Companion to the University-Style Practice Set大学风格练习题配套解答
Sections 1 to 7: critical points, the second-derivative test, absolute extrema on closed regions, Lagrange multipliers (one and two constraints), applications第 1 至 7 节：极值点、二阶导数判别法、闭区域上的绝对极值、拉格朗日乘数法（单约束与双约束）及应用CALC III
$f(x,y)=x^{3}+y^{3}-3x-12y+20$: (a) compute $f_x,f_y$ and write the system; (b) solve for all critical points; (c) apply the second-derivative test to classify each.(a) 计算 $f_x,f_y$ 并写出方程组；(b) 求所有极值候选点；(c) 应用二阶导数判别法对每个点分类。
Differentiate with respect to each variable: (M1)分别对每个变量求偏导数：(M1)
$$ f_x = 3x^{2}-3, \qquad f_y = 3y^{2}-12. $$Setting $f_x=0$ and $f_y=0$ gives the independent equations $3x^{2}-3=0$ and $3y^{2}-12=0$. (A1)令 $f_x=0$ 且 $f_y=0$，得到相互独立的方程 $3x^{2}-3=0$ 与 $3y^{2}-12=0$。(A1)
From $3x^{2}=3$ we get $x=\pm 1$, and from $3y^{2}=12$ we get $y=\pm 2$. (M1) Since the equations in $x$ and $y$ decouple, all four combinations are critical points:由 $3x^{2}=3$ 得 $x=\pm 1$，由 $3y^{2}=12$ 得 $y=\pm 2$。(M1) 由于 $x$ 和 $y$ 的方程互相独立，所有四种组合均为极值候选点：
$$ (1,2),\quad (1,-2),\quad (-1,2),\quad (-1,-2). \quad\text{(A1)}$$Compute the second partial derivatives: $f_{xx}=6x$, $f_{yy}=6y$, $f_{xy}=0$. The discriminant is (M1)计算二阶偏导数：$f_{xx}=6x$，$f_{yy}=6y$，$f_{xy}=0$。判别式为 (M1)
$$ D = f_{xx}f_{yy} - f_{xy}^{2} = (6x)(6y) - 0 = 36xy. $$Evaluate at each point: (A1)在每个点处求值：(A1)
$g(x,y)=x^{3}+y^{3}-3xy$: (a) find all critical points; (b) compute $D$ at each; (c) classify each point.(a) 求所有极值候选点；(b) 在每个点处计算 $D$；(c) 对每个点分类。
$g_x=3x^{2}-3y=0$ and与 $g_y=3y^{2}-3x=0$. (M1)
From $g_x=0$: $y=x^{2}$. Substitute into $g_y=0$: $3(x^{2})^{2}-3x=3x^{4}-3x=3x(x^{3}-1)=0$. (M1)由 $g_x=0$ 得 $y=x^{2}$。代入 $g_y=0$：$3(x^{2})^{2}-3x=3x^{4}-3x=3x(x^{3}-1)=0$。(M1)
So $x=0$ or $x=1$. When $x=0$: $y=0$. When $x=1$: $y=1$. The critical points are $(0,0)$ and $(1,1)$. (A1)故 $x=0$ 或 $x=1$。$x=0$ 时 $y=0$；$x=1$ 时 $y=1$。极值候选点为 $(0,0)$ 和 $(1,1)$。(A1)
$g_{xx}=6x$, $g_{yy}=6y$, $g_{xy}=-3$. Hence $D=g_{xx}g_{yy}-g_{xy}^{2}=(6x)(6y)-9=36xy-9$. (M1)故 $D=g_{xx}g_{yy}-g_{xy}^{2}=(6x)(6y)-9=36xy-9$。(M1)
(M1) Applying the second-derivative test: (A1)(M1) 应用二阶导数判别法：(A1)
Note that $D=0$ does not occur here; had it occurred at any point, the test would be inconclusive and higher-order analysis (or restriction to curves through the point) would be required.本题中 $D=0$ 的情形不出现；若某点处 $D=0$，则判别法失效，需要高阶分析（或沿过该点的曲线进行限制分析）。
$h(x,y)=x^{2}-xy+y^{2}-2x$ on the closed triangular region $R$ with vertices $(0,0)$, $(4,0)$, $(0,4)$: (a) interior critical points; (b) extreme values on each edge; (c) absolute max and min on $R$.在顶点为 $(0,0)$、$(4,0)$、$(0,4)$ 的闭三角形区域 $R$ 上：(a) 求内部极值候选点；(b) 求各边上的极值；(c) 求 $R$ 上的绝对最大值和最小值。
$h_x=2x-y-2=0$ and与 $h_y=-x+2y=0$. From $h_y=0$: $x=2y$. (M1) Substitute: $2(2y)-y-2=3y-2=0$, so $y=\tfrac{2}{3}$ and $x=\tfrac{4}{3}$. The interior critical point is $\bigl(\tfrac{4}{3},\tfrac{2}{3}\bigr)$, which lies inside $R$ since $\tfrac{4}{3}+\tfrac{2}{3}=2<4$. (A1)由 $h_y=0$ 得 $x=2y$。(M1) 代入：$2(2y)-y-2=3y-2=0$，故 $y=\tfrac{2}{3}$，$x=\tfrac{4}{3}$。内部极值候选点为 $\bigl(\tfrac{4}{3},\tfrac{2}{3}\bigr)$，由于 $\tfrac{4}{3}+\tfrac{2}{3}=2<4$，该点在 $R$ 内部。(A1)
$$ h\!\left(\tfrac{4}{3},\tfrac{2}{3}\right)=\tfrac{16}{9}-\tfrac{8}{9}+\tfrac{4}{9}-\tfrac{8}{3}=\tfrac{12}{9}-\tfrac{8}{3}=\tfrac{4}{3}-\tfrac{8}{3}=-\tfrac{4}{3}. $$Edge 1:边 1： $y=0$, $0\le x\le 4$. Then $h=x^{2}-2x$, so $\tfrac{d}{dx}(x^{2}-2x)=2x-2=0$ at $x=1$. (M1) The value is $h(1,0)=1-2=-1$.此时 $h=x^{2}-2x$，令 $\tfrac{d}{dx}(x^{2}-2x)=2x-2=0$，得 $x=1$。(M1) 取值 $h(1,0)=1-2=-1$。
Edge 2:边 2： $x=0$, $0\le y\le 4$. Then $h=y^{2}$, so $\tfrac{d}{dy}(y^{2})=2y=0$ only at $y=0$ (an endpoint). (M1) No interior critical point on this edge.此时 $h=y^{2}$，$\tfrac{d}{dy}(y^{2})=2y=0$ 仅在 $y=0$（端点）处成立。(M1) 该边无内部极值候选点。
Edge 3:边 3（斜边）： Hypotenuse $x+y=4$. Parametrize $x=t$, $y=4-t$, $0\le t\le 4$. (A1)斜边 $x+y=4$。参数化：$x=t$，$y=4-t$，$0\le t\le 4$。(A1)
$$ h(t,4-t)=t^{2}-t(4-t)+(4-t)^{2}-2t=t^{2}-4t+t^{2}+16-8t+t^{2}-2t=3t^{2}-14t+16. $$Setting $\tfrac{d}{dt}(3t^{2}-14t+16)=6t-14=0$ gives $t=\tfrac{7}{3}$, so the critical point on Edge 3 is $\bigl(\tfrac{7}{3},\tfrac{5}{3}\bigr)$. (A1)令 $\tfrac{d}{dt}(3t^{2}-14t+16)=6t-14=0$，得 $t=\tfrac{7}{3}$，故边 3 上的极值候选点为 $\bigl(\tfrac{7}{3},\tfrac{5}{3}\bigr)$。(A1)
$$ h\!\left(\tfrac{7}{3},\tfrac{5}{3}\right)=3\cdot\tfrac{49}{9}-14\cdot\tfrac{7}{3}+16=\tfrac{49}{3}-\tfrac{98}{3}+\tfrac{48}{3}=-\tfrac{1}{3}. $$Vertices: $h(0,0)=0$; $h(4,0)=16-8=8$; $h(0,4)=16$. (M1) Collect all candidate values:顶点处：$h(0,0)=0$；$h(4,0)=16-8=8$；$h(0,4)=16$。(M1) 汇总所有候选值：
| Point点 | $h$ value值 | Source来源 |
|---|---|---|
| $(\tfrac{4}{3},\tfrac{2}{3})$ | $-\tfrac{4}{3}\approx -1.33$ | Interior CP内部极值候选点 |
| $(1,0)$ | $-1$ | Edge 1 CP边 1 极值候选点 |
| $(\tfrac{7}{3},\tfrac{5}{3})$ | $-\tfrac{1}{3}\approx -0.33$ | Edge 3 CP边 3 极值候选点 |
| $(0,0)$ | $0$ | Vertex顶点 |
| $(4,0)$ | $8$ | Vertex顶点 |
| $(0,4)$ | $16$ | Vertex顶点 |
Absolute minimum:绝对最小值： $-\tfrac{4}{3}$ at $\bigl(\tfrac{4}{3},\tfrac{2}{3}\bigr)$ (interior). Absolute maximum: $16$ at $(0,4)$ (vertex). (A1)在 $\bigl(\tfrac{4}{3},\tfrac{2}{3}\bigr)$ 处（内部点）。绝对最大值：$16$ 在 $(0,4)$ 处（顶点）。(A1)
$\phi(x,y)=3x+4y$ on the closed disk $x^{2}+y^{2}\le 25$: (a) no interior CPs; (b) extreme values on the boundary circle; (c) absolute max and min.在闭圆盘 $x^{2}+y^{2}\le 25$ 上：(a) 无内部极值候选点；(b) 在边界圆上的极值；(c) 绝对最大值和最小值。
$\phi_x=3$ and $\phi_y=4$ are never simultaneously zero. (M1) A linear function has no stationary points, so $\phi$ attains its extreme values on the closed disk only on the boundary circle $x^{2}+y^{2}=25$. (A1)$\phi_x=3$ 与 $\phi_y=4$ 不能同时为零。(M1) 线性函数无驻点，故 $\phi$ 在闭圆盘上的极值只能在边界圆 $x^{2}+y^{2}=25$ 上取得。(A1)
Parametrize: $x=5\cos t$, $y=5\sin t$, $t\in[0,2\pi)$. Then $\phi=15\cos t+20\sin t$. (M1)参数化：$x=5\cos t$，$y=5\sin t$，$t\in[0,2\pi)$。则 $\phi=15\cos t+20\sin t$。(M1)
$$ \frac{d\phi}{dt}=-15\sin t+20\cos t=0\implies\tan t=\frac{4}{3}. $$In $[0,2\pi)$ this gives $\sin t=\tfrac{4}{5},\cos t=\tfrac{3}{5}$ (maximum) and $\sin t=-\tfrac{4}{5},\cos t=-\tfrac{3}{5}$ (minimum).在 $[0,2\pi)$ 上，解为 $\sin t=\tfrac{4}{5},\cos t=\tfrac{3}{5}$（最大值）和 $\sin t=-\tfrac{4}{5},\cos t=-\tfrac{3}{5}$（最小值）。
Maximum point: $(5\cdot\tfrac{3}{5},5\cdot\tfrac{4}{5})=(3,4)$, giving $\phi=9+16=25$.最大值点：$(5\cdot\tfrac{3}{5},5\cdot\tfrac{4}{5})=(3,4)$，$\phi=9+16=25$。
Minimum point: $(-3,-4)$, giving $\phi=-9-16=-25$. (A1)最小值点：$(-3,-4)$，$\phi=-9-16=-25$。(A1)
Since $\phi$ has no interior extrema, the absolute maximum is $25$ at $(3,4)$ and the absolute minimum is $-25$ at $(-3,-4)$, both on the boundary. (M1·A1)由于 $\phi$ 无内部极值，绝对最大值为 $25$，在 $(3,4)$ 处取得；绝对最小值为 $-25$，在 $(-3,-4)$ 处取得，均在边界上。(M1·A1)
Let $f,g:\mathbb{R}^{2}\to\mathbb{R}$ be smooth and let $\mathbf{x}^{*}$ be a constrained extremum of $f$ subject to $g=c$ with $\nabla g(\mathbf{x}^{*})\ne\mathbf{0}$. (a) Show $\nabla g\perp$ every constraint-tangent vector; (b) deduce $\nabla f=\lambda\nabla g$; (c) interpret geometrically.设 $f,g:\mathbb{R}^{2}\to\mathbb{R}$ 光滑，$\mathbf{x}^{*}$ 是 $f$ 在约束 $g=c$ 下的极值点，且 $\nabla g(\mathbf{x}^{*})\ne\mathbf{0}$。(a) 证明 $\nabla g$ 与每个约束切向量正交；(b) 推导 $\nabla f=\lambda\nabla g$；(c) 给出几何解释。
Let $\mathbf{r}(t)$ be any smooth curve in $\mathbb{R}^{2}$ with $g(\mathbf{r}(t))=c$ for all $t$ and $\mathbf{r}(0)=\mathbf{x}^{*}$. Differentiate the identity $g(\mathbf{r}(t))=c$ with respect to $t$ using the chain rule: (M1)设 $\mathbf{r}(t)$ 是 $\mathbb{R}^{2}$ 中满足对所有 $t$ 有 $g(\mathbf{r}(t))=c$ 且 $\mathbf{r}(0)=\mathbf{x}^{*}$ 的任意光滑曲线。利用链式法则对恒等式 $g(\mathbf{r}(t))=c$ 关于 $t$ 求导：(M1)
$$ \nabla g(\mathbf{r}(t))\cdot\mathbf{r}'(t)=0. $$At $t=0$ this gives $\nabla g(\mathbf{x}^{*})\cdot\mathbf{r}'(0)=0$. (A1) Since $\mathbf{r}$ was an arbitrary smooth curve on the constraint, $\mathbf{r}'(0)$ is an arbitrary vector tangent to the level set $g=c$ at $\mathbf{x}^{*}$; the equation says $\nabla g(\mathbf{x}^{*})$ is orthogonal to every such tangent vector. (R1)在 $t=0$ 处得 $\nabla g(\mathbf{x}^{*})\cdot\mathbf{r}'(0)=0$。(A1) 由于 $\mathbf{r}$ 是约束上的任意光滑曲线，$\mathbf{r}'(0)$ 是等位集 $g=c$ 在 $\mathbf{x}^{*}$ 处的任意切向量；该方程表明 $\nabla g(\mathbf{x}^{*})$ 与每个这样的切向量正交。(R1)
Since $\mathbf{x}^{*}$ is an extremum of $f$ restricted to $g=c$, the directional derivative of $f$ in any direction $\mathbf{v}$ tangent to the constraint must vanish: $\nabla f(\mathbf{x}^{*})\cdot\mathbf{v}=0$ for every constraint-tangent $\mathbf{v}$. (M1)由于 $\mathbf{x}^{*}$ 是 $f$ 限制在 $g=c$ 上的极值点，$f$ 沿任意约束切线方向 $\mathbf{v}$ 的方向导数必须为零：对每个约束切向量 $\mathbf{v}$，$\nabla f(\mathbf{x}^{*})\cdot\mathbf{v}=0$。(M1)
In $\mathbb{R}^{2}$, the tangent space to a smooth curve at $\mathbf{x}^{*}$ is one-dimensional (assuming $\nabla g\ne\mathbf{0}$). Both $\nabla f(\mathbf{x}^{*})$ and $\nabla g(\mathbf{x}^{*})$ are orthogonal to this one-dimensional tangent space, so they must be parallel. (M1) Therefore there exists a scalar $\lambda$ such that在 $\mathbb{R}^{2}$ 中，光滑曲线在 $\mathbf{x}^{*}$ 处的切空间是一维的（假设 $\nabla g\ne\mathbf{0}$）。$\nabla f(\mathbf{x}^{*})$ 和 $\nabla g(\mathbf{x}^{*})$ 都与这个一维切空间正交，故它们必须平行。(M1) 因此存在标量 $\lambda$，使得
$$ \nabla f(\mathbf{x}^{*}) = \lambda\,\nabla g(\mathbf{x}^{*}). \quad\text{(A1)}$$The condition $\nabla f=\lambda\nabla g$ says the gradients are parallel, which means the gradient of $f$ points in the same (or opposite) direction as the gradient of $g$. Since the gradient of each function is perpendicular to its own level curves, two gradients being parallel implies the two families of level curves are tangent to each other at $\mathbf{x}^{*}$. (A1)条件 $\nabla f=\lambda\nabla g$ 表明梯度平行，即 $f$ 的梯度与 $g$ 的梯度方向相同或相反。由于每个函数的梯度都与其等位曲线垂直，两个梯度平行意味着两族等位曲线在 $\mathbf{x}^{*}$ 处相切。(A1)
At a generic point on the constraint the level curve of $f$ crosses the constraint curve transversally, so moving along the constraint changes $f$; at a constrained extremum the level curve of $f$ can only touch (be tangent to) the constraint, which is exactly when no motion along the constraint produces a first-order change in $f$. (R1)在约束上的一般点处，$f$ 的等位曲线横截约束曲线，故沿约束移动会改变 $f$；在约束极值点处，$f$ 的等位曲线只能与约束相切，恰好是沿约束的运动不产生 $f$ 的一阶变化的情形。(R1)
Let $(a,b)$ be a critical point of $f$ with Hessian $H$ and $D=\det H=f_{xx}f_{yy}-f_{xy}^{2}$. (a) Write the Taylor expansion and identify the controlling quadratic form; (b) show $D>0$ forces $Q$ to be definite; (c) explain the saddle case $D<0$ and the inconclusive case $D=0$.设 $(a,b)$ 是 $f$ 的极值候选点，黑塞矩阵为 $H$，$D=\det H=f_{xx}f_{yy}-f_{xy}^{2}$。(a) 写出泰勒展开式并识别控制符号的二次型；(b) 证明 $D>0$ 使 $Q$ 定号；(c) 解释鞍点情形 $D<0$ 与失效情形 $D=0$。
Since $(a,b)$ is a critical point, $f_x(a,b)=f_y(a,b)=0$. The second-order Taylor expansion about $(a,b)$ is: (M1)由于 $(a,b)$ 是极值候选点，$f_x(a,b)=f_y(a,b)=0$。在 $(a,b)$ 处的二阶泰勒展开式为：(M1)
$$ f(a+h,b+k)=f(a,b)+\underbrace{f_x\cdot h+f_y\cdot k}_{=\,0}+\tfrac{1}{2}\bigl(f_{xx}h^{2}+2f_{xy}hk+f_{yy}k^{2}\bigr)+O\!\left(h^{2}+k^{2}\right)^{3/2}. $$The sign of $f(a+h,b+k)-f(a,b)$ near $(a,b)$ is therefore controlled (for small $(h,k)$) by the quadratic form (A1)故 $(a,b)$ 附近 $f(a+h,b+k)-f(a,b)$ 的符号（对较小的 $(h,k)$）由二次型控制 (A1)
$$ Q(h,k) = f_{xx}h^{2}+2f_{xy}hk+f_{yy}k^{2}. $$If $Q>0$ for all $(h,k)\ne(0,0)$, the function is larger than $f(a,b)$ in every direction, so $(a,b)$ is a local min; if $Q<0$ everywhere, it is a local max; if $Q$ takes both signs, it is a saddle. (R1)若对所有 $(h,k)\ne(0,0)$ 均有 $Q>0$，函数在每个方向上都大于 $f(a,b)$，故 $(a,b)$ 为极小值；若处处 $Q<0$，则为极大值；若 $Q$ 既取正值又取负值，则为鞍点。(R1)
Assume $f_{xx}\ne 0$ (if $f_{xx}=0$ the argument is symmetric in $h,k$). Complete the square in $h$: (M1)设 $f_{xx}\ne 0$（若 $f_{xx}=0$，则论证关于 $h,k$ 对称）。对 $h$ 配方：(M1)
$$ Q = f_{xx}\!\left(h+\frac{f_{xy}}{f_{xx}}k\right)^{2}+\left(f_{yy}-\frac{f_{xy}^{2}}{f_{xx}}\right)k^{2} = f_{xx}\!\left(h+\frac{f_{xy}}{f_{xx}}k\right)^{2}+\frac{D}{f_{xx}}k^{2}. $$(M1) If $D>0$, then $\tfrac{D}{f_{xx}}$ has the same sign as $f_{xx}$. Both terms in the completed-square form are non-negative (when $f_{xx}>0$) or non-positive (when $f_{xx}<0$), and at least one is strictly so for $(h,k)\ne(0,0)$. Therefore: (A1)(M1) 若 $D>0$，则 $\tfrac{D}{f_{xx}}$ 与 $f_{xx}$ 同号。配方后的两项在 $f_{xx}>0$ 时均非负、在 $f_{xx}<0$ 时均非正，且对 $(h,k)\ne(0,0)$ 至少有一项严格成立。因此：(A1)
(M1) If $D<0$, then $\tfrac{D}{f_{xx}}$ has the opposite sign from $f_{xx}$. Setting $k=0$ shows the first term of the completed-square form can be positive, while choosing $k\ne 0$ and $h=-\tfrac{f_{xy}}{f_{xx}}k$ makes the first term vanish, leaving a term of opposite sign. Hence $Q$ takes both signs, confirming a saddle point.(M1) 若 $D<0$，则 $\tfrac{D}{f_{xx}}$ 与 $f_{xx}$ 异号。令 $k=0$，配方后第一项可为正；令 $k\ne 0$ 且 $h=-\tfrac{f_{xy}}{f_{xx}}k$，则第一项消失，留下异号项。故 $Q$ 既取正值又取负值，确认为鞍点。
If $D=0$, the completed square has $\tfrac{D}{f_{xx}}=0$: one squared term vanishes entirely, leaving $Q=f_{xx}\bigl(h+\tfrac{f_{xy}}{f_{xx}}k\bigr)^{2}$, which is zero along the line $h=-\tfrac{f_{xy}}{f_{xx}}k$. The second-order information alone cannot distinguish a flat direction from a local extremum or saddle along that direction; higher-order terms must be examined. The test is genuinely inconclusive. (A1)若 $D=0$，配方后 $\tfrac{D}{f_{xx}}=0$：一个平方项完全消失，剩余 $Q=f_{xx}\bigl(h+\tfrac{f_{xy}}{f_{xx}}k\bigr)^{2}$，沿直线 $h=-\tfrac{f_{xy}}{f_{xx}}k$ 为零。仅凭二阶信息无法区分平坦方向与沿该方向的极值或鞍点，必须检验高阶项。判别法真正失效。(A1)
(a) Optimize $f=xy$ subject to $x+y=1$ with full case analysis; (b) state the two-constraint Lagrange condition and count equations and unknowns; (c) write the Lagrange system for $f=x+y+z$ on the intersection of $x^{2}+y^{2}=2$ and $z=xy$.(a) 对 $f=xy$ 在约束 $x+y=1$ 下进行完整的分情况讨论求解；(b) 写出双约束拉格朗日条件并计算方程数与未知数数；(c) 写出 $f=x+y+z$ 在 $x^{2}+y^{2}=2$ 与 $z=xy$ 交线上的拉格朗日方程组。
$\nabla f=(y,x)$ and与 $\nabla g=(1,1)$. The Lagrange system is: (M1)拉格朗日方程组为：(M1)
$$ y=\lambda,\quad x=\lambda,\quad x+y=1. $$Case $\lambda=0$:情形 $\lambda=0$： Then $y=0$ and $x=0$, but $0+0=0\ne 1$, contradicting the constraint. There is no solution with $\lambda=0$. (M1)则 $y=0$，$x=0$，但 $0+0=0\ne 1$，与约束矛盾。$\lambda=0$ 时无解。(M1)
Case $\lambda\ne 0$:情形 $\lambda\ne 0$： The first two equations give $x=y=\lambda$. Substituting into the constraint: $2\lambda=1$, so $\lambda=\tfrac{1}{2}$. (A1) The unique critical point is $\bigl(\tfrac{1}{2},\tfrac{1}{2}\bigr)$, at which $f=\tfrac{1}{4}$. (A1)前两个方程给出 $x=y=\lambda$。代入约束：$2\lambda=1$，故 $\lambda=\tfrac{1}{2}$。(A1) 唯一极值候选点为 $\bigl(\tfrac{1}{2},\tfrac{1}{2}\bigr)$，此处 $f=\tfrac{1}{4}$。(A1)
(By the extreme value theorem on the closed segment this is the global maximum on $\{x+y=1,\,x,y\ge 0\}$, but on all of the constraint line $f$ is unbounded above.)（根据闭区间上的极值定理，这是 $\{x+y=1,\,x,y\ge 0\}$ 上的全局最大值，但在整条约束线上 $f$ 无上界。）
For $g_1(\mathbf{x})=c_1$ and $g_2(\mathbf{x})=c_2$ in $\mathbb{R}^{3}$, the condition at a constrained extremum is (M1)对于 $\mathbb{R}^{3}$ 中的 $g_1(\mathbf{x})=c_1$ 和 $g_2(\mathbf{x})=c_2$，约束极值点处的条件为 (M1)
$$ \nabla f = \lambda_1\nabla g_1 + \lambda_2\nabla g_2. $$Geometrically: the intersection of the two constraint surfaces is a curve in $\mathbb{R}^{3}$; its tangent direction at $\mathbf{x}^{*}$ is perpendicular to both $\nabla g_1$ and $\nabla g_2$. At an extremum, $\nabla f$ is also perpendicular to the curve's tangent, hence $\nabla f$ lies in the plane spanned by $\nabla g_1$ and $\nabla g_2$, which is exactly the condition above. (A1)从几何角度看：两个约束曲面的交线是 $\mathbb{R}^{3}$ 中的一条曲线；其在 $\mathbf{x}^{*}$ 处的切线方向与 $\nabla g_1$ 和 $\nabla g_2$ 都垂直。在极值点处，$\nabla f$ 也与曲线切线垂直，故 $\nabla f$ 落在 $\nabla g_1$ 和 $\nabla g_2$ 所张成的平面内，这正是上述条件。(A1)
The full system consists of 3 gradient equations ($\nabla f=\lambda_1\nabla g_1+\lambda_2\nabla g_2$) plus 2 constraint equations ($g_1=c_1$, $g_2=c_2$), giving 5 equations in 5 unknowns ($x,y,z,\lambda_1,\lambda_2$). (R1)完整方程组由 3 个梯度方程（$\nabla f=\lambda_1\nabla g_1+\lambda_2\nabla g_2$）加上 2 个约束方程（$g_1=c_1$，$g_2=c_2$）组成，共5 个方程，5 个未知数（$x,y,z,\lambda_1,\lambda_2$）。(R1)
Take $g_1=x^{2}+y^{2}-2$ and $g_2=z-xy$. Then: (M1)取 $g_1=x^{2}+y^{2}-2$，$g_2=z-xy$。则：(M1)
$\nabla f=(1,1,1)$, $\nabla g_1=(2x,2y,0)$, $\nabla g_2=(-y,-x,1)$. The three gradient equations are: (M1)三个梯度方程为：(M1)
$$ 1 = 2\lambda_1 x - \lambda_2 y,\quad 1 = 2\lambda_1 y - \lambda_2 x,\quad 1 = \lambda_2, $$together with the constraint equations $x^{2}+y^{2}=2$ and $z=xy$. This is 5 equations in 5 unknowns $(x,y,z,\lambda_1,\lambda_2)$. (A1)加上约束方程 $x^{2}+y^{2}=2$ 和 $z=xy$，共 5 个方程，5 个未知数 $(x,y,z,\lambda_1,\lambda_2)$。(A1)
Alternative approach: the constraint $x^{2}+y^{2}=2$ is a cylinder, so parametrize $x=\sqrt{2}\cos\theta$, $y=\sqrt{2}\sin\theta$, $z=xy=2\sin\theta\cos\theta=\sin 2\theta$, reducing $f=x+y+z$ to a single-variable function of $\theta$ that can be differentiated directly. This bypasses the multipliers but requires a parametrization to be recognizable.另一种方法：约束 $x^{2}+y^{2}=2$ 是柱面，参数化为 $x=\sqrt{2}\cos\theta$，$y=\sqrt{2}\sin\theta$，$z=xy=2\sin\theta\cos\theta=\sin 2\theta$，将 $f=x+y+z$ 化为关于 $\theta$ 的单变量函数，可直接求导。此法绕过乘数，但需能识别出合适的参数化。
Box with $V=8xyz$ inscribed in $\tfrac{x^{2}}{4}+\tfrac{y^{2}}{9}+z^{2}=1$, $x,y,z>0$: (a) state the Lagrange system; (b) solve for the optimal dimensions; (c) state the maximum volume and justify it.体积 $V=8xyz$ 的长方体内接于 $\tfrac{x^{2}}{4}+\tfrac{y^{2}}{9}+z^{2}=1$，$x,y,z>0$：(a) 写出拉格朗日方程组；(b) 求最优尺寸；(c) 给出最大体积并说明理由。
Objective: maximize $V=8xyz$. Constraint: $g(x,y,z)=\tfrac{x^{2}}{4}+\tfrac{y^{2}}{9}+z^{2}-1=0$. (M1)目标：最大化 $V=8xyz$。约束：$g(x,y,z)=\tfrac{x^{2}}{4}+\tfrac{y^{2}}{9}+z^{2}-1=0$。(M1)
$\nabla V=(8yz,\,8xz,\,8xy)$ and与 $\nabla g=\bigl(\tfrac{x}{2},\,\tfrac{2y}{9},\,2z\bigr)$. Setting $\nabla V=\lambda\nabla g$ gives: (A1)令 $\nabla V=\lambda\nabla g$ 得：(A1)
$$ 8yz = \tfrac{\lambda x}{2},\quad 8xz = \tfrac{2\lambda y}{9},\quad 8xy = 2\lambda z. \quad\text{(A1)}$$Since $x,y,z>0$, multiply each gradient equation by the matching variable and use the constraint. From the first equation: $16yz\cdot x=\lambda x^{2}$, giving $\lambda=16yz/x$. Similarly from the second: $\lambda=36xz/y$. (M1)由于 $x,y,z>0$，将每个梯度方程乘以对应的变量并利用约束条件。由第一个方程：$16yz\cdot x=\lambda x^{2}$，得 $\lambda=16yz/x$。同理由第二个方程：$\lambda=36xz/y$。(M1)
Equating the two expressions for $\lambda$: $16yz/x=36xz/y$, and since $z>0$ we cancel $z$: $16y^{2}=36x^{2}$, so $y/x=3/2$, i.e. $y=\tfrac{3x}{2}$. (M1)令两个 $\lambda$ 的表达式相等：$16yz/x=36xz/y$，由于 $z>0$ 可约去 $z$：$16y^{2}=36x^{2}$，故 $y/x=3/2$，即 $y=\tfrac{3x}{2}$。(M1)
From the first and third expressions: $16yz/x=4xy/z$, cancel $y>0$: $16z^{2}=4x^{2}$, so $z=x/2$. (A1)由第一个和第三个表达式：$16yz/x=4xy/z$，约去 $y>0$：$16z^{2}=4x^{2}$，故 $z=x/2$。(A1)
Substitute $y=\tfrac{3x}{2}$ and $z=\tfrac{x}{2}$ into the constraint:将 $y=\tfrac{3x}{2}$ 和 $z=\tfrac{x}{2}$ 代入约束：
$$ \frac{x^{2}}{4}+\frac{(3x/2)^{2}}{9}+\left(\frac{x}{2}\right)^{2}=\frac{x^{2}}{4}+\frac{9x^{2}/4}{9}+\frac{x^{2}}{4}=\frac{x^{2}}{4}+\frac{x^{2}}{4}+\frac{x^{2}}{4}=\frac{3x^{2}}{4}=1. $$So $x^{2}=\tfrac{4}{3}$, $x=\tfrac{2}{\sqrt{3}}$, $y=\sqrt{3}$, $z=\tfrac{1}{\sqrt{3}}$. (A1) Verify: $\tfrac{4/3}{4}+\tfrac{3}{9}+\tfrac{1}{3}=\tfrac{1}{3}+\tfrac{1}{3}+\tfrac{1}{3}=1$. Constraint satisfied.故 $x^{2}=\tfrac{4}{3}$，$x=\tfrac{2}{\sqrt{3}}$，$y=\sqrt{3}$，$z=\tfrac{1}{\sqrt{3}}$。(A1) 验证：$\tfrac{4/3}{4}+\tfrac{3}{9}+\tfrac{1}{3}=\tfrac{1}{3}+\tfrac{1}{3}+\tfrac{1}{3}=1$。约束满足。
(M1) $V=8xyz=8\cdot\tfrac{2}{\sqrt{3}}\cdot\sqrt{3}\cdot\tfrac{1}{\sqrt{3}}=8\cdot\tfrac{2}{\sqrt{3}}=\dfrac{16}{\sqrt{3}}=\dfrac{16\sqrt{3}}{3}$. (A1)
This critical point is a maximum rather than a minimum because: as $x,y,z\to 0^{+}$ (any side collapses) the volume $V=8xyz\to 0$, while as the point approaches any face of the ellipsoid the volume also tends to zero (the ellipsoid is bounded). The Extreme Value Theorem guarantees a maximum exists on the compact set, and the unique interior critical point (with $x,y,z>0$) must be it. (R1)此极值候选点是极大值而非极小值：当 $x,y,z\to 0^{+}$（任一边长趋于零）时，$V=8xyz\to 0$；当点趋近椭球面的任一面时，体积也趋于零（椭球面有界）。极值定理保证紧集上存在最大值，而唯一的内部极值候选点（满足 $x,y,z>0$）必然是它。(R1)
Minimize $d^{2}=x^{2}+y^{2}+z^{2}$ subject to $2x+2y+z=9$ using Lagrange multipliers: (a) state the system; (b) solve; (c) compute the distance and confirm via the formula.用拉格朗日乘数法在约束 $2x+2y+z=9$ 下最小化 $d^{2}=x^{2}+y^{2}+z^{2}$：(a) 写出方程组；(b) 求解；(c) 计算距离并用公式验证。
Minimizing $d^{2}=x^{2}+y^{2}+z^{2}$ subject to $g=2x+2y+z-9=0$ is equivalent to minimizing $d$ because the square root is monotone increasing on $[0,\infty)$; squaring avoids a radical in the gradient calculation. (M1)在约束 $g=2x+2y+z-9=0$ 下最小化 $d^{2}=x^{2}+y^{2}+z^{2}$ 与最小化 $d$ 等价，因为平方根在 $[0,\infty)$ 上单调递增；平方处理避免了梯度计算中出现根号。(M1)
$\nabla(d^{2})=(2x,2y,2z)$ and与 $\nabla g=(2,2,1)$. Setting $\nabla(d^{2})=\lambda\nabla g$: (A1)令 $\nabla(d^{2})=\lambda\nabla g$：(A1)
$$ 2x=2\lambda,\quad 2y=2\lambda,\quad 2z=\lambda, $$together with the constraint $2x+2y+z=9$. (R1)加上约束 $2x+2y+z=9$。(R1)
From the gradient equations: $x=\lambda$, $y=\lambda$, $z=\tfrac{\lambda}{2}$. (M1) Substitute into the constraint:由梯度方程：$x=\lambda$，$y=\lambda$，$z=\tfrac{\lambda}{2}$。(M1) 代入约束：
$$ 2\lambda+2\lambda+\tfrac{\lambda}{2}=\tfrac{9\lambda}{2}=9\implies\lambda=2. $$(A1) The closest point is $x=2$, $y=2$, $z=1$. (A1) Verify: $2(2)+2(2)+1=4+4+1=9$. Constraint satisfied.(A1) 最近点为 $x=2$，$y=2$，$z=1$。(A1) 验证：$2(2)+2(2)+1=4+4+1=9$。约束满足。
$d^{2}=4+4+1=9$, so $d=3$. (M1)故 $d=3$。(M1)
The point-to-plane distance formula from the origin $(0,0,0)$ to $2x+2y+z=9$ is:从原点 $(0,0,0)$ 到平面 $2x+2y+z=9$ 的点到平面距离公式为：
$$ d=\frac{|2(0)+2(0)+(0)-9|}{\sqrt{2^{2}+2^{2}+1^{2}}}=\frac{9}{\sqrt{9}}=\frac{9}{3}=3. $$Both methods agree: minimum distance is $3$. (A1)两种方法结果一致：最短距离为 $3$。(A1)
Extrema of $f=x^{2}+y^{2}+z^{2}$ on the curve of intersection of $x^{2}+y^{2}+z^{2}+2z=0$ and $x+y+z=0$: (a) state both constraints and the Lagrange system; (b) show $x=y$; (c) solve for the candidate points; (d) evaluate $f$ and state the extrema.求 $f=x^{2}+y^{2}+z^{2}$ 在 $x^{2}+y^{2}+z^{2}+2z=0$ 与 $x+y+z=0$ 交线上的极值：(a) 写出两个约束和拉格朗日方程组；(b) 证明 $x=y$；(c) 求候选点；(d) 计算 $f$ 并给出极值。
Rewrite: $g_1=x^{2}+y^{2}+z^{2}+2z=0$ (sphere, equivalently $x^{2}+y^{2}+(z+1)^{2}=1$) and $g_2=x+y+z=0$ (plane through the origin). (M1)改写：$g_1=x^{2}+y^{2}+z^{2}+2z=0$（球面，等价为 $x^{2}+y^{2}+(z+1)^{2}=1$）和 $g_2=x+y+z=0$（过原点的平面）。(M1)
$\nabla f=(2x,2y,2z)$, $\nabla g_1=(2x,2y,2z+2)$, $\nabla g_2=(1,1,1)$. The condition $\nabla f=\lambda_1\nabla g_1+\lambda_2\nabla g_2$ yields: (A1)条件 $\nabla f=\lambda_1\nabla g_1+\lambda_2\nabla g_2$ 给出：(A1)
$$ 2x=2\lambda_1 x+\lambda_2,\quad 2y=2\lambda_1 y+\lambda_2,\quad 2z=\lambda_1(2z+2)+\lambda_2, $$together with $g_1=0$ and $g_2=0$, giving 5 equations in 5 unknowns $(x,y,z,\lambda_1,\lambda_2)$. (A1)加上 $g_1=0$ 和 $g_2=0$，共 5 个方程，5 个未知数 $(x,y,z,\lambda_1,\lambda_2)$。(A1)
From equation (i): $\lambda_2=2x(1-\lambda_1)$. From equation (ii): $\lambda_2=2y(1-\lambda_1)$. (M1) Setting these equal:由方程 (i)：$\lambda_2=2x(1-\lambda_1)$。由方程 (ii)：$\lambda_2=2y(1-\lambda_1)$。(M1) 令两者相等：
$$ 2x(1-\lambda_1)=2y(1-\lambda_1). $$If $\lambda_1\ne 1$, we may cancel $(1-\lambda_1)$ to get $x=y$. (A1)若 $\lambda_1\ne 1$，可约去 $(1-\lambda_1)$ 得 $x=y$。(A1)
If $\lambda_1=1$: equation (i) gives $2x=2x+\lambda_2$, so $\lambda_2=0$. Equation (iii) then gives $2z=2z+2+0$, i.e. $0=2$, a contradiction. Therefore $\lambda_1\ne 1$ and we must have $x=y$. (R1)若 $\lambda_1=1$：方程 (i) 给出 $2x=2x+\lambda_2$，故 $\lambda_2=0$。方程 (iii) 则给出 $2z=2z+2+0$，即 $0=2$，矛盾。因此 $\lambda_1\ne 1$，必有 $x=y$。(R1)
With $x=y$, substitute into $g_2=0$: $2x+z=0$, so $z=-2x$. (M1)令 $x=y$，代入 $g_2=0$：$2x+z=0$，故 $z=-2x$。(M1)
Substitute $y=x$ and $z=-2x$ into $g_1=0$:将 $y=x$ 和 $z=-2x$ 代入 $g_1=0$：
$$ x^{2}+x^{2}+4x^{2}+2(-2x)=6x^{2}-4x=2x(3x-2)=0. $$So $x=0$ (giving the point $(0,0,0)$) or $x=\tfrac{2}{3}$ (giving $y=\tfrac{2}{3}$, $z=-\tfrac{4}{3}$, the point $\bigl(\tfrac{2}{3},\tfrac{2}{3},-\tfrac{4}{3}\bigr)$). (A1)故 $x=0$（对应点 $(0,0,0)$）或 $x=\tfrac{2}{3}$（对应 $y=\tfrac{2}{3}$，$z=-\tfrac{4}{3}$，点 $\bigl(\tfrac{2}{3},\tfrac{2}{3},-\tfrac{4}{3}\bigr)$）。(A1)
Verify both points satisfy $g_1=0$ and $g_2=0$:验证两个点均满足 $g_1=0$ 和 $g_2=0$：
(M1) $f(0,0,0)=0$ and与 $f\bigl(\tfrac{2}{3},\tfrac{2}{3},-\tfrac{4}{3}\bigr)=\tfrac{4}{9}+\tfrac{4}{9}+\tfrac{16}{9}=\tfrac{24}{9}=\tfrac{8}{3}$. (A1)
Minimum value of $f$: $0$ at $(0,0,0)$.$f$ 的最小值：$0$，在 $(0,0,0)$ 处取得。 The origin lies on both the sphere and the plane and has zero squared distance from itself.原点同时在球面和平面上，与自身的距离平方为零。
Maximum value of $f$: $\tfrac{8}{3}$ at $\bigl(\tfrac{2}{3},\tfrac{2}{3},-\tfrac{4}{3}\bigr)$.$f$ 的最大值：$\tfrac{8}{3}$，在 $\bigl(\tfrac{2}{3},\tfrac{2}{3},-\tfrac{4}{3}\bigr)$ 处取得。