C.6 · Multivariable extrema and constrained optimisation

Undergraduate GRE preparation. Local objectives within the reviewed ETS scope; this is original teaching, not an official test or score predictor.

Prerequisites: Partial derivatives, quadratic forms, compactness and constraints.

• Classify two-variable critical points with the Hessian
• Use Lagrange multipliers with a regular constraint
• Compare extrema and attainable values under regular constraints

Hessian 海森矩阵: The matrix of second partial derivatives used to study local curvature.

Lagrange multiplier 拉格朗日乘子: A scalar relating objective and regular constraint gradients at a constrained extremum.

For a differentiable function on an open domain, an interior local extremum has zero gradient. This condition is necessary, not sufficient. For two variables at a critical point, let D=f_xx f_yy−(f_xy)². If D>0, f_xx>0 gives a strict local minimum and f_xx<0 gives a strict local maximum; D<0 gives a saddle. These tests require suitable second-derivative regularity near the point.

When D=0 the second-derivative test is inconclusive, not evidence of a saddle. The functions x⁴+y⁴ and x⁴−y⁴ have the same zero Hessian at the origin, but one has a strict minimum and the other changes sign. Evaluate along contrasting directions or use a direct inequality. For a quadratic form, positive definiteness gives a more general Hessian interpretation.

On a smooth equality constraint g(x,y)=c with nonzero gradient g, solve gradient f=lambda gradient g together with g=c. These equations generate candidates; they do not classify or guarantee a global extremum. If the constraint gradient vanishes, the regularity condition fails and the multiplier equations can miss a constrained extremum. Treat such points separately. For x²+y²=a and xy=b>0, (x−y)²=a−2b shows necessity a≥2b. It is also sufficient: take s=√(a+2b), d=√(a−2b), x=(s+d)/2 and y=(s−d)/2. These real values have xy=b and x²+y²=a. A minimum condition alone needs this attainment check to establish solvability.

A continuous function on a compact feasible set attains global extrema. To find them, compare all interior candidates and all boundary pieces, including corners and endpoints. For a rectangle, optimise the restrictions on each edge. For a disk, the circular boundary may use a parameter or multiplier. Solving only the unconstrained gradient ignores possible boundary winners.

Minimise x²+y² subject to x+y=6. The constraint gradient is (1,1), so 2x=lambda and 2y=lambda. Thus x=y and the constraint gives x=y=3, with value 18. Completing the square on the line gives x²+(6−x)²=2(x−3)²+18, proving a global minimum. There is no maximum on this unbounded line, despite the existence of a multiplier candidate.

A Lagrange multiplier solution is a candidate, not automatically a maximum or minimum. Check regularity and the complete feasible set.

Classify the origin for $f(x,y)=x^2-y^2$ and for $g(x,y)=x^4+y^4$. State what a zero Hessian determinant permits you to conclude.

Worked solution

For f the Hessian is $\operatorname{diag}(2,-2)$ with determinant -4; the origin is a saddle. Along the axes the values have opposite signs. For g the Hessian at zero is zero, so the second-derivative test is inconclusive. Nevertheless $x^4+y^4\ge0$, with equality only at the origin; it is a strict global minimum. An inconclusive test is not a classification of the point.

Find all extrema of $f(x,y)=xy$ subject to $x^2+y^2=2$. Compare with minimising $x^2+y^2$ on the line $x+y=4$, including whether maxima exist.

Check after attempting

The circle is compact and its constraint gradient $(2x,2y)$ never vanishes there. Multipliers give $y=2\lambda x$, $x=2\lambda y$. Neither coordinate can be zero at a stationary point; hence $y=\pm x$. The four candidates $(1,1),(-1,-1),(1,-1),(-1,1)$ give global maximum 1 and minimum -1. The bound $2|xy|\le x^2+y^2=2$ checks completeness. On the line, $x^2+(4-x)^2=2(x-2)^2+8$, so the minimum is 8 at $(2,2)$. There is no maximum because the expression is unbounded. Multiplier candidates alone do not establish compactness.