exactly what $\nabla f=\lambda\nabla g$ says.
2
Solve each equation for $\lambda$ and set the results equal, or divide one equation by the other, to get a relation between $x$ and $y$ alone.

Example

From $1=2\lambda x$ : $\lambda=\dfrac{1}{2x}$ (valid since $x=0$ would force $2(0)=2\lambda y$ , giving $y=0$ too, which fails the constraint). From $2=2\lambda y$ : $\lambda=\dfrac{1}{y}$ . Setting these equal: $\dfrac{1}{2x}=\dfrac1y$ , so $y=2x$ .
- neither variable is forced to be zero by the constraint — divide freely, since dividing by zero is never in play
- one coordinate could plausibly be zero — check that case directly in the original equations first, since dividing by a quantity that might be zero would silently throw the solution away
Why: $\lambda$ itself is almost never the quantity being asked for, and collapsing two equations into one relation between $x$ and $y$ is what makes the system solvable by substitution.
3
Plug the relation from Step 2 into $g=k$ and solve for the remaining variable, then recover every full point.

Example

Substituting $y=2x$ into $x^2+y^2=5$ : $x^2+4x^2=5x^2=5$ , so $x^2=1$ and $x=\pm1$ . With $y=2x$ , that gives the two candidates $(1,2)$ and $(-1,-2)$ .

Why: the constraint is the one equation not yet used, and it is exactly what pins down actual numbers instead of a whole family of $(x,y)$ pairs satisfying only the gradient condition.
4
Compute $f$ at each candidate point; the largest and smallest values are the constrained max and min, provided the constraint set is closed and bounded.

Example

$f(1,2)=1+2(2)=5$ and $f(-1,-2)=-1+2(-2)=-5$ . The circle $x^2+y^2=5$ is closed and bounded, so the multivariable extreme value theorem guarantees these two candidates hold the true extrema: maximum $5$ at $(1,2)$ , minimum $-5$ at $(-1,-2)$ .

Why: Lagrange's condition only narrows the search to candidates, the same way $f'=0$ only narrows a one-variable search; comparing actual $f$ -values is what turns a list of candidates into an answer.

Common mistakes

Dividing by a variable to eliminate $\lambda$ without first checking whether that variable could be zero on the constraint.
Solving $\nabla f=\lambda\nabla g$ and stopping there, reporting $\lambda$ or an $(x,y)$ pair without ever substituting into the constraint.
Forgetting to check that the constraint set is actually closed and bounded before trusting that the largest candidate value is the true maximum.
Mixing up which equation is $f_x=\lambda g_x$ and which is $f_y=\lambda g_y$ , and dividing them the wrong way around.