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Lagrange Multipliers Reference cheat sheet - grade college

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Calculus Grade college

Lagrange Multipliers Reference Cheat Sheet

A printable reference covering Lagrange multiplier equations, constraints, gradients, boundary checks, and second-order conditions for college.

TopicsCalculusDerivativesFunctionsVectorsStudy Guide
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This reference covers how to optimize a multivariable function subject to one or more constraints using Lagrange multipliers. Students need it because constrained optimization problems appear in calculus, economics, physics, engineering, and data science. The method replaces a difficult geometric condition with a system of equations involving gradients.

A clear step-by-step reference helps prevent algebra errors and missed candidates.

Key Facts

  • For one constraint g(x,y)=cg(x,y)=c, solve ∇f(x,y)=λ∇g(x,y)\nabla f(x,y)=\lambda \nabla g(x,y) together with g(x,y)=cg(x,y)=c.
  • For one constraint in 33 variables, solve ∇f(x,y,z)=λ∇g(x,y,z)\nabla f(x,y,z)=\lambda \nabla g(x,y,z) together with g(x,y,z)=cg(x,y,z)=c.
  • With two constraints, use ∇f=λ∇g+μ∇h\nabla f=\lambda \nabla g+\mu \nabla h together with g=cg=c and h=dh=d.
  • The Lagrangian for one constraint is L(x,y,λ)=f(x,y)−λ(g(x,y)−c)\mathcal{L}(x,y,\lambda)=f(x,y)-\lambda(g(x,y)-c).
  • Critical candidates satisfy ∂L∂x=0\frac{\partial \mathcal{L}}{\partial x}=0, ∂L∂y=0\frac{\partial \mathcal{L}}{\partial y}=0, and ∂L∂λ=0\frac{\partial \mathcal{L}}{\partial \lambda}=0 for a two-variable one-constraint problem.
  • The geometric meaning of ∇f=λ∇g\nabla f=\lambda \nabla g is that the level curve of ff is tangent to the constraint curve g=cg=c.
  • Lagrange multipliers require regular constraint points, meaning ∇g≠0⃗\nabla g \neq \vec{0} for a single constraint.
  • To find absolute extrema on a closed and bounded constraint set, evaluate ff at every valid candidate and compare the resulting values.

Vocabulary

Objective function
The objective function is the function ff that you want to maximize or minimize.
Constraint
A constraint is an equation such as g(x,y)=cg(x,y)=c that limits which input points are allowed.
Gradient
The gradient ∇f\nabla f is the vector of partial derivatives that points in the direction of greatest increase of ff.
Lagrange multiplier
A Lagrange multiplier such as λ\lambda is a scalar that relates the gradient of the objective function to the gradient of the constraint.
Lagrangian
The Lagrangian is an auxiliary function, often written L=f−λ(g−c)\mathcal{L}=f-\lambda(g-c), whose critical points encode the constrained optimization equations.
Regular point
A regular point on a constraint is a point where the constraint gradient is nonzero, such as ∇g≠0⃗\nabla g \neq \vec{0}.

Common Mistakes to Avoid

  • Forgetting to include the constraint equation is wrong because ∇f=λ∇g\nabla f=\lambda \nabla g alone does not guarantee that the point lies on g=cg=c.
  • Using ∇f=λg\nabla f=\lambda g instead of ∇f=λ∇g\nabla f=\lambda \nabla g is wrong because the multiplier must scale a vector, not the scalar constraint function.
  • Dividing by a variable that might be 00 can remove valid solutions, so always check cases such as x=0x=0 before simplifying.
  • Stopping after finding candidate points is wrong because absolute maxima and minima require comparing the values of ff at all valid candidates.
  • Ignoring boundary or singular cases is wrong because Lagrange multipliers apply cleanly only where the constraint is regular, such as ∇g≠0⃗\nabla g \neq \vec{0}.

Practice Questions

  1. 1 Use Lagrange multipliers to find the maximum and minimum of f(x,y)=xyf(x,y)=xy subject to x2+y2=10x^2+y^2=10.
  2. 2 Find the point on the plane x+2y+2z=9x+2y+2z=9 closest to the origin by minimizing f(x,y,z)=x2+y2+z2f(x,y,z)=x^2+y^2+z^2.
  3. 3 Maximize f(x,y)=3x+4yf(x,y)=3x+4y subject to the ellipse x2+4y2=16x^2+4y^2=16.
  4. 4 Explain why the condition ∇f=λ∇g\nabla f=\lambda \nabla g means the level curve of ff is tangent to the constraint curve g=cg=c at a regular constrained extremum.