C.6 · Multivariable extrema and constrained optimisation
GRE Subject Test · GRE Mathematics Topic 17 2:55 English narration · English + 中文 subtitles burned in
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This unit classifies two-variable critical points with the Hessian, uses Lagrange multipliers with a regular constraint, and compares extrema and attainable values under regular constraints.
本单元用黑塞矩阵对二元函数的临界点分类，在正则约束下使用拉格朗日乘数法，并比较约束下的极值与可取到的值。
An interior local extremum has zero gradient, but that is not sufficient.
内部局部极值点的梯度为零，但这不充分。
With D equal to f x x times f y y minus f x y squared: positive D with positive f x x gives a minimum, with negative f x x a maximum, and negative D a saddle.
记 D = f_xx f_yy − f_xy²：D > 0 且 f_xx > 0 为极小值，f_xx < 0 为极大值，D < 0 为鞍点。
D equal to zero is inconclusive: x to the 4 plus y to the 4 and x to the 4 minus y to the 4 share a zero Hessian.
D = 0 时无法判断：x⁴ + y⁴ 与 x⁴ − y⁴ 的黑塞矩阵都为零。
On a smooth constraint with nonzero gradient, solve grad f equals lambda grad g with g equals c.
在梯度非零的光滑约束上，联立 ∇f = λ∇g 与 g = c。
These give candidates; they do not classify them.
这样得到的是候选点，并不对它们分类。
For x squared plus y squared equals a and x y equals b, the square x minus y squared equals a minus 2b forces a at least 2b, and explicit real values show that it is enough.
对 x² + y² = a 与 xy = b，(x − y)² = a − 2b 迫使 a ≥ 2b，而显式的实数解说明这个条件也是充分的。
A continuous function on a compact feasible set attains global extrema.
紧可行集上的连续函数能取到全局最值。
Compare all interior candidates with every boundary piece, including corners and endpoints.
把所有内部候选点与每段边界（包括角点和端点）一起比较。
Solving only the unconstrained gradient ignores boundary winners.
只解无约束梯度方程会漏掉边界上的最值。
Minimise x squared plus y squared subject to x+y=6.
在 x + y = 6 约束下求 x² + y² 的最小值。
The multiplier equations give x=y=3, with value 18: there the level circles just touch the line.
乘数方程给出 x = y = 3，值为 18：此处等值圆恰好与直线相切。
Completing the square proves a global minimum.
配方证明这是全局最小值。
There is no maximum on this unbounded line.
在这条无界直线上没有最大值。
A Lagrange multiplier solution is a candidate, not automatically a maximum or minimum.
拉格朗日乘数法的解只是候选点，不一定是最大值或最小值。
Check regularity and the complete feasible set.
要检查正则性和整个可行集。
For x squared minus y squared the Hessian has determinant minus 4, so the origin is a saddle.
对 x² − y²，黑塞行列式为 −4，所以原点是鞍点。
For x to the 4 plus y to the 4 the Hessian is zero and the test is inconclusive; but the function is never negative and is zero only at the origin, a strict global minimum.
对 x⁴ + y⁴，黑塞矩阵为零，判别法失效；但函数从不为负，且只在原点为零，是严格全局最小值。
The circle is compact and the constraint gradient never vanishes.
圆是紧的，约束梯度处处非零。
The multiplier equations force y equals plus or minus x, giving maximum 1 and minimum minus 1, confirmed by 2 x y at most 2.
乘数方程迫使 y = ±x，得最大值 1、最小值 −1，由 2|xy| ≤ 2 验证。
On the line x plus y equals 4, the minimum is 8, and there is no maximum.
在直线 x + y = 4 上，最小值是 8，没有最大值。
Check yourself.
自我检测。
Negative D means a saddle; zero D is inconclusive.
D < 0 是鞍点；D = 0 无法判断。
The minimum of x squared plus y squared on x plus y equals 6 is 18.
x² + y² 在 x + y = 6 上的最小值是 18。
The maximum of x y on the circle is 1.
xy 在圆上的最大值是 1。
And a multiplier solution is only a candidate.
乘数法的解只是候选点。