Lagrange Multipliers Optimizer

Math Calculus • Multivariable Calculus

Written by STEM Calculators Team Published December 30, 2025 Updated February 24, 2026

Objective \(f(x,y,z)\): Constraint \(g(x,y,z)\): Constraint value \(c\): Variables:

Solves constrained extrema using Lagrange multipliers: \(\nabla f = \lambda \nabla g\) with \(g=c\). Type pi, e, sqrt(2), powers like x^2. Graph shows contour map of \(f\) and the constraint curve \(g=c\) (2D). Drag to pan, wheel to zoom.

Search domain (box)

x min: x max: y min: y max: Unconstrained compare:

Solver & graph settings (optional)

Seeds per axis: Newton iterations: Equation tolerance: Dedup radius: Graph samples: Contour levels: Show grid: Shaded heatmap:

Ready

Enter \(f\), \(g\), \(c\), and a domain, then click “Calculate”.

Graph

Contours of \(f\) and constraint \(g=c\)

Drag to pan, mouse wheel to zoom, double-click to reset. Hover to read \((x,y)\), \(f\), and \(g\). The constraint curve \(g=c\) is drawn thicker; candidate points are marked and labeled.

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Theory

Lagrange multipliers (one constraint)

To optimize \(f(x,y,z)\) subject to a constraint \(g(x,y,z)=c\), we introduce a scalar multiplier \(\lambda\) and solve the system:

\[ \nabla f(x,y,z)=\lambda\,\nabla g(x,y,z), \qquad g(x,y,z)=c. \]

Intuition: on the constraint surface \(g=c\), the direction of steepest increase of \(f\) must be tangent to the constraint. At an optimum on the constraint, \(\nabla f\) is normal to the constraint surface, and \(\nabla g\) is also normal—so they are parallel.

This calculator solves the Lagrange system numerically (multi-seed Newton) and reports all distinct solutions found in the chosen domain.

2D case (x, y): the equation system

For \(f(x,y)\) with constraint \(g(x,y)=c\), the Lagrange conditions are:

\[ \begin{cases} f_x(x,y) - \lambda\,g_x(x,y)=0 \\ f_y(x,y) - \lambda\,g_y(x,y)=0 \\ g(x,y) - c = 0 \end{cases} \]

Solving gives candidate points on the constraint. You then evaluate \(f\) at each candidate to identify maxima and minima.

3D case (x, y, z): one constraint

For \(f(x,y,z)\) with one constraint \(g(x,y,z)=c\), you solve four equations in four unknowns \((x,y,z,\lambda)\):

\[ \begin{cases} f_x - \lambda\,g_x=0 \\ f_y - \lambda\,g_y=0 \\ f_z - \lambda\,g_z=0 \\ g - c = 0 \end{cases} \]

With multiple constraints \(g_1=c_1, g_2=c_2,\dots\), you use multiple multipliers \(\lambda_1,\lambda_2,\dots\) and solve \(\nabla f = \sum_k \lambda_k \nabla g_k\) plus all constraints.

Worked example: maximize \(f=xy\) on \(x^2+y^2=1\)

Let \(f(x,y)=xy\) and \(g(x,y)=x^2+y^2\) with \(c=1\).

\[ \nabla f = \langle y,\ x\rangle,\qquad \nabla g = \langle 2x,\ 2y\rangle. \]

Solve \(\nabla f=\lambda\nabla g\):

\[ y=2\lambda x,\qquad x=2\lambda y. \]

If \(x\neq 0\) and \(y\neq 0\), combine the equations to get \(y=2\lambda x\) and substitute into \(x=2\lambda y\):

\[ x=2\lambda(2\lambda x)=4\lambda^2 x \Rightarrow 4\lambda^2=1 \Rightarrow \lambda=\pm \tfrac12. \]

With \(\lambda=\tfrac12\), we have \(y=x\). With \(\lambda=-\tfrac12\), we have \(y=-x\). Enforce the constraint \(x^2+y^2=1\):

\[ y=\pm x \Rightarrow 2x^2=1 \Rightarrow x=\pm \frac{1}{\sqrt{2}},\quad y=\pm \frac{1}{\sqrt{2}}. \]

Evaluate \(f=xy\):

\[ f=\frac12 \text{ when } (x,y) \text{ have the same sign},\qquad f=-\frac12 \text{ when they have opposite signs}. \]

So the maximum is \(1/2\) at \(\left(\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}\right)\) and \(\left(-\tfrac{1}{\sqrt{2}},-\tfrac{1}{\sqrt{2}}\right)\), and the minimum is \(-1/2\) at \(\left(\tfrac{1}{\sqrt{2}},-\tfrac{1}{\sqrt{2}}\right)\) and \(\left(-\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}\right)\).

Notes and extensions (KKT preview idea)

Lagrange multipliers handle equality constraints \(g=c\). For inequality constraints (e.g., \(g\le c\)), the correct extension is the Karush–Kuhn–Tucker (KKT) conditions:

\[ \nabla f = \lambda \nabla g,\quad g\le c,\quad \lambda\ge 0,\quad \lambda\,(g-c)=0. \]

(This calculator currently focuses on equality constraints; a KKT preview can be added as an advanced mode.)

Frequently Asked Questions

What does the Lagrange multipliers method solve?

It finds candidate maxima and minima of an objective function f subject to an equality constraint g = c. The method solves ∇f = λ∇g together with the constraint to locate points on the constraint where f can be extremal.

Why do I need to set a search domain for the solver?

The calculator uses a multi-seed Newton method that searches for solutions within the bounds you provide. Restricting the domain helps locate relevant solutions and avoid missing or duplicating points in large regions.

What does the unconstrained compare option do?

When enabled, the tool also searches for unconstrained critical points by solving ∇f = 0. This helps compare constrained candidates on g = c with stationary points of f that may lie off the constraint.

How should I interpret the contour graph and the constraint curve?

The contour map shows level sets of f, while the thicker curve represents the constraint g = c. Candidate points are marked where the constraint intersects contours in a way consistent with tangency implied by ∇f being parallel to ∇g.