Constrained extrema and Lagrange multipliers

Try

A rectangle has side lengths x,y≥0 and fixed perimeter 12. The constraint is x+y=6 and the area is f=xy. Substitution gives f=x(6−x)=9−(x−3)², so the maximum area is 9 at a square. This concrete calculation already illustrates constrained optimization: most directions in the plane are forbidden because they change the perimeter. The unconstrained gradient of area is not zero at the constrained maximum.

Predict

Predict how the gradient relates to the allowed direction. Along x+y=6, a tangent direction is (1,−1). At the square, ∇f=(3,3), perpendicular to that tangent. The constraint gradient ∇g=(1,1), for g=x+y, is also perpendicular. Thus ∇f=3∇g. The scalar 3 is a Lagrange multiplier; it records proportional normal directions, not a claim that either gradient vanishes.

Experiment

Solve the same problem with ∇f=λ∇g: y=λ, x=λ and x+y=6. These equations give x=y=3. Include the nonnegative boundary possibilities x=0 or y=0, which have zero area. The feasible segment is compact and the area continuous, so a global maximum exists. Comparing every candidate establishes the answer.

Now replace the perimeter by 2x+4y=12. The equations y=2λ and x=4λ yield x=3,y=3/2, area 9/2. More costly y-direction length changes the optimal shape. Verify the result by substituting y=3−x/2 into xy.

Understand

For a differentiable constraint g(x)=c, suppose ∇g is nonzero at a constrained local extremum. The level set is locally regular: allowed tangent directions are perpendicular to ∇g. At a constrained extremum, the derivative of f along every such tangent must vanish. Therefore ∇f is also normal to the tangent space and must be a scalar multiple of ∇g.

The multiplier equations are necessary under this regularity hypothesis. They are not sufficient, do not establish existence, and do not include irregular points automatically. For several equality constraints, the relevant regularity condition is independence of their gradients; then the objective gradient is a linear combination of those constraint gradients. Inequality constraints need additional active-boundary reasoning beyond this introductory equality method.

Apply

Examine f(x,y)=x subject to x²+y²=1. The equations (1,0)=λ(2x,2y) give y=0 and x=±1. Compactness ensures extrema exist; the values ±1 classify the two candidates. No parameter was divided away without checking whether it could vanish.

Now diagnose the invalid argument “if the multiplier equations have no solution, no optimum exists.” Under x²+y²=0 the only feasible point is the origin. Every objective has both a maximum and minimum on that one-point set, but the constraint gradient is zero there. The regularity condition fails. Singular feasible points must be examined directly, not discarded as impossible.

Continue

For optional depth, the multiplier can sometimes describe sensitivity of an optimal value to a constraint level, provided the optimum and multiplier vary smoothly and the chosen sign convention is tracked. This interpretation requires more than solving one system of equations; it is not a general shadow-price guarantee.

Save a constrained solution with four parts: feasible domain, regular candidates, exceptional or boundary cases, and the argument establishing global or merely local status. The constraint optimization investigation offers a bounded computational comparison. A successful numerical iteration suggests a feasible candidate; it does not replace the conditions and completeness reasoning developed here.