--- title: "Lagrange Multipliers" description: "Find constrained extrema using proportional gradients at regular points. Compare objective values and check endpoints or singular points separately." canonical: https://lightmysky.com/learn/mathematics/lagrange-multipliers-mt_0RPOBWZ1gs source: https://lightmysky.com/learn/mathematics/lagrange-multipliers-mt_0RPOBWZ1gs.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Lagrange Multipliers Find constrained extrema using proportional gradients at regular points. Compare objective values and check endpoints or singular points separately. Subject: Mathematics · Area: Calculus & Analysis · Ages 19 to 20 Page: https://lightmysky.com/learn/mathematics/lagrange-multipliers-mt_0RPOBWZ1gs ## Ready when they can - Set up and solve multiplier equations, checking regularity and exceptional feasible points - Explain regular tangency and interpret a multiplier using the stated constraint scale - Compare the multiplier method with substituting the constraint directly ## Lesson: Best point on a fixed curve You have solved gradient-zero equations to find unconstrained candidates. For smooth f and g, a local extremum on g=c with grad g nonzero satisfies grad f=lambda grad g. In two variables solve fx=lambda gx, fy=lambda gy and g=c together. These equations give candidates; they do not by themselves classify them. At a regular constraint point, grad g is normal to the constraint. At an extremum, the derivative of f along the constraint's tangent is zero. If grad f is also nonzero, the two level curves share a tangent. Their gradients are parallel, possibly in opposite directions. If grad f=0, lambda=0 satisfies the equation even when f has no regular level curve there. **Example.** For positive rectangle sides, maximise f=xy with g=2x+2y=P. The equations y=2lambda and x=2lambda give x=y=P/4. At P=24, the sides are 6 and the area is 36. Substitution gives xy=P²/16-(x-P/4)². The square is nonnegative, proving the maximum area is P²/16. The area tends to 0 as either side tends to 0. *(drawing: The multiplier equations find equal sides. The completed-square formula proves their area is largest.)* For f=ax+by on x²+y²=R², take R>0 and (a,b) nonzero. Solve a=2lambda x, b=2lambda y and the circle equation. Lambda cannot be zero. The two candidates are R(a,b)/sqrt(a²+b²) and its negative. Evaluate f at both. This smooth circle is closed and bounded, so its global maximum and minimum occur among these candidates. On the line x+y=b, substitute y=b-x. For f=x²+y², this gives x²+(b-x)²=2(x-b/2)²+b²/2. The square term is nonnegative, so the minimum is b²/2 at x=y=b/2. Multipliers give 2x=lambda and 2y=lambda, hence the same candidate. Completing the square proves it is a global minimum. On x²+y²=R², solving for y gives y=+sqrt(R²-x²) and y=-sqrt(R²-x²), with -R<=x<=R. A substitution solution must cover both branches and their endpoints. Multiplier equations can avoid this branch split. Neither method removes the need to check all feasible cases. *(drawing: The squared term cannot be negative. The least objective value occurs when this term is zero.)* For f=x²+y²+z² on ax+by+cz=d, with (a,b,c) nonzero, solve 2x=lambda a, 2y=lambda b and 2z=lambda c. The candidate is p=d(a,b,c)/(a²+b²+c²). Its value is d²/(a²+b²+c²). For any feasible point, expanding the sum of squared differences from p gives f minus this value. That sum is nonnegative, proving a global minimum. A smooth objective has global extrema on a closed, bounded constraint curve. If grad g is nonzero everywhere, each extremum satisfies the multiplier equations. Find every solution and compare objective values. Comparing lambda values does not classify them. The unconstrained Hessian test does not settle behaviour along the constraint. On an unbounded feasible set, also examine behaviour at infinity. Check endpoints, nonsmooth points and points with grad g=0 separately. On x+y=b with x,y>=0 and b>0, f=x has its minimum at endpoint (0,b). For g=(x-1)²+(y+1)²=0, the only feasible point is (1,-1). There grad g=0 and f=x has both extrema, but no lambda satisfies (1,0)=lambda(0,0). *(drawing: With x,y nonnegative, the endpoints give the minimum and maximum of f=x.)* Fix g=c and use grad f=lambda grad g. If a regular optimal point varies smoothly with c, its value V(c) satisfies V'(c)=lambda. A small level change h changes V by about lambda h. Lambda can be negative, so this need not be an improvement. Rewriting the same constraint as 2g=2c doubles its gradient and halves its multiplier. State the constraint's scale when interpreting lambda. **Recap.** Solve the gradient equations with the constraint. Compare objective values, check exceptional points, and interpret lambda using the stated constraint scale. *Lesson, questions, worked solutions and flashcards written for this dataset. External records were studied for mathematical facts and question design.* ## Practice 14 questions on this page, each with its working shown. ## Needs first - [Critical Points and Optimisation in Two Variables](https://lightmysky.com/learn/mathematics/critical-points-and-optimisation-in-two-variables-mt_4qB5pAGeuR) ## Optional connections - [Directional Derivatives and the Gradient](https://lightmysky.com/learn/mathematics/directional-derivatives-and-the-gradient-mt_LW_KckY5Ad)