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Constrained Extrema and Lagrange Multipliers

Mathematics Algebra Difficulty 2/5

Beginning of the lesson

1. The Problem Reframed: Constrained vs. Free Extrema

Recall (Callback): In the previous topic you learned to find free extrema — the absolute or local maxima and minima of a function $f(x,y)$ with no restrictions on $(x,y)$. You set $\nabla f = \mathbf{0}$, classified critical points, and checked boundaries of closed regions. Now we add a new twist: what if you are forced to stay on a particular curve or surface?

A constrained extremum is the largest or smallest value a function $f$ attains not over the whole plane, but only among points that satisfy an extra equation $g(x,y)=c$. That equation is the constraint, and the set of points satisfying it — a curve in the plane, or a surface in space — is the constraint set.

Motivating Picture

Consider the plane $z = 2x + 3y$. As a whole it has no highest or lowest point — it extends to $\pm\infty$. But if we restrict attention to the circle $x^2+y^2=9$, then the plane's height on that circle does have a maximum and a minimum. Those are constrained extrema.

The Substitution Route — Honest Acknowledgement

Your first instinct might be: solve the constraint for one variable and substitute. If $g(x,y)=c$ can be rewritten as $y=h(x)$, then you reduce the problem to a single-variable calculus problem: extremize $\tilde{f}(x)=f(x,h(x))$.

When it works: For simple constraints (e.g., $2x+y=6 \Rightarrow y=6-2x$), substitution is fast and clean.

When it fails or hurts:

  • Unsolvable constraints: Many constraints cannot be solved explicitly for a variable. Try solving $x^3 + \sin(xy) = 1$ for $y$ — you can't.

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