Lagrange Multipliers Calculator

The Lagrange multiplier method in a nutshell

Suppose you want to optimize a differentiable function f(x, y) subject to a differentiable constraint g(x, y) = 0. The method introduces an auxiliary variable λ and forms the Lagrangian

L(x, y, λ) = f(x, y) + λ g(x, y).

At a constrained extremum, the gradients of f and g are parallel. This leads to the system of equations

together with the original constraint g(x, y) = 0. Equivalently, we set the partial derivatives of the Lagrangian to zero:

• ∂L/∂x = 0
• ∂L/∂y = 0
• ∂L/∂λ = g(x, y) = 0

Solving this system gives candidate points that may be maxima, minima, or saddle points along the constraint curve.

How to use the calculator

Inputs

Fill in the fields using the variables x and y:

• Objective function f(x, y) – the function you want to maximize or minimize.
  Examples: x*y^2, x^2 + y^2, sin(x) + y^2.
• Constraint g(x, y) = 0 – write the left-hand side only; the calculator sets it equal to zero.
  Examples: x + y - 3, x^2 + y^2 - 1.
• Initial x, Initial y – optional starting guesses (seeds) for the numerical solver. Try different guesses to explore multiple solutions when they exist.

Acceptable syntax follows standard programming-style notation:

• Use * for multiplication (e.g., x*y).
• Use ^ for powers (e.g., y^2, x^3).
• Use parentheses to control order, e.g., (x - 1)^2 + (y + 2)^2.

After entering your expressions, click the Solve button. The tool uses symbolic differentiation and a numerical root-finder to solve the first-order conditions.

Worked example (matches the calculator)

Consider maximizing or minimizing f(x, y) = x*y^2 subject to the linear constraint g(x, y) = x + y - 3 = 0.

1. Enter the functions.
   In the calculator fields, type
   • f(x, y): x*y^2
   • g(x, y): x + y - 3
2. Choose initial guesses.
   For example, set Initial x = 1 and Initial y = 2, which already satisfy x + y ≈ 3.
3. Form the Lagrangian.
   L(x, y, λ) = x*y^2 + λ (x + y - 3).
4. Compute partial derivatives.
   • ∂L/∂x = y^2 + λ
   • ∂L/∂y = 2*x*y + λ
   • ∂L/∂λ = x + y - 3
   The calculator does this differentiation automatically.
5. Solve the system.
   Setting the partial derivatives equal to zero yields
   • y^2 + λ = 0
   • 2*x*y + λ = 0
   • x + y - 3 = 0
   Solving gives two candidate solutions on the line x + y = 3. One of them is (x*, y*) = (2, 1) with λ = -1. At this point, f(2, 1) = 2 * 1^2 = 2.
6. Interpret the result.
   If you evaluate f at other candidate points on the same constraint, you can see whether f(2, 1) is larger or smaller. Comparing values along the constraint allows you to classify each stationary point as a constrained maximum, constrained minimum, or saddle point.

In practice, the calculator performs steps 3–5 internally. You only need to provide the functions and, optionally, the initial guesses.

Interpreting the calculator output

A successful run typically returns:

• Stationary point (x*, y*) – the point on the constraint where the necessary conditions are satisfied.
• Lagrange multiplier λ – measures how sensitive the optimal value of f is to small changes in the constraint.
• Objective value f(x*, y*) – the value of f at the stationary point.

To decide whether each point is a maximum or a minimum along the constraint, you can:

• Compare f(x*, y*) across multiple solutions returned by the solver.
• Optionally test a few nearby points on the constraint curve and compare their f-values.
• For more advanced users, apply second-order conditions using the Hessian and the constraint geometry.

Remember that the method finds local stationary points. The largest f(x*, y*) among all feasible candidates is the constrained maximum, and the smallest is the constrained minimum on the specified constraint.

Comparison: manual method vs. calculator

Assumptions, limitations, and tips

To use this calculator effectively, keep these points in mind:

• Differentiability: Both f(x, y) and g(x, y) should be differentiable in the region of interest so that gradients exist.
• Single equality constraint: The current interface is designed for exactly one constraint of the form g(x, y) = 0 in two variables.
• Local vs. global extrema: Lagrange multipliers give necessary conditions for local extrema. The solver may find local maxima, minima, or saddle points, but it does not guarantee a global optimum unless you analyze the problem further.
• Gradient non-degeneracy: The standard theory assumes that ∇g(x, y) is nonzero at the solution. If the constraint gradient vanishes there, the method may fail or return misleading results.
• Numerical issues: For highly nonlinear or ill-conditioned problems, the numerical solver might not converge or may be sensitive to initial guesses. If a run fails, try different seeds or simplify the expressions.
• Checking work: For homework and exams, use the calculator to verify your analytical solution, not as a substitute for understanding the method. Show algebraic steps separately.

For deeper study, you may also want to review related topics such as gradients, unconstrained optimization, and second-order conditions for maxima and minima.