University Calculus | Unit C5: Optimization and Lagrange Multipliers

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

Critical Points and the Second-Derivative Test极值点与二阶导数判别法

Show all partial derivatives. A critical point must satisfy both $f_x=0$ and $f_y=0$ simultaneously; state this system explicitly before solving it. When applying the second-derivative test, compute $D=f_{xx}f_{yy}-f_{xy}^{2}$ at each critical point and state the conclusion.写出所有偏导数。极值点必须同时满足 $f_x=0$ 与 $f_y=0$；在求解之前请明确写出该方程组。应用二阶导数判别法时，在每个极值候选点计算 $D=f_{xx}f_{yy}-f_{xy}^{2}$ 并给出结论。

Q1MEDIUM CORE finding critical points via $\nabla f = \mathbf{0}$通过 $\nabla f = \mathbf{0}$ 求极值候选点 [6 marks]

Let $f(x,y)=x^{3}+y^{3}-3x-12y+20$.

(a) Compute $f_x$ and $f_y$, and write the system $f_x=0,\ f_y=0$.计算 $f_x$ 和 $f_y$，并写出方程组 $f_x=0,\ f_y=0$。 [2]

(b) Solve the system to find all critical points of $f$.求解该方程组，找出 $f$ 的所有极值候选点。 [2]

(c) For each critical point, apply the second-derivative test: compute $D=f_{xx}f_{yy}-f_{xy}^{2}$ and classify the point as a local maximum, local minimum, or saddle point.对每个极值候选点应用二阶导数判别法：计算 $D=f_{xx}f_{yy}-f_{xy}^{2}$，并将该点分类为极大值、极小值或鞍点。 [2]

Q2MEDIUM CORE second-derivative test with mixed partial含混合偏导数的二阶导数判别法 [8 marks]

Let $g(x,y)=x^{3}+y^{3}-3xy$.

(a) Find all critical points of $g$ by solving $g_x=0$ and $g_y=0$ simultaneously.联立 $g_x=0$ 与 $g_y=0$，求 $g$ 的所有极值候选点。 [3]

(b) Compute the discriminant $D=g_{xx}g_{yy}-g_{xy}^{2}$ at each critical point.在每个极值候选点计算判别式 $D=g_{xx}g_{yy}-g_{xy}^{2}$。 [3]

(c) Classify each critical point. If $D=0$ at any point, state explicitly that the test is inconclusive and explain what additional analysis would be needed.对每个极值候选点进行分类。若某点处 $D=0$，请明确指出判别法失效，并说明需要进行哪些补充分析。 [2]

Q3HARD CORE absolute extrema on a closed triangular region闭三角形区域上的绝对极值 [8 marks]

Let $h(x,y)=x^{2}-xy+y^{2}-2x$ on the closed triangular region $R$ with vertices $(0,0)$, $(4,0)$, and $(0,4)$.

(a) Find any interior critical points of $h$ in $R$ and evaluate $h$ there.求 $h$ 在 $R$ 内部的极值候选点，并计算 $h$ 在该点的值。 [2]

(b) Parametrize each edge of $R$ and find the extreme values of $h$ on each edge. Show your parametrization and the resulting single-variable optimization for each edge.对 $R$ 的每条边进行参数化，求 $h$ 在每条边上的极值。写出参数化表达式以及由此导出的单变量优化过程。 [4]

(c) Evaluate $h$ at all vertices of $R$, and state the absolute maximum and absolute minimum of $h$ on $R$, giving the point(s) where each occurs.计算 $h$ 在 $R$ 各顶点处的值，给出 $h$ 在 $R$ 上的绝对最大值和绝对最小值，并注明各自发生的位置。 [2]

Q4MEDIUM CORE absolute extrema on a closed disk闭圆盘上的绝对极值 [6 marks]

Find the absolute maximum and minimum values of $\phi(x,y)=3x+4y$ on the closed disk $D=\{(x,y):x^{2}+y^{2}\le 25\}$.

(a) Show that $\phi$ has no interior critical points in $D$, and explain why the extreme values must occur on the boundary circle $x^{2}+y^{2}=25$.证明 $\phi$ 在 $D$ 内部没有极值候选点，并解释为何极值必然出现在边界圆 $x^{2}+y^{2}=25$ 上。 [2]

(b) Parametrize the boundary circle and find the extreme values of $\phi$ on it.对边界圆进行参数化，求 $\phi$ 在其上的极值。 [2]

(c) State the absolute maximum and minimum of $\phi$ on $D$, giving the point(s) where each occurs.给出 $\phi$ 在 $D$ 上的绝对最大值和绝对最小值，并注明各自发生的位置。 [2]

Geometry of Lagrange Multipliers and the Hessian拉格朗日乘数法的几何意义与黑塞矩阵

Geometric and theoretical arguments must be complete. For the Lagrange condition, work from the definition of a regular constraint surface and the gradient interpretation. For the Hessian, connect the algebraic discriminant to the quadratic form; do not merely quote the test.几何与理论论证须完整。对于拉格朗日条件，须从正则约束曲面的定义和梯度的几何意义出发加以推导。对于黑塞矩阵，须将代数判别式与二次型联系起来，不得仅仅引用判别法结论。

Q5HARD PROOF geometric derivation of the Lagrange condition拉格朗日条件的几何推导 [8 marks]

Let $f,g:\mathbb{R}^{2}\to\mathbb{R}$ be smooth, and let $\mathbf{x}^{*}$ be a constrained extremum of $f$ subject to $g(\mathbf{x})=c$. Assume $\nabla g(\mathbf{x}^{*})\ne\mathbf{0}$.

(a) Let $\mathbf{r}(t)$ be any smooth curve lying entirely on the level set $g=c$ with $\mathbf{r}(0)=\mathbf{x}^{*}$. By differentiating the constraint $g(\mathbf{r}(t))=c$ with respect to $t$ at $t=0$, show that $\nabla g(\mathbf{x}^{*})$ is perpendicular to every tangent vector of the constraint curve at $\mathbf{x}^{*}$.设 $\mathbf{r}(t)$ 是完全落在等位集 $g=c$ 上且满足 $\mathbf{r}(0)=\mathbf{x}^{*}$ 的任意光滑曲线。对约束条件 $g(\mathbf{r}(t))=c$ 在 $t=0$ 处关于 $t$ 求导，证明 $\nabla g(\mathbf{x}^{*})$ 与约束曲线在 $\mathbf{x}^{*}$ 处的每个切向量正交。 [3]

(b) Since $\mathbf{x}^{*}$ is an extremum of $f$ on the constraint, the directional derivative of $f$ along any constraint-tangent direction must vanish. Use this and part (a) to show that $\nabla f(\mathbf{x}^{*})$ is also perpendicular to every tangent vector of the constraint curve at $\mathbf{x}^{*}$. Hence deduce that $\nabla f(\mathbf{x}^{*})=\lambda\,\nabla g(\mathbf{x}^{*})$ for some scalar $\lambda$.由于 $\mathbf{x}^{*}$ 是 $f$ 在约束上的极值点，$f$ 沿任意约束切线方向的方向导数必须为零。利用这一点及第 (a) 部分，证明 $\nabla f(\mathbf{x}^{*})$ 同样与约束曲线在 $\mathbf{x}^{*}$ 处的每个切向量正交，从而推出存在标量 $\lambda$，使得 $\nabla f(\mathbf{x}^{*})=\lambda\,\nabla g(\mathbf{x}^{*})$。 [3]

(c) Explain geometrically why the condition $\nabla f=\lambda\nabla g$ means the level curves of $f$ and $g$ are tangent to each other at $\mathbf{x}^{*}$. Why does this tangency fail to hold at a generic point on the constraint?从几何角度解释：条件 $\nabla f=\lambda\nabla g$ 为何意味着 $f$ 和 $g$ 的等位曲线在 $\mathbf{x}^{*}$ 处相切。为何在约束上的一般点处此切线关系不成立？ [2]

Q6HARD PROOF Hessian and the second-derivative test黑塞矩阵与二阶导数判别法 [8 marks]

Let $f:\mathbb{R}^{2}\to\mathbb{R}$ be twice continuously differentiable, and let $(a,b)$ be a critical point of $f$, so $f_x(a,b)=f_y(a,b)=0$. The Hessian matrix at $(a,b)$ is $H=\begin{pmatrix}f_{xx}&f_{xy}\\f_{xy}&f_{yy}\end{pmatrix}$ and $D=\det(H)=f_{xx}f_{yy}-f_{xy}^{2}$.

(a) Write the second-order Taylor expansion of $f(a+h,b+k)$ about $(a,b)$. Hence show that the sign of $f(a+h,b+k)-f(a,b)$ near $(a,b)$ is controlled by the quadratic form $Q(h,k)=f_{xx}h^{2}+2f_{xy}hk+f_{yy}k^{2}$.写出 $f(a+h,b+k)$ 在 $(a,b)$ 处的二阶泰勒展开式，并由此证明在 $(a,b)$ 附近 $f(a+h,b+k)-f(a,b)$ 的符号由二次型 $Q(h,k)=f_{xx}h^{2}+2f_{xy}hk+f_{yy}k^{2}$ 决定。 [3]

(b) By completing the square in $h$, show that if $D>0$ then $Q(h,k)$ has the same sign as $f_{xx}$ for all $(h,k)\ne(0,0)$. Deduce the classification: local minimum when $D>0,\,f_{xx}>0$; local maximum when $D>0,\,f_{xx}<0$.通过对 $h$ 配方，证明若 $D>0$，则对所有 $(h,k)\ne(0,0)$，$Q(h,k)$ 与 $f_{xx}$ 同号。由此推出分类结论：$D>0,\,f_{xx}>0$ 时为极小值；$D>0,\,f_{xx}<0$ 时为极大值。 [3]

(c) Explain why $D<0$ implies $Q$ takes both positive and negative values (i.e. the critical point is a saddle). What does $D=0$ tell you, and why does the second-derivative test fail in that case?解释为何 $D<0$ 意味着 $Q$ 取正值和负值（即该极值候选点为鞍点）。$D=0$ 说明了什么？为何二阶导数判别法在该情况下失效？ [2]

Q7HARD PROOF Lagrange system: case analysis and two constraints拉格朗日方程组：分情况讨论与双约束 [10 marks]

This question examines the algebraic structure of Lagrange systems.本题考查拉格朗日方程组的代数结构。

(a) Consider optimizing $f(x,y)=xy$ subject to $g(x,y)=x+y-1=0$. Set up the Lagrange system $\nabla f=\lambda\nabla g$, $g=0$. Solve completely, being careful not to divide by an expression that could be zero; instead, treat the cases $\lambda=0$ and $\lambda\ne 0$ separately and show that $\lambda=0$ leads to no point on the constraint.考虑在约束 $g(x,y)=x+y-1=0$ 下优化 $f(x,y)=xy$。建立拉格朗日方程组 $\nabla f=\lambda\nabla g$，$g=0$。完整求解，注意不得用可能为零的表达式做除法；应分 $\lambda=0$ 和 $\lambda\ne 0$ 两种情况讨论，并证明 $\lambda=0$ 在约束上无解。 [4]

(b) For two constraints $g_1(\mathbf{x})=c_1$ and $g_2(\mathbf{x})=c_2$ in $\mathbb{R}^{3}$, state the two-constraint Lagrange condition $\nabla f=\lambda_1\nabla g_1+\lambda_2\nabla g_2$ and explain geometrically why $\nabla f$ must lie in the plane spanned by $\nabla g_1$ and $\nabla g_2$ at a constrained extremum. How many equations does the full system provide, and how many unknowns ($x,y,z,\lambda_1,\lambda_2$) must be solved?对于 $\mathbb{R}^{3}$ 中的双约束 $g_1(\mathbf{x})=c_1$ 与 $g_2(\mathbf{x})=c_2$，写出双约束拉格朗日条件 $\nabla f=\lambda_1\nabla g_1+\lambda_2\nabla g_2$，并从几何角度解释为何在约束极值点处 $\nabla f$ 必须落在 $\nabla g_1$ 和 $\nabla g_2$ 所张成的平面内。完整方程组提供多少个方程？需要求解多少个未知数（$x,y,z,\lambda_1,\lambda_2$）？ [3]

(c) A student attempts to find the extrema of $f(x,y,z)=x+y+z$ on the curve of intersection of $x^{2}+y^{2}=2$ and $z=xy$. Write the Lagrange system (5 equations in 5 unknowns). You do not need to solve the system, but identify every equation and explain why simply parametrizing the curve and differentiating is an alternative approach for this particular setup.一名学生尝试求 $f(x,y,z)=x+y+z$ 在 $x^{2}+y^{2}=2$ 与 $z=xy$ 交线上的极值。写出拉格朗日方程组（5 个方程，5 个未知数）。无需求解，但须逐一说明每个方程，并解释为何对此特定结构也可通过直接参数化曲线再求导的方法来替代。 [3]

Constrained Optimization in Context有约束优化的实际应用

Set up each problem from the problem statement. Clearly define your objective function and constraint(s). Carry exact values and verify that candidate points actually satisfy the constraint. A final answer without a check that the constraint is satisfied will not earn the answer mark.根据题目描述建立模型。清晰定义目标函数和约束条件。全程保持精确值，并验证候选点确实满足约束条件。最终答案若未验证约束是否满足，将不得答案分。

Q8HARD APPLIED maximum volume box inscribed in an ellipsoid内接于椭球面的最大体积长方体 [10 marks]

A rectangular box with sides parallel to the coordinate planes is inscribed in the ellipsoid $\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}+z^{2}=1$, so one corner is at $(x,y,z)$ with $x,y,z>0$ and the opposite corner at $(-x,-y,-z)$. The box has volume $V=8xyz$.

(a) State the objective function and constraint. Set up the Lagrange system $\nabla V=\lambda\nabla g$ where $g(x,y,z)=\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}+z^{2}-1$. Write all four equations of the system explicitly.写出目标函数和约束条件。建立拉格朗日方程组 $\nabla V=\lambda\nabla g$，其中 $g(x,y,z)=\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}+z^{2}-1$。明确写出方程组的全部四个方程。 [3]

(b) Solve the Lagrange system (noting $x,y,z>0$) to find the dimensions of the optimal box. Verify that the constraint is satisfied.求解拉格朗日方程组（注意 $x,y,z>0$），求出最优长方体的尺寸，并验证约束条件满足。 [4]

(c) State the maximum volume and briefly justify (without a formal second-derivative test) why this critical point is indeed a maximum rather than a minimum.给出最大体积，并简要说明（无需正式的二阶导数判别）为何此极值候选点确实是极大值而非极小值。 [3]

Q9HARD APPLIED closest point on a plane to the origin平面上距原点最近的点 [8 marks]

Find the point on the plane $2x+2y+z=9$ that is closest to the origin. Use Lagrange multipliers throughout.用拉格朗日乘数法，求平面 $2x+2y+z=9$ 上距原点最近的点。

(a) Explain why minimizing the squared distance $d^{2}=x^{2}+y^{2}+z^{2}$ subject to $2x+2y+z=9$ is equivalent to finding the closest point, and state the Lagrange system.解释为何在约束 $2x+2y+z=9$ 下最小化距离平方 $d^{2}=x^{2}+y^{2}+z^{2}$ 与求最近点等价，并写出拉格朗日方程组。 [3]

(b) Solve the Lagrange system to find the closest point, verifying that it satisfies the plane equation.求解拉格朗日方程组，找出最近点，并验证其满足平面方程。 [3]

(c) Compute the minimum distance. Then explain how your answer relates to the point-to-plane distance formula $d=\dfrac{|ax_0+by_0+cz_0-d|}{\sqrt{a^{2}+b^{2}+c^{2}}}$, confirming the two methods agree.计算最短距离，并解释所得结果如何与点到平面距离公式 $d=\dfrac{|ax_0+by_0+cz_0-d|}{\sqrt{a^{2}+b^{2}+c^{2}}}$ 相对应，确认两种方法结果一致。 [2]

Q10HARD APPLIED two-constraint Lagrange: extrema on a curve of intersection双约束拉格朗日：交线上的极值 [10 marks]

Find the extreme values of $f(x,y,z)=x^{2}+y^{2}+z^{2}$ on the curve of intersection of the sphere $x^{2}+y^{2}+z^{2}+2z=0$ (equivalently $x^{2}+y^{2}+(z+1)^{2}=1$) and the plane $x+y+z=0$.

(a) Rewrite both constraints in the form $g_1=0$ and $g_2=0$. Then state the two-constraint Lagrange system $\nabla f=\lambda_1\nabla g_1+\lambda_2\nabla g_2$ together with the two constraint equations, giving 5 equations in 5 unknowns.将两个约束分别改写为 $g_1=0$ 和 $g_2=0$ 的形式，然后写出双约束拉格朗日方程组 $\nabla f=\lambda_1\nabla g_1+\lambda_2\nabla g_2$ 及两个约束方程，共 5 个方程，5 个未知数。 [3]

(b) From the three gradient equations, express $x$, $y$, and $z$ in terms of $\lambda_1$ and $\lambda_2$. Show that the system forces $x=y$.由三个梯度方程将 $x$、$y$、$z$ 用 $\lambda_1$ 和 $\lambda_2$ 表示，证明方程组要求 $x=y$。 [3]

(d) Evaluate $f$ at each candidate and state the maximum and minimum values of $f$ on the curve of intersection.计算 $f$ 在每个候选点处的值，给出 $f$ 在交线上的最大值和最小值。 [2]