Lagrange multipliers
Objective. Solve smooth equality-constrained problems.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
Lagrange multipliers find stationary points of a function subject to equality constraints by introducing extra variables that balance gradients. To optimise f(x,y) subject to g(x,y)=0, we solve ∇f = λ∇g together with g=0, because at a constrained extremum the gradient of f must be parallel to the gradient of the constraint surface. Example: maximise f(x,y)=xy on x+y=10 gives ∇f=(y,x)=λ(1,1), so x=y, and with x+y=10 this gives x=y=5, product 25. The multiplier λ itself measures the sensitivity of the optimal value to relaxing the constraint.
Worked example
Use a Lagrange multiplier to find the maximum of f(x,y) = xy subject to x + 2y = 8.
- 1Form ∇f = (y,x) and ∇g = (1,2), so y = λ and x = 2λ
- 2Hence x = 2y
- 3Substitute into constraint: 2y + 2y = 8, giving y = 2 and x = 4
- 4Evaluate f(4,2) = 4×2 = 8
Answer: 8
Concept mastery
Based on your past attempts at this lesson — the question types to practise again come first.
Complete a practice set to see your concept breakdown here.
Practice
Practice happens on its own screen, one question at a time. Answers stay hidden until you submit yours, and you can stop and pick up at the same question later.
Start practisingCommon mistake. Students forget to include the constraint equation itself when solving the system, treating ∇f=λ∇g as the whole answer; without substituting g=0 the solution is underdetermined.
Your progress
Tick each step as you finish it. Your exact reading position is saved automatically, so Continue drops you back on the same line.
0 of 5 steps complete
Your place saves automatically as you read.