6: Exercises

Exercises

[exer:1] Find the point on the plane $2x+3y+z=7$ closest to $(1,-2,3)$.

[exer:2] Find the extreme values of $f(x,y)=2x+y$ subject to $x^{2}+y^{2}=5$.

[exer:3] Suppose that $a,b>0$ and $a\alpha^{2}+b\beta^{2}=1$. Find the extreme values of $f(x,y)=\beta x+\alpha y$ subject to $ax^{2}+by^{2}=1$.

[exer:4] Find the points on the circle $x^{2}+y^{2}=320$ closest to and farthest from $(2,4)$.

[exer:5] Find the extreme values of $$f(x,y,z)=2x+3y+z\text{\quad subject to\quad} x^{2}+2y^{2}+3z^{2}=1. \nonumber$$

[exer:6] Find the maximum value of $f(x,y)=xy$ on the line $ax+by=1$, where $a,b>0$.

[exer:7] A rectangle has perimeter $p$. Find its largest possible area.

[exer:8] A rectangle has area $A$. Find its smallest possible perimeter.

[exer:9] A closed rectangular box has surface area $A$. Find it largest possible volume.

[exer:10] The sides and bottom of a rectangular box have total area $A$. Find its largest possible volume.

[exer:11] A rectangular box with no top has volume $V$. Find its smallest possible surface area.

[exer:12] Maximize $f(x,y,z)=xyz$ subject to $$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1, \nonumber$$ where $a$, $b$, $c>0$.

[exer:13] Two vertices of a triangle are $(-a,0)$ and $(a,0)$, and the third is on the ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1. \nonumber$$ Find its largest possible area.

[exer:14] Show that the triangle with the greatest possible area for a given perimeter is equilateral, given that the area of a triangle with sides $x$, $y$, $z$ and perimeter $s$ is $$A= \sqrt{s(s-x)(s-y)(s-z)}. \nonumber$$

[exer:15] A box with sides parallel to the coordinate planes has its vertices on the ellipsoid $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1. \nonumber$$ Find its largest possible volume.

[exer:16] Derive a formula for the distance from $(x_{1},y_{1},z_{1})$ to the plane $$ax+by+cz=\sigma. \nonumber$$

[exer:17] Let $\mathbf{X}_{i}=(x_{i},y_{i},z_{i})$, $1 \le i \le n$. Find the point in the plane $$ax+by+cz=\sigma \nonumber$$ for which $\sum_{i=1}^{n}|\mathbf{X}-\mathbf{X}_{i}|^{2}$ is a minimum. Assume that none of the ${\bf X}_{i}$ are in the plane.

[exer:18] Find the extreme values of $f({\bf X})=\displaystyle{\sum_{i=1}^{n}(x_{i}-c_{i})^{2}}$ subject to $\displaystyle{\sum_{i=1}^{n}x_{i}^{2}}=1$.

[exer:19] Find the extreme values of $$f(x,y,z)=2xy+2xz+2yz\text{\quad subject to\quad} x^{2}+y^{2}+z^{2}=1. \nonumber$$

[exer:20] Find the extreme values of $$f(x,y,z)=3x^{2}+2y^{2}+3z^{2}+2xz\text{\quad subject to\quad} x^{2}+y^{2}+z^{2}=1. \nonumber$$

[exer:21] Find the extreme values of $$f(x,y)=x^{2}+8xy+4y^{2} \text{\quad subject to\quad} x^{2}+2xy+4y^{2}=1. \nonumber$$

[exer:22] Find the extreme value of $f(x,y)=\alpha+\beta xy$ subject to $(ax+by)^{2}=1$. Assume that $ab\ne0$.

[exer:23] Find the extreme values of $f(x,y,z)=x+y^{2}+2z$ subject to $$4x^{2}+9y^{2}-36z^{2}=36. \nonumber$$

[exer:24] Find the extreme values of $f(x,y,z,w)=(x+z)(y+w)$ subject to $$x^{2}+y^{2}+z^{2}+w^{2}=1. \nonumber$$

[exer:25] Find the extreme values of $f(x,y,z,w)=(x+z)(y+w)$ subject to $$x^{2}+y^{2}=1 \text{\;and \;\;} z^{2}+w^{2}=1. \nonumber$$

[exer:26] Find the extreme values of $f(x,y,z,w)=(x+z)(y+w)$ subject to $$x^{2}+z^{2}=1 \text{\;and \;\;} y^{2}+w^{2}=1. \nonumber$$

[exer:27] Find the distance between the circle $x^{2}+y^{2}=1$ the hyperbola $xy=1$.

[exer:28] Minimize $f(x,y,x)=\displaystyle{\frac{x^{2}}{\alpha^{2}}+\frac{y^{2}}{\beta^{2}} +\frac{z^{2}}{\gamma^{2}}}$ subject to $ax+by+cz=d$ and $x$, $y$, $z>0$.

[exer:29] Find the distance from $(c_{1},c_{2},\dots,c_{n})$ to the plane $$a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}=d. \nonumber$$

[exer:30] Find the maximum value of $f({\bf X})=\displaystyle{\sum_{i=1}^{n}a_{i}x_{i}^{2}}$ subject to $\displaystyle{\sum_{i=1}^{n}b_{i}x_{i}^{4}}=1$, where $p,$ $q>0$ and $a_{i}$, $b_{i}$ $x_{i}>0$, $1\le i\le n$.

[exer:31] Find the extreme value of $f({\bf X})=\displaystyle{\sum_{i=1}^{n}a_{i}x_{i}^{p}}$ subject to $\displaystyle{\sum_{i=1}^{n}b_{i}x_{i}^{q}}=1$, where $p$, $q$>0 and $a_{i}$, $b_{i}$, $x_{i}>0$, $1\le i\le n$.

[exer:32] Find the minimum value of $$f(x,y,z,w)=x^{2}+2y^{2}+z^{2}+w^2 \nonumber$$ subject to $$\begin{aligned} x+y+\phantom{2}z+3w&=&1\\ x+y+2z+\phantom{3}w&=&2.\end{aligned} \nonumber$$

[exer:33] Find the minimum value of $$f(x,y,z)= \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}} \nonumber$$ subject to $p_{1}x+p_{2}y+p_{3}z=d$, assuming that at least one of $p_{1}$, $p_{2}$, $p_{3}$ is nonzero.

[exer:34] Find the extreme values of $f(x,y,z)= p_{1}x+p_{2}y+p_{3}z$ subject to $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1, \nonumber$$ assuming that at least one of $p_{1}$, $p_{2}$, $p_{3}$ is nonzero.

[exer:35] Find the distance from $(-1,2,3)$ to the intersection of the planes
$x+2y-3z=4$ and $2x-y+2z=5$.

[exer:36] Find the extreme values of $f(x,y,z)=2x+y+2z$ subject to $x^{2}+y^{2}=4$ and $x+z=2$.

[exer:37] Find the distance between the parabola $y=1+x^{2}$ and the line $x+y=-1$.

[exer:38] Find the distance between the ellipsoid $$3x^{2}+9y^{2}+6z^{2}=10 \nonumber$$ and the plane $$3x+3y+6z=70. \nonumber$$

[exer:39] Show that the extreme values of $f(x,y,z)=xy+yz+zx$ subject to $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1 \nonumber$$ are the largest and smallest eigenvalues of the matrix $$\left[\begin{array}{ccccccc} 0&a^{2}&a^{2}\\ b^{2}&0&b^{2}\\c^{2}&c^{2}&0 \end{array}\right]. \nonumber$$

[exer:40] Show that the extreme values of $f(x,y,z)=xy+2yz+2zx$ subject to $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1 \nonumber$$ are the largest and smallest eigenvalues of the matrix $$\left[\begin{array}{ccccccc} 0&a^{2}/2&a^{2}\\ b^{2}/2&0&b^{2}\\c^{2}&c^{2}&0 \end{array}\right]. \nonumber$$

[exer:41] Find the extreme values of $x(y+z)$ subject to $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1. \nonumber$$

[exer:42] Let $a$, $b$, $c$, $p$, $q$, $r$, $\alpha$, $\beta$, and $\gamma$ be positive constants. Find the maximum value of $f(x,y,z)=x^{\alpha}y^{\beta}z^{\gamma}$ subject to $$ax^{p}+by^{q}+cz^{r}=1 \text{\; and\;\;} x,y,z>0 . \nonumber$$

[exer:43] Find the extreme values of $$f(x,y,z,w)=xw-yz \text{\quad subject to\quad} x^{2}+2y^{2}=4\text{\quad and\quad} 2z^{2}+w^{2}=9. \nonumber$$

[exer:44] Let $a$, $b$, $c$,and $d$ be positive. Find the extreme values of $$f(x,y,z,w)=xw-yz \nonumber$$ subject to $$ax^{2}+by^{2}=1, \quad cz^{2}+dw^{2}=1, \nonumber$$ if (a) $ad\ne bc$; (b) $ad=bc.$

[exer:45] Minimize $f(x,y,z)=\alpha x^{2}+\beta y^{2}+\gamma z^{2}$ subject to $$a_{1}x+a_{2}y+a_{3}z=c\text{\; and\;\;} b_{1}x+b_{2}y+b_{3}z=d. \nonumber$$ Assume that $$\alpha,\beta,\gamma>0,\quad a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\ne0, \text{\; and\;\;} b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\ne 0. \nonumber$$ Formulate and apply a required additional assumption.

[exer:46] Minimize $f({\bf X},{\bf Y})=\displaystyle{\sum_{i=1}^{n}(x_{i}-\alpha_{i})^{2}}$ subject to $$\displaystyle{\sum_{i=1}^{n}a_{i}x_{i}=c} \text{\; and\;\;} \displaystyle{\sum_{i=1}^{n}b_{i}x_{i}=d}, \nonumber$$ where $$\sum_{i=1}^{n}a_{i}^{2}=\sum_{i=1}^{n}b_{i}^{2}=1 \text{\; and\;\;} \sum_{i=1}^{n}a_{i}b_{i}=0. \nonumber$$

[exer:47] Find $(x_{10,x_{20}},\dots,x_{n0})$ to minimize $$Q(\mathbf{X})=\sum_{i=1}^{n}x_{i}^{2} \nonumber$$ subject to $$\sum_{i=1}^{n}x_{i}=1\text{\quad and\quad} \sum_{i=1}^{n}ix_{i}=0. \nonumber$$ Prove explicitly that if $$\sum_{j=1}^{n}y_{i}=1,\quad \sum_{i=1}^{n}iy_{i}=0 \nonumber$$ and $y_{i}\ne x_{i0}$ for some $i\in\{1,2,\dots,n\}$, then $$\sum_{i=1}^{n}y_{i}^{2}>\sum_{i=1}^{n}x_{i0}^{2}. \nonumber$$

[exer:48] Let $p_{1}$, $p_{2}$, …, $p_{n}$ and $s$ be positive numbers. Maximize $$f({\bf X})= (s-x_{1})^{p_{1}}(s-x_{2})^{p_{2}}\cdots(s-x_{n})^{p_{n}} \nonumber$$ subject to $x_{1}+x_{2}+\cdots+x_{n}=s$.

[exer:49] Maximize $f({\bf X})=x_{1}^{p_{1}}x_{2}^{p_{2}}\cdots x_{n}^{p_{n}}$ subject to $x_{i}>0$, $1\le i\le n$, and $$\sum_{i=1}^{n}\frac{x_{i}}{\sigma_{i}} = S, \nonumber$$ where $p_{1}$, $p_{2}$,…, $p_{n}$, $\sigma_{1}$, $\sigma_{2}$, …, $\sigma_{n}$, and $V$ are given positive numbers.

[exer:50] Maximize $$f({\bf X})=\sum_{i=1}^{n}\frac{x_{i}}{\sigma_{i}} \nonumber$$ subject to $x_{i}>0$, $1\le i\le n$, and $$x_{1}^{p_{1}}x_{2}^{p_{2}}\cdots x_{n}^{p_{n}}=V, \nonumber$$ where $p_{1}$, $p_{2}$,…, $p_{n}$, $\sigma_{1}$, $\sigma_{2}$, …, $\sigma_{n}$, and $S$ are given positive numbers.

[exer:51] Suppose that $\alpha_{1}$, $\alpha_{2}$, …$\alpha_{n}$ are positive and at least one of $a_{1}$, $a_{2}$, …, $a_{n}$ is nonzero. Let $(c_{1},c_{2},\dots,c_{n})$ be given. Minimize $$Q({\bf X})=\sum_{i=1}^{n}\frac{(x_{i}-c_{i})^{2}}{\alpha_{i}} \nonumber$$ subject to $$a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}=d. \nonumber$$

[exer:52] Schwarz’s inequality says that $(a_{1},a_{2},\dots,a_{n})$ and $(x_{1},x_{2},\dots,x_{n})$ are arbitrary $n$-tuples of real numbers, then $$|a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}|\le (a_{1}^{2}+a_{2}^{2}+ \cdots+ a_{n}^{2})^{1/2} (x_{1}^{2}+x_{2}^{2}+ \cdots+ x_{n}^{2})^{1/2}. \nonumber$$ Prove this by finding the extreme values of $f({\bf X})=\displaystyle{\sum_{i=1}^{n}a_{i}x_{i}}$ subject to $\displaystyle{\sum_{i=1}^{n}x_{i}^{2}}~=~\sigma^{2}$.

[exer:53] Let $x_{1}$, $x_{2}$, …, $x_{m}$, $r_{1}$, $r_{2}$, …, $r_{m}$ be positive and $$r_{1}+r_{2}+\cdots+r_{m}=r. \nonumber$$ Show that $$\left(x_{1}^{r_{1}}x_{2}^{r_{2}}\cdots x_{m}^{r_{m}}\right)^{1/r} \le \frac{r_{1}x_{1}+r_{2}x_{2}+\cdots r_{m}x_{m}}{r}, \nonumber$$ and give necessary and sufficient conditions for equality. (Hint: Maximize $x_{1}^{r_{1}}x_{2}^{r_{2}}\cdots x_{m}^{r_{m}}$ subject to $\sum_{j=1}^{m}r_{j}x_{j}=\sigma>0$, $x_{1}>0$, $x_{2}>0$, …, $x_{m}>0$.)

[exer:54] Let $\mathbf{A}=[a_{ij}]$ be an $m\times n$ matrix. Suppose that $p_{1}$, $p_{2}$, …, $p_{m}>0$ and $$\sum_{j=1}^{m}\frac{1}{p_{j}}=1, \nonumber$$ and define $$\sigma_{i}=\sum_{j=1}^{n}|a_{ij}|^{p_{i}}, \quad 1 \le i \le m. \nonumber$$ Use Exercise [exer:53] to show that $$\left|\sum_{j=1}^{n}a_{ij}a_{2j}\cdots a_{mj}\right| \le \sigma_{1}^{1/p_{1}}\sigma_{2}^{1/p_{2}}\cdots \sigma_{m}^{1/p_{m}}. \nonumber$$ (With $m=2$ this is Hölder’s inequality, which reduces to Schwarz’s inequality if $p_{1}=p_{2}=2$.)

[exer:55] Let $c_{0}$, $c_{1}$, …, $c_{m}$ be given constants and $n\ge m+1$. Show that the minimum value of $$Q({\bf X})=\sum_{r=0}^{n}x_{r}^{2} \nonumber$$ subject to $$\sum_{r=0}^{n}x_{r}r^{s}=c_{s},\quad 0\le s \le m, \nonumber$$ is attained when $$x_{r}=\sum_{s=0}^{m}\lambda_{s}r^{s},\quad 0\le r\le n, \nonumber$$ where $$\sum_{\ell=0}^{m}\sigma_{s+\ell}\lambda_{\ell}=c_{s} \text{\; and\;\;} \sigma_{s}= \sum_{r=0}^{n}r^{s},\quad 0\le s\le m. \nonumber$$ Show that if $\{x_{r}\}_{r=0}^{n}$ satisfies the constraints and $x_{r}\ne x_{r0}$ for some $r$, then $$\sum_{r=0}^{n}x_{r}^{2}>\sum_{r=0}^{n}x_{r0}^{2}. \nonumber$$

[exer:56] Suppose that $n> 2k$. Show that the minimum value of $f({\bf W})=\displaystyle{\sum_{i=-n}^{n}w_{i}^{2}}$, subject to the constraint $$\sum_{i=-n}^{n}w_{i}P(r-i)=P(r) \nonumber$$ whenever $r$ is an integer and $P$ is a polynomial of degree $\le 2k$, is attained with $$w_{i0}=\sum_{r=0}^{2k}\lambda_{r}i^{r},\quad 1\le i\le n, \nonumber$$ where $$\sum_{r=0}^{2k}\lambda_{r}\sigma_{r+s}= \begin{cases} 1& \text{if } s=0,\\ 0&\text{if }1\le s\le 2k, \end{cases} \text{\; and\;\;} \sigma_{s}=\sum_{j=-n}^{n}j^{s}. \nonumber$$ Show that if $\{w_{i}\}_{i=-n}^{n}$ satisfies the constraint and $w_{i}\ne w_{i0}$ for some $i$, then $$\sum_{i=-n}^{n}w_{i}^{2}>\sum_{i=-n}^{n}w_{i0}^{2}. \nonumber$$

[exer:57] Suppose that $n\ge k$. Show that the minimum value of $f\displaystyle{\sum_{i=0}^{n}w_{i}^{2}}$, subject to the constraint $$\sum_{i=0}^{n}w_{i}P(r-i)=P(r+1) \nonumber$$ whenever $r$ is an integer and $P$ is a polynomial of degree $\le k$, is attained with $$w_{i0}=\sum_{r=0}^{k}\lambda_{r}i^{r},\quad 0\le i\le n, \nonumber$$ where $$\sum_{r=0}^{k}\sigma_{r+s}\lambda_{r}=(-1)^{s},\quad 0\le s \le k, \text{\quad and\quad } \sigma_{\ell}=\sum_{i=0}^{n}i^{\ell},\quad 0\le \ell\le 2k. \nonumber$$ Show that if $$\sum_{i=0}^{n}u_{i}P(r-i)=P(r+1) \nonumber$$ whenever $r$ is an integer and $P$ is a polynomial of degree $\le k$, and $u_{i}\ne w_{i0}$ for some $i$, then $$\sum_{i=0}^{n}u_{i}^{2}>\sum_{i=0}^{n}w_{i0}^{2}. \nonumber$$

[exer:58] Minimize $$f({\bf X})=\sum_{i=1}^{n}\frac{(x_{i}-c_{i})^{2}}{\alpha_{i}} \nonumber$$ subject to $$\sum_{i=1}^{n}a_{ir}x_{i}=d_{r},\quad 1\le r \le m \nonumber$$ Assume that $m>1$, $\alpha_{1}$, $\alpha_{2}$, … $\alpha_{m}>0$, and $$\sum_{i=1}^{n}\alpha_{i}a_{ir}a_{is}= \begin{cases} 1 & \text{ if } r=s,\\0 & \text{ if }r\ne s. \end{cases} \nonumber$$

Answers to selected exercises

[exer:1]. $\left(\frac{15}{7} -\frac{2}{7},\frac{25}{7}\right)$ $\pm5$ $1/\sqrt{ab}$, $-1/\sqrt{ab}$

[exer:4]. $(8,16)$ is closest, $(-8,-16)$ is farthest. $\pm\sqrt{53/6}$ $1/4ab$ $p^{2}/4$

$4\sqrt{A}$ $A^{3/2}/6\sqrt{6}$ $A^{3/2}/6\sqrt{3}$ $3(2V)^{2/3}$ $abc/27$

$ab$ $8abc/3\sqrt{3}$

[exer:18]. $(1-\mu)^{2}$ and $(1+\mu)^{2}$, where $\mu =\displaystyle{\left(\sum_{j=1}^{n}c_{j}^{2}\right)^{1/2}}$ $-1$, $2$ $2$, $4$

$-2/3$, $2$ $\alpha\pm|\beta|/4|ab|$ $-\sqrt{5}$, $73/16$ $\pm1$ $\pm2$

$\pm2$ $\sqrt2-1$ $\displaystyle{\frac{d^{2}}{(a\alpha)^{2}+(b\beta^{2})+(c\gamma)^{2}}}$

$\displaystyle{\frac{|d-a_{1}c_{1}-a_{2}c_{2}-\cdots-a_{n}c_{n})a_{i}|} {\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots a_{n}^{2}}}}$ $\displaystyle{\left(\sum_{i=1}^{n}\frac{a_{i}^{2}}{b_{i}}\right)^{1/2}}$

[exer:31]. $\displaystyle{\left(\sum_{i=1}^{n}a_{i}^{q/(q-p)} b_{i}^{p/(p-q)}\right)^{1-p/q}}$ is a constrained maximum if $p<q$, a constrained minimum if $p>q$

$689/845$ $\displaystyle{\frac{d^{2}}{p_{1}^{2}a^{2}+p_{2}^{2}b^{2}+p_{3}^{2}c^{2}}}$ $\pm (p_{1}^{2}a^{2}+p_{2}^{2}b^{2}+p_{3}^{2}c^{2})^{1/2}$

[exer:35]. $\sqrt{693/45}$ [exer:36]. $2$, $6$ $7/4\sqrt{2}$ $10\sqrt{6}/3$ $\pm|c|\sqrt{a^{2}+b^{2}}/2$

$\displaystyle{\frac{\alpha\beta\gamma}{pqr} \left(\frac{\alpha}{p}+\frac{\beta}{q}+\frac{\gamma}{r}\right)^{-3}}$ [exer:43]. $\pm3$ $\pm1/\sqrt{bc}$ (b) $\pm1/\sqrt{ad}=\pm1/\sqrt{bc}$

[exer:46]. $\displaystyle{\left(c-\sum_{i=1}^{n}a_{i}\alpha_{i}\right)^{2} +\left(d-\sum_{i=1}^{n}b_{i}\alpha_{i}\right)^{2}}$ $x_{i0}=(4n+2-6i)/n(n-1)$

[exer:48]. $\left[\frac{(n-1)s}{P}\right]^{P}p_{1}^{p_{1}}p_{2}^{p_{2}}\cdots p_{n}^{p_{n}}$

[exer:49]. $\displaystyle{\left(\frac{S}{p_{1}+p_{2}+\cdots+ p_{n}}\right)^{p_{1}+p_{2}+\cdots+p_{n}} (p_{1}\sigma_{1})^{p_{1}} (p_{2}\sigma_{2})^{p_{2}} \cdots (p_{n}\sigma_{n})^{p_{n}}}$

[exer:50]. $\displaystyle{(p_{1}+p_{2}+\cdots+p_{n}) \left(\frac{V}{(\sigma_{1}p_{1})^{p_{1}}(\sigma_{2}p_{2})^{p_{2}} \cdots (\sigma_{ n}p_{n})^{p_{n}}}\right)^{\frac{1}{p_{1}+p_{2}+\cdots+p_{n}}}}$

[exer:51]. $\displaystyle{\left(d-\sum_{i=1}^{n}a_{i}c_{i}\right))^{2}/ \left(\sum_{i=1}^{n}a_{i}^{2}\alpha_{i}\right)}$ $\displaystyle{\pm\left(\sum_{i=1}^{n}a_{i}^{2}\right)^{1/2} \left(\sum_{i=1}^{n}x_{i0}^{2}\right)^{1/2}}$

$\displaystyle\sum_{r=1}^{m} \left(d_{r}-\sum_{i=1}^{n}a_{ir}c_{i}\right)^{2}$

INSTRUCT0R’S SolutionS MANUAL

SolutionS OF EXERCISES

$L=\displaystyle{\frac{(x-1)^{2}+(y+2)^{2}+(z-3)^{2}}{2}-\lambda(2x+3y+z)}$ $$L_{x}=x-1-2\lambda,\quad L_{y}=y+2-3\lambda, \quad L_{z}=z-3-\lambda \nonumber$$ $$x_{0}=1+2\lambda, \quad y_{0}=-2+3\lambda, \quad z_{0}=3+\lambda \quad \nonumber$$ $$2(1+2\lambda)+3(-2+3\lambda)+(3+\lambda)=7, \quad \lambda=\displaystyle{\frac{4}{7}} \nonumber$$ $$x_{0}=\displaystyle{\frac{15}{7}}, \quad y_{0}=-\displaystyle{\frac{2}{7}},\quad z_{0}=\displaystyle{\frac{25}{7}} \nonumber$$ The distance from $(x_{01},y_{01},z_{01})$ to the plane is $$\sqrt{(x_{0}-1)^{2}+(y_{0}+2)^{2}+(z_{0}-3)^{2}}= \sqrt{4\lambda^{2}+9\lambda^{2}+\lambda^{2}}=4\sqrt{\frac{2}{7}}. \nonumber$$

$L=2x+y-\displaystyle{\frac{\lambda}{2}}(x^{2}+y^{2})$, $L_{x}=2-\lambda x$, $L_{y}=1-\lambda y$

$$x_{0}=2y_{0},\quad 5y_{0}^{2}=5, \quad (x_{0},y_{0})=\pm(2,1) \nonumber$$ Constrained minimum $=-5$, constrained maximum $=5$.

$L=\beta x+\alpha y-\displaystyle{\frac{\lambda}{2}}(ax^{2}+by^{2})$

$$L_{x}=\beta-\lambda ax,\quad L_{y}=\alpha-\lambda by,\quad x_{0}=\displaystyle{\frac{\beta}{\lambda a}}, \quad y_{0}=\displaystyle{\frac{\alpha}{\lambda b}} \nonumber$$ $$1=ax_{0}^{2}+by_{0}^{2}=\frac{1}{\lambda^{2}} \left(\frac{\beta^{2}}{a}+\frac{\alpha^{2}}{b}\right) =\frac{1}{ab\lambda^{2}}(a\alpha^{2}+b\beta^{2})=\frac{1}{ab\lambda^{2}}. \nonumber$$ $\displaystyle{\frac{1}{\lambda}}=\pm\sqrt{ab}$; $(x_{0},y_{0})=\pm\displaystyle{\left(\beta\sqrt{\frac{b}{a}},\alpha\sqrt\frac{a}{b}\right)}$. Choosing “$+$” yields the constrained maximum $$f(x_{0},y_{0})=\beta^{2}\sqrt{\frac{b}{a}}+\alpha^{2}\sqrt{\frac{a}{b}} =\frac{b\beta^{2}}{\sqrt{ab}}+\frac{a\alpha^{2}}{\sqrt{ab}}=\frac{1}{\sqrt{ab}}. \nonumber$$ Choosing “$-$” yields the constrained minimum $-\displaystyle{\frac{1}{\sqrt{ab}}}$.

$L=\displaystyle{\frac{(x-2)^{2}+(y-4)^{2}-\lambda(x^{2}+y^{2})}{2}}$

$$L_{x}(x,y)=(x-2)-\lambda x,\quad L_{y}(x,y)=(y-4)-\lambda y \nonumber$$ $$\frac{x_{0}-2}{x_{0}}=\frac{y_{0}-4}{y_{0}}=\lambda, \text{\; so\;\;} y_{0}=2x_{0}. \nonumber$$ Therefore, $x_{0}^{2}+y_{0}^{2}=5x_{0}^{2}=320$, $(x_{0},y_{0})=\pm(8,16)$ so the constrained critical points are $(8,16)$ and $(-8,-16)$; $(8,16)$ is closest to $(2,4)$ and $(-8,-16)$ is farthest.

$L=2x+3y+z-\displaystyle{\frac{\lambda}{2}}(x^{2}+2y^{2}+3z^{2})$

$$L_{x}=2-\lambda x,\quad L_{y}=3-2\lambda y,\quad L_{z}=1-3\lambda z \nonumber$$ $$x_{0}=\frac{2}{\lambda}\quad y_{0}=\frac{3}{2\lambda}, \quad z_{0}=\frac{1}{3\lambda}, \quad x_{0}^{2}+2y_{0}^{2}+3z_{0}^{2}=\displaystyle{\frac{53}{6\lambda^{2}}}=1,\quad \lambda=\pm\sqrt{53/6}. \nonumber$$ Since $f(2/\lambda,3/2\lambda,1/3\lambda)=\displaystyle{\frac{53}{6\lambda}}=\pm \lambda$, the constrained extreme values are $\pm\sqrt{53/6}$.

$L=xy-\lambda (ax+by)$, $L_{x}(x,y)=y-\lambda a$, $L_{y}=x-\lambda b$

$$x_{0}=\lambda b,\quad y_{0}=\lambda a, \quad ax_{0}+by_{0}=2\lambda ab=1,\quad \lambda=\frac{1}{2ab} \nonumber$$ $$x_{0}=\frac{1}{2a},\quad y_{0}=\frac{1}{2b},\quad x_{0}y_{0}=\frac{1}{4ab}=\text{constrained maximum\;\;} \nonumber$$

$p=2x+2y$, $A=xy$, $L=xy-\lambda(x+y)$, $L_{x}=y-\lambda$, $L_{y}=x-\lambda$, $y_{0}=x_{0}$, $x_{0}=p/4$, $A_{\text max}=p^{2}/4$.

Let $x$ and $y$ denote lengths of sides. We must mimimize $x+y$ subject to $xy=A$. $$L=x+y-\lambda xy,\quad L_{x}=1-\lambda y,\; L_{y}=1-\lambda x, \; x_{0}=y_{0},\; x_{0}y_{0}=A,\; x_{0}=\sqrt{A}. \nonumber$$ The minimum perimeter is $4\sqrt{A}$.

Denote the vertices of the box by $(0,0,0)$, $(x,0,0)$, $(0,y,0)$, and $(0,0,z)$. $$V=xyz,\quad A=2xz+2yz +2xy,\quad L=xyz-\lambda(xz+yz+xy) \nonumber$$ $$L_{x}=yz-\lambda(z+y),\quad L_{y}=xz-\lambda(z+x), \quad L_{z}=xy-\lambda(x+y) \nonumber$$ $$y_{0}z_{0}=\lambda(z_{0}+y_{0}),\quad x_{0}z_{0}=\lambda(z_{0}+x_{0}), \quad x_{0}y_{0}=\lambda(x_{0}+y_{0}) \nonumber$$ $$x_{0}z_{0}+x_{0}y_{0}=z_{0}y_{0}+x_{0}y_{0} =x_{0}z_{0}+y_{0}z_{0},\quad x_{0}=y_{0}=z_{0} \nonumber$$ $$A=6z_{0}^{2}, \quad z=\sqrt{\frac{A}{6}},\quad V_{\text{max}}=z_{0}^{3}=\displaystyle{\frac{A^{3/2}}{6\sqrt{6}}}. \nonumber$$

Denote the vertices of the box by $(0,0,0)$, $(x,0,0)$, $(0,y,0)$, and $(0,0,z)$. $$V=xyz,\quad A=2xz+2yz+xy,\quad L=xyz-\lambda(2xz+2yz+xy) \nonumber$$ $$L_{x}=yz-\lambda(2z+y),\quad L_{y}=xz-\lambda(2z+x), \quad L_{z}=xy-\lambda(2x+2y) \nonumber$$ $$y_{0}z_{0}=\lambda(2z_{0}+y_{0}),\quad x_{0}z_{0}=\lambda(2z_{0}+x_{0}),\quad x_{0}y_{0}=\lambda(2x_{0}+2y_{0}) \nonumber$$ $$x_{0}y_{0}z_{0}=\lambda x_{0}(2z_{0}+y_{0}),\quad x_{0}y_{0}z_{0}=\lambda y_{0}(2z_{0}+x_{0}),\quad x_{0}y_{0}z_{0}=\lambda z_{0}(2x_{0}+2y_{0}) \nonumber$$ $$2x_{0}z_{0}+x_{0}y_{0}=2y_{0}z_{0}+x_{0}y_{0}=2x_{0}z_{0}+2y_{0}z_{0} \nonumber$$ $$x_{0}=y_{0}=2z_{0}, \quad A=12z_{0}^{2}, \quad z_{0}=\sqrt{\frac{A}{12}},\quad V_{\text{max}}=z_{0}^{3}=\displaystyle{\frac{A^{3/2}}{6\sqrt{3}}}. \nonumber$$

Denote the vertices of the box by $(0,0,0)$, $(x,0,0)$, $(0,y,0)$, and $(0,0,z)$. $$V=xyz,\quad A=2xz+2yz+xy, \quad L=2xz+2yz+xy-\lambda xyz,\quad \nonumber$$ $$L_{x}=2z+y-\lambda yz, \quad L_{y}=2z+x-\lambda xz, \quad L_{z}=2x+2y-\lambda xy \nonumber$$ $$2z_{0}+y_{0}=\lambda y_{0}z_{0}, \quad 2z_{0}+x_{0}=\lambda x_{0}z_{0}, \quad 2x_{0}+2y_{0}-\lambda x_{0}y_{0} \nonumber$$ $$2x_{0}z_{0}+x_{0}y_{0}=2y_{0}z_{0}+x_{0}y_{0}=2x_{0}z_{0}+2y_{0}z_{0} \nonumber$$ $$x_{0}=y_{0}=2z_{0},\; V=4z_{0}^{3}, \; z_{0}=\frac{(2V)^{1/3}}{2},\; x_{0}=y_{0}=(2V)^{1/3},\; A_\text{min}=3(2V)^{2/3} \nonumber$$

$L=xyz-\lambda\displaystyle{\left(\displaystyle{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}}\right)}$, $L_{x}=yz-\displaystyle{\frac{\lambda}{a}}$, $L_{y}=xz-\displaystyle{\frac{\lambda}{b}}$, $L_{z}=xy-\displaystyle{\frac{\lambda}{c}}$ $$y_{0}z_{0}=\displaystyle{\frac{\lambda}{a}},\quad x_{0}z_{0}=\displaystyle{\frac{\lambda}{b}}, \quad x_{0}y_{0}=\displaystyle{\frac{\lambda}{c}},\quad \displaystyle{\frac{x_{0}}{a}} = \displaystyle{\frac{y_{0}}{b}}=\displaystyle{\frac{z_{0}}{c}}=\displaystyle{\frac{1}{3}},\quad V_{\text{max}}=\frac{abc}{27}. \nonumber$$

We may assume without loss of generality that $y>0$, so $A=ay$. $$L=\displaystyle{ay-\frac{\lambda}{2}\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right)},\quad \displaystyle{L_{x}=\frac{\lambda x}{a^{2}}},\quad x_{0}=0,\quad y_{0}=b,\quad A_{\text{max}}=ab. \nonumber$$

We must maximize $A^{2}=s(s-x)(s-y)(s-z)$ subject to $x+y+z=s$. $$L=-s(s-x)(s-y)(s-z)-\lambda(x+y+z) \nonumber$$ $$L_{x}=s(s-y)(s-z)-\lambda,\quad L_{y}=s(s-x)(s-z)-\lambda,\quad L_{z}=s(s-x)(s-y)-\lambda\quad \nonumber$$ $$s(s-y_{0})(s-z_{0})= s(s-x_{0})(s-z_{0})= s(s-x_{0})(s-y_{0})=\lambda,\quad x_{0}=y_{0}=z_{0}=\frac{s}{3}. \nonumber$$

We must maximize $V=8xyz$ subject to $\displaystyle{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}}=1.$ $$L=xyz-\frac{\lambda}{2}\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}\right) \nonumber$$ $$L_{x}=yz-\frac{\lambda x}{a^{2}}, \quad L_{y}=xz-\frac{\lambda y}{b^{2}}, \quad L_{z}=xy-\frac{\lambda z}{c^{2}} \nonumber$$ $$y_{0}z_{0}=\frac{\lambda x_{0}}{a^{2}}, \quad x_{0}z_{0}=\frac{\lambda y_{0}}{b^{2}}, \quad x_{0}y_{0}=\frac{\lambda z}{c^{2}}\quad \nonumber$$ $$\displaystyle{\frac{x_{0}^{2}}{a^{2}}}=\displaystyle{\frac{y_{0}^{2}}{b^{2}}}=\displaystyle{\frac{z_{0}^{2}}{c^{2}}} =\lambda x_{0}y_{0}z_{0} \nonumber$$ To satisfy the constraint, $x_{0}=\displaystyle{\frac{a}{\sqrt{3}}}$, $y_{0}=\displaystyle{\frac{b}{\sqrt{3}}}$, $z_{0}=\displaystyle{\frac{c}{\sqrt{3}}}$, so $V_{\text max}=\displaystyle{\frac{8abc}{3\sqrt{3}}}$.

Let $(x_{0},y_{0},z_{0})$ be the point on the plane closest to $(x_{1},y_{1},z_{1})$, so $$\tag{A} ax_{0}+by_{0}+cz_{0}=\sigma.$$ $$L=\displaystyle{\frac{(x-x_{1})^{2}+(y-y_{1})^{2}+(z-z_{1})^{2}}{2}}-\lambda(ax+by+cz) \nonumber$$ $$L_{x}=(x-x_{1})-\lambda a,\quad L_{y}=(y-y_{1})-\lambda b,\quad L_{z}=(z-z_{1})-\lambda c \nonumber$$ $$x_{0}=x_{1}+\lambda a,\quad y_{0}=y_{1}+\lambda b,\text{\quad and\quad} z_{0}=z_{1}+\lambda c, \tag{B}$$ $$\tag{C} d^{2}=\lambda^{2}(a^{2}+b^{2}+c^{2})$$ (A) and (B) imply that $$ax_{1}+by_{1}+cz_{1}+\lambda(a^{2}+b^{2}+c^{2})=\sigma, \nonumber$$ so $$\lambda=\frac{\sigma-ax_{1}-by_{1}-cz_{1}}{a^{2}+b^{2}+c^{2}}, \nonumber$$ and (C) implies that $$d=\frac{|\sigma-ax_{1}-by_{1}-cz_{1}|}{\sqrt{a^{2}+b^{2}+c^{2}}}. \nonumber$$

$\displaystyle{ L=\frac{1}{2}\sum_{i=1}^{n}\left[(x-x_{i})^{2}+(y-y_{i})^{2}+(z-z_{i})^{2}\right] -\lambda(ax+by+cz)}$ $$L_{x}=nx - \lambda a -\sum_{i=1}^{n}x_{i},\quad L_{y}=ny - \lambda b -\sum_{i=1}^{n}y_{i},\quad L_{z}=nz - \lambda b -\sum_{i=1}^{n}z_{i}. \nonumber$$ $$x_{0}=\displaystyle{\frac{1}{n}\left[\lambda a+\sum_{i=1}^{n}x_{i}\right]},\quad y_{0}=\displaystyle{\frac{1}{n}\left[\lambda b+\sum_{i=1}^{n}y_{i}\right]},\quad z_{0}=\displaystyle{\frac{1}{n}\left[\lambda c+\sum_{i=1}^{n}z_{i}\right]} \nonumber$$ $$ax_{0}+by_{0}+cz_{0}=\frac{1}{n} \left[\lambda(a^{2}+b^{2}+c^{2})+\sum_{i=1}^{n}(ax_{i}+by_{i}+cz_{i})\right] \nonumber$$ Since $ax_{0}+by_{0}+cz_{0}=\sigma$, $$\lambda=(a^{2}+b^{2}+c^{2})^{-1}\displaystyle{\sum_{i=1}^{n} (\sigma-ax_{i}-by_{i}-cz_{i})}. \nonumber$$

$L=\displaystyle{\frac{1}{2}}\displaystyle\left({\sum_{i=1}^{n}(x_{i}-c_{i})^{2}- \lambda\sum_{i=1}^{n}x_{i}^{2}}\right)$,  $L_{x_{i}}=x_{i}-c_{i}-\lambda x_{i}$,  $x_{i0}=(1-\lambda)^{-1} c_{i}$

$\displaystyle{\sum_{i=1}^{n}x_{i0}^{2}=(1-\lambda)^{-2}\sum_{j=1}^{n}c_{j}^{2}}=1$, so $\lambda =1\pm \mu$ where $\mu =\displaystyle{\left(\sum_{j=1}^{n}c_{j}^{2}\right)^{1/2}}$ Since $x_{i0}=c_{i}+\lambda x_{i0}$ and $\displaystyle{\sum_{i=1}^{n}x_{i0}^{2}}=1$, $\displaystyle{\sum_{i=1}^{n}(x_{i0}-c_{i})^{2}=\lambda^{2}}$. Since $x_{i0}=(1-\lambda)^{-1}c_{i}$, the constrained maximum is $(1+\mu)^{2}$, attained with $x_{i0}=-c_{i}/\mu$, $1\le i\le n$, and the constrained minimum is $(1-\mu)^{2}$, attained with $x_{i0}=c_{i}/\mu$, $1\le i\le n$.

[exer:19].

$L=xy+xz+yz-\displaystyle{\frac{\lambda}{2}(x^{2}+y^{2}+z^{2})}$

$$L_{x}=y+z-\lambda x,\quad L_{y}=x+z-\lambda y,\quad L_{z}=x+y-\lambda z \nonumber$$

$$\left[\begin{array}{ccccccc} 0&1&1\\1&0&1\\1&1&0 \end{array}\right] \left[\begin{array}{ccccccc} x_{0}\\y_{0}\\z_{0} \end{array}\right]=\lambda \left[\begin{array}{ccccccc} x_{0}\\y_{0}\\z_{0} \end{array}\right]. \nonumber$$ The eigenvalues of the matrix are $2$ and $-1$, which are therefore the extremes of $Q$ subject to the constraint.

$L=\displaystyle{\frac{3x^{2}+2y^{2}+3z^{2}+2xz}{2}}-\displaystyle{\frac{\lambda}{2}(x^{2}+y^{2}+z^{2})}$

$$L_{x}=3x+z-\lambda x,\quad L_{y}=2y-\lambda y,\quad L_{z}=3z+x-\lambda z \nonumber$$

$$\left[\begin{array}{ccccccc} 3&0&1\\0&2&0\\1&0&3 \end{array}\right] \left[\begin{array}{ccccccc} x_{0}\\y_{0}\\z_{0} \end{array}\right]=\lambda \left[\begin{array}{ccccccc} x_{0}\\y_{0}\\z_{0} \end{array}\right] \nonumber$$ The largest and smallest eigenvalues of the matrix are $4$ and $2$, which are therefore the extremes of $Q$ subject to the constraint.

$L=\displaystyle{\frac{x^{2}+8xy+4y^{2}-\lambda(x^{2}+2xy+4y^{2})}{2}}$ $$\begin{aligned} L_{x}&=&(x+4y)-\lambda(x+y)=(1-\lambda)x+(4-\lambda)y \\ L_{y}&=& (4x+4y)-\lambda(x+4y)=(4-\lambda)x+4(1-\lambda y)\end{aligned} \nonumber$$ $$\left[\begin{array}{ccccccc} 1-\lambda & 4-\lambda \\ 4-\lambda &4(1-\lambda) \end{array}\right] \left[\begin{array}{ccccccc} x_{0}\\ y_{0} \end{array}\right]= \left[\begin{array}{ccccccc} 0\\0 \end{array}\right], \nonumber$$ so $$4(\lambda-1)^{2}-(\lambda-4)^{2}=3(\lambda-2)(\lambda+2)=0. \nonumber$$ If $\lambda=2$, then $x_{0}=2y_{0}$. To satisfy the constraint, $(x_{0},y_{0})=\pm\left(\frac{1}{\sqrt3},\frac{1}{2\sqrt3}\right)$ and $f(x_{0},y_{0})=2$. If $\lambda=-2$, then $x_{0}=-2y_{0}$. To satisfy the constraint, $(x_{0},y_{0})=\pm\left(-\frac{1}{\sqrt3},\frac{1}{2\sqrt3}\right)$ and $f(x_{0},y_{0})=-\frac{2}{3}$.

$L=\alpha+\beta xy-\displaystyle{\frac{\lambda}{2}(ax+by)^{2}}$,  $L_{x}=\beta y-\lambda a(ax+by)$,  $L_{y}=\beta x-\lambda b(ax+by)$

$x_{0}=\displaystyle{\frac{\lambda b(ax_{0}+by_{0})}{\beta}}$, $y_{0}=\displaystyle{\frac{\lambda a(ax_{0}+by_{0})}{\beta}}$, $(x_{0},y_{0})=\displaystyle{\pm\left(\frac{\lambda b}{\beta},\frac{\lambda a}{\beta}\right)}$

$ax_{0}+by_{0}=\displaystyle{\frac{2\lambda ab}{\beta}=\pm1}$, $\lambda=\displaystyle{\pm\displaystyle{\frac{\beta}{2ab}}}$, $(x_{0},y_{0})=\displaystyle{\pm\left(\frac{1}{2a},\frac{1}{2b}\right)}$

$(\alpha+\beta x_{0}y_{0})_\text{max}=\displaystyle{\alpha+\frac{|\beta|}{4|ab|}}$, $(\alpha+\beta x_{0}y_{0})_\text{min}=\alpha-\displaystyle{\frac{|\beta|}{4|ab|}}$

$L=x+y^{2}+2z-\displaystyle{\frac{\lambda}{2}(4x^{2}+9y^{2}-36z^{2})}$

$$L_{x}=1-4\lambda x,\; L_{y}=2y-9\lambda y, \; L_{z}=2+36\lambda z \nonumber$$ $$x_{0}=\displaystyle{\frac{1}{4\lambda}},\; z_{0}=-\displaystyle{\frac{1}{18\lambda}}=-\frac{2}{9}x_{0}, \nonumber$$ and either $y_{0}=0$ or $\lambda=\displaystyle{\frac{2}{9}}$. If $y_{0}=0$, then $$36=4x_{0}^{2}-36z_{0}^{2}=\left(4-36\left(\frac{2}{9}\right)^{2}\right)x_{0}^{2} =\frac{20}{9}x_{0}^{2},\text{\quad so\quad} (x_{0},z_{0})=\pm \left(\frac{9}{\sqrt{5}},-\frac{2}{\sqrt{5}}\right) \nonumber$$ and $f(x_{0},0,z_{0})=x_{0}+2z_{0}=\pm\sqrt{5}$. If $\lambda=\displaystyle{\frac{2}{9}}$, then $x_{0}=\displaystyle{\frac{9}{8}}$ and $z_{0}=-\displaystyle{\frac{1}{4}}$, so $$9y_{0}^{2}=36(1+z_{0}^{2})-4x_{0}^{2}=36\left(1+\frac{1}{16}\right)-4\left(\frac{81}{64}\right) =\frac{531}{16}, \text{\; so\;\;}y_{0}=\pm\displaystyle{\frac{\sqrt{59}}{4}}. \nonumber$$ Therefore, the constrained maximum is $f\left(\frac{9}{8},\frac{\sqrt{59}}{4},-\frac{1}{4}\right)= f\left(\frac{9}{8},-\frac{\sqrt{59}}{4},-\frac{1}{4}\right)=\frac{73}{16}$ and the constrained minimum is $f\left(-\frac{9}{\sqrt{5}},0,\frac{2}{\sqrt{5}}\right)=-\sqrt{5}$.

$L=(x+z)(y+w)-\displaystyle{\frac{\lambda}{2}(x^{2}+y^{2}+z^{2}+w^{2})}$

$$L_{x}=y+w-\lambda x,\, L_{y}=x+z-\lambda y,\, L_{z}=y+w -\lambda z, \, L_{w}=x+z-\lambda w \nonumber$$ $$x_{0}+z_{0}=\lambda y_{0}=\lambda w_{0}, \quad y_{0}+w_{0}=\lambda x_{0}=\lambda z_{0} \nonumber$$

If $\lambda=0$, then all $(x_{0},y_{0},-x_{0},w_{0})$ with $2x_{0}^{2}+y_{0}^{2}+w_{0}^{2}=1$ and all $(x_{0},y_{0},z_{0},-y_{0})$ with $x_{0}^{2}+2y_{0}^{2}+z_{0}^{2}=1$ are constrained critical points, with $f(x_{0},y_{0},-x_{0},w_{0})=0$ and $f(x_{0},y_{0},z_{0},-y_{0})$.

If $\lambda \ne0$, then $y_{0}=w_{0}$ and $x_{0}=z_{0}$, so $$2x_{0}=\lambda y_{0},\quad 2y_{0}=\lambda x_{0},\quad 2z_{0}=\lambda w_{0},\quad 2w_{0} =\lambda z_{0}, \nonumber$$ and $$2x_{0}=\lambda y_{0} =\frac{\lambda}{2}(2y_{0})=\frac{\lambda^{2}}{2}x_{0} \text{\; and \;\;}2z_{0}=\lambda w_{0}=\frac{\lambda}{2}(2w_{0})= \frac{\lambda^{2}}{2}z_{0}. \nonumber$$ If $\lambda\ne2$, then $x_{0}=y_{0}=z_{0}=w_{0}$, which does not satisfy the constraint. If $\lambda=2$, then $$x_{0}=y_{0}=z_{0}=w_{0}=\pm\frac{1}{2}\text{\; and\;\;} (x_{0}+z_{0})(y_{0}+w_{0})=1. \nonumber$$ If $\lambda=-2$, then $$x_{0}=-y_{0}=z_{0}=-w_{0}=\pm\frac{1}{2} \text{\; and\;\;} (x_{0}+z_{0})(y_{0}+w_{0})=-1. \nonumber$$ Therefore, the constrained maximum is $1$, attained at $\pm\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$ the constrained minimum is $-1$, attained at $\pm\left(\frac{1}{2},-\frac{1}{2},\frac{1}{2},-\frac{1}{2}\right)$.

$L=(x+z)(y+w)-\displaystyle{\frac{\lambda}{2}(x^{2}+y^{2})} -\displaystyle{\frac{\mu}{2}(z^{2}+w^{2})}$

$$L_{x}=y+w-\lambda x,\, L_{y}=x+z-\lambda y,\, L_{z}=y+w -\mu z, \, L_{w}=x+z-\mu w \nonumber$$ $$\tag{A} x_{0}+z_{0}=\lambda y_{0}=\mu w_{0}, \quad y_{0}+w_{0}=\lambda x_{0}=\mu z_{0}$$

If $\lambda=\mu=0$, then $z_{0}=-x_{0}$ and $w_{0}=-y_{0}$, $(x_{0},y_{0},-x_{0},-y_{0})$ satisfies the constraints and $f(x_{0},y_{0},-x_{0},-y_{0})=0$ for all $(x_{0},y_{0})$ such that $x_{0}^{2}+y_{0}^{2}=1$.

If $\lambda=0$ and $\mu\ne0$, then $z_{0}=w_{0}=0$, which does not satisfy the constraint $z^{2}+y^{2}=1$. If $\mu=0$ and $\lambda\ne0$, then $x_{0}=y_{0}=0$, which does not satisfy the constraint $x^{2}+y^{2}=1$.

Now assume that $\lambda$, $\mu\ne0$. From (A), $\lambda(x_{0}^{2}+y_{0}^{2})=\mu^{2}(z_{0}^{2}+w_{0}^{2})$, so $\lambda=\pm\mu$. If $\lambda=-\mu$, (A) implies that $y_{0}=-w_{0}$ and $x_{0}=-z_{0}$, so again $(x_{0},y_{0},-x_{0},-y_{0})$ satisfies the constraints and $f(x_{0},y_{0},-x_{0},-y_{0})=0$ for all $(x_{0},y_{0})$ such that $x_{0}^{2}+y_{0}^{2}=1$.

If $\lambda=\mu$, (A) becomes $$x_{0}+z_{0}=\lambda y_{0}= \lambda w_{0},\quad y_{0}+w_{0}=\lambda x_{0}=\lambda z_{0}, \nonumber$$ so $y_{0}=w_{0}$, $x_{0}=z_{0}$, $2x_{0}=\lambda y_{0}$, and $2y_{0}=\lambda x_{0}$, $4x_{0}=2\lambda y_{0}=\lambda^{2}x_{0}$, so $\lambda=\pm2$.

If $\lambda=2$, $x_{0}=y_{0}=z_{0}=w_{0}$. To satisfy the constraints,

$$(x_{0},y_{0},z_{0},w_{0})=\pm\left( \frac{1}{\sqrt2}, \frac{1}{\sqrt2}, \frac{1}{\sqrt2}, \frac{1}{\sqrt2} \right), \text{\; so\;\;} \nonumber$$ and the constrained maximum is $f(x_{0},y_{0},z_{0},w_{0})=2$.

If $\lambda=-2$, $x_{0}=-y_{0}=z_{0}=-w_{0}$. To satisfy the constraints, $$(x_{0},y_{0},z_{0},w_{0})=\pm\left( \frac{1}{\sqrt2}, - \frac{1}{\sqrt2}, \frac{1}{\sqrt2}, -\frac{1}{\sqrt2} \right), \text{\; so\;\;} \nonumber$$ and the constrained minimum is $f(x_{0},y_{0},z_{0},w_{0})=-2$.

$L=(x+z)(y+w)-\displaystyle{\frac{\lambda}{2}}(x^{2}+z^{2}) -\displaystyle{\frac{\mu}{2}}(y^{2}+w^{2})$

$$L_{x}=y+w-\lambda x,\; L_{y}=x+z-\mu y,\; L_{w}=x+z-\mu w,\; L_{z}=y+w-\lambda z \nonumber$$

$y_{0}+w_{0}=\lambda x_{0}$, $x_{0}+z_{0}=\mu y_{0}$, $x_{0}+z_{0}=\mu w_{0}$, $y_{0}+w_{0}=\lambda z_{0}$

If $\mu=0$, then $x_{0}=-z_{0}$, so the constrained critical points are $\pm\left(\frac{1}{\sqrt2},y_{0},-\frac{1}{\sqrt2},w_{0}\right)$ for all $(y_{0},w_{0})$ such that $y_{0}^{2}+w_{0}^{2}=1$; $f=0$ at all such points.

If $\lambda=0$, then $y_{0}=-w_{0}$, so the constrained critical points are $\pm\left(x_{0},\frac{1}{\sqrt2},z_{0},-\frac{1}{\sqrt2}\right)$ for all $(x_{0},z_{0})$ such that $x_{0}^{2}+z_{0}^{2}=1$; $f=0$ at all such points.

Now suppose that $\lambda\mu\ne0$. Since $\lambda x_{0}=\lambda z_{0}$ and $\mu y_{0}=\mu w_{0}$, $x_{0}=z_{0}$ and $y_{0}=w_{0}$. Therefore, $(x_{0},z_{0})=\pm\left(\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)$ and $(y_{0},w_{0})=\pm\left(\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)$, so the constrained maximum is $2$, attained at $\pm(\frac{1}{\sqrt2},\frac{1}{\sqrt2},\frac{1}{\sqrt2},\frac{1}{\sqrt2})$, and constrained minimum is $-2$, attained $\pm\left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2},\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right)$,

$L=\displaystyle{\frac{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}{2}}- \displaystyle{\frac{\lambda}{2}}(x_{1}^{2}+y_{1}^{2})-\mu x_{2}y_{2}$

$$L_{x_{1}}=x_{1}-x_{2}-\lambda x_{1},\; L_{x_{2}}=x_{2}-x_{1}-\mu y_{2},\; L_{y_{1}}=y_{1}-y_{2}-\lambda y_{1},\; L_{y_{2}}=y_{2}-y_{1}-\mu x_{2} \nonumber$$

(i) $x_{10}-x_{20}=\lambda x_{10}$,(ii) $y_{10}-y_{20}=\lambda y_{10}$

(iii) $x_{20}-x_{10}=\mu y_{20}$, (iv) $y_{20}-y_{10}=\mu x_{20}$

Since $0<x_{10}<x_{20}$ and $0<y_{10}<y_{20}$, $\lambda<0$ and $\mu>0$ Since $x_{20}\ne0$, $\lambda \ne 1$, (i) and (ii) imply that (v) $x_{10}y_{20}=y_{10}x_{20}$. From (i) and (iii), (vi) $\lambda x_{10}=-\mu y_{20}$; from (ii) and (iv), (vii) $\lambda y_{10}=-\mu x_{20}$. Since $x_{20}y_{20}=1$, (vi) and (vii) imply that $x_{10}x_{20}=y_{10}y_{20}$. This and (v) imply that $$\frac{x_{10}}{y_{10}}=\frac{x_{20}}{y_{20}}=\frac{y_{20}}{x_{20}}. \nonumber$$ Therefore, $x_{20}=y_{20}=1$ and $x_{10}=y_{10}=\frac{1}{\sqrt2}$, so $$(x_{10}-x_{20})^{2}+(y_{10}-y_{20})^{2}= 2\left(1-\frac{1}{\sqrt 2}\right)^{2} \nonumber$$ and the distance between the curves is $\sqrt2-1$.

$L=\displaystyle{\frac{1}{2}}\left(\displaystyle{\frac{x^{2}}{\alpha^{2}}+\frac{y^{2}}{\beta^{2}}} +\frac{z^{2}}{\gamma^{2}}\right) -\lambda (ax+by+cz)$

$$L_{x}=\frac{x}{\alpha^{2}}-\lambda a,\quad L_{y}=\frac{y}{\beta^{2}}-\lambda b,\quad L_{z}=\frac{z}{\gamma^{2}}-\lambda c \nonumber$$ $$x_{0}=\lambda a \alpha^{2},\quad y_{0}=\lambda b \beta^{2},\quad z_{0}=\lambda c \gamma^{2} \nonumber$$ $$ax_{0}+by_{0}+cz_{0}= \lambda[(a \alpha)^{2}+(b\beta)^{2}+(c\gamma^{2})]=d, \quad \lambda=\frac{d}{(a\alpha)^{2}+(b\beta^{2})+(c\gamma)^{2}} \nonumber$$ $$\frac{x_{0}^{2}}{\alpha^{2}}+\frac{y_{0}^{2}}{\beta^{2}} +\frac{z_{0}^{2}}{\gamma^{2}}= \lambda^{2}[(a\alpha)^{2}+(b\beta)^{2}+(c\gamma)^{2}] = \frac{d^{2}}{(a\alpha)^{2}+(b\beta^{2})+(c\gamma)^{2}}. \nonumber$$

$\displaystyle{L(x_{1},x_{2},\dots,x_{n})=\frac{(x_{1}-c_{1})^{2}+(x_{2}-c_{2})^2+ \cdots+(x_{n}-c_{n})^2}{2}}$ $$-\lambda(a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}) \nonumber$$ $L_{x_{i}}=x_{i}-c_{i}-\lambda a_{i}$, $1\le i\le n$. We must choose $\lambda$ so that if $x_{i0}=c_{i}+\lambda a_{i}$, $1\le i\le n$, then $$\begin{aligned} a_{1}x_{10}+a_{2}x_{20}+\cdots+a_{n}x_{n0}&=&a_{1}c_{1}+a_{2}c_{2}+\dots + a_{n}c_{n}\\ &+& \lambda (a_{1}^{2}+a_{2}^{2}+\cdots+ a_{n}^{2})=d,\end{aligned} \nonumber$$ which holds if and only if $$\lambda=\frac{d-a_{1}c_{1}-a_{2}c_{2}-\cdots-a_{n}c_{n}} {a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}}. \nonumber$$ Therefore, $$x_{i0}=c_{i}+ \frac{(d-a_{1}c_{1}-a_{2}c_{2}-\cdots-a_{n}c_{n})a_{i}} {a_{1}^{2}+a_{2}^{2}+\cdots a_{n}^{2}},\quad 1\le i\le n, \nonumber$$ and the distance from $(x_{10},x_{10},\dots,x_{n0})$ to $(c_{1},c_{2},\dots,c_{n})$ is $$\frac{|(d-a_{1}c_{1}-a_{2}c_{2}-\cdots-a_{n}c_{n})a_{i}|} {\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots a_{n}^{2}}}. \nonumber$$

$L=\displaystyle{\frac{1}{2}\sum_{i=1}^{n}a_{i}x_{i}^{2}- \frac{\lambda}{4}\sum_{i=1}^{n}b_{i}x_{i}^{4}}$, $L_{x_{i}}=a_{i}x_{i}-\lambda b_{i}x_{i}^{3}$, $a_{i}x_{i0}^{2}=\lambda b_{i}x_{i0}^{4}$

$\displaystyle{\sum_{i=1}^{n}a_{i}x_{i0}^{2}=\lambda \sum_{i=1}^{n} b_{i}x_{i0}^{4}=\lambda}$, $x_{i0}^{2}=\displaystyle{\frac{a_{i}}{\lambda b_{i}}}$, $\lambda=\displaystyle{\sum_{i=1}^{n}a_{i}x_{i0}^{2}= \frac{1}{\lambda}\sum_{i=1}^{n}\frac{a_{i}^{2}}{b_{i}}}$, $\lambda=\displaystyle{\left(\sum_{i=1}^{n}\frac{a_{i}^{2}}{b_{i}}\right)^{1/2}}$

[exer:31].

$L=\displaystyle{\frac{1}{p}\sum_{i=1}^{n}a_{i}x_{i}^{p}- \frac{\lambda}{q}\sum_{i=1}^{n}b_{i}x_{i}^{q}}$, $L_{x_{i}}=a_{i}x_{i}^{p}-\lambda b_{i}x_{i}^{q}$, $a_{i}x_{i0}^{p}=\lambda b_{i}x_{i0}^{q}$

$\displaystyle{\sum_{i=1}^{n}a_{i}x_{i0}^{p}=\lambda \sum_{i=1}^{n} b_{i}x_{i0}^{q}=\lambda}$, $x_{i0}^{q-p}=\displaystyle{\frac{a_{i}}{\lambda b_{i}}}$, $x_{i0}=\displaystyle{\left(\frac{a_{i}}{\lambda b_{i}}\right)^{1/(q-p)}}$, $x_{i0}^{p}=\displaystyle{\left(\frac{a_{i}}{\lambda b_{i}}\right)^{p/(q-p)}}$

$$\lambda=\sum_{i=1}^{n}a_{i}x_{i0}^{p}=\lambda^{p/(p-q)} \sum_{i=1}^{n}a_{i}^{q/(q-p)} b_{i}^{p/(p-q)},\quad \nonumber$$ $$\lambda^{q/(q-p)}= \sum_{i=1}^{n}a_{i}^{q/(q-p)} b_{i}^{p/(p-q)},\quad \lambda= \left(\sum_{i=1}^{n}a_{i}^{q/(q-p)} b_{i}^{p/(p-q)}\right)^{1-p/q}= \sum_{i=1}^{n}a_{i}x_{i0}^{p} \nonumber$$ $\lambda$ is the constrained maximum if $p<q$, the constrained minimum if $p>q$, undefined if $p=q$.

[exer:32]. $L=\displaystyle{\frac{x^{2}+2y^{2}+z^{2}+w^{2}}{2}}-\lambda(x+y+z+3w)-\mu(x+y+2z+w)$ $$L_{x}=x-\lambda-\mu, \quad L_{y}=2y-\lambda-\mu, \quad L_{z}=z-\lambda-2\mu, \quad L_{w}=w-3\lambda-\mu \nonumber$$ $$x_{0}=\lambda+\mu, \quad y_{0}=\frac{\lambda+\mu}{2}, \quad z_{0}=\lambda+2\mu, \quad w_{0}=3\lambda+\mu \nonumber$$ $$x_{0}+y_{0}+z_{0}+3w_{0}=\frac{23}{2}\lambda+\frac{13}{2}\mu=1, \quad x_{0}+y_{0}+2z_{0}+w_{0}=\frac{13}{2}\lambda+\frac{13}{2}\mu=2 \nonumber$$ $$\lambda=-\frac{1}{5},\, \mu=\frac{33}{65},\, x_{0}=\frac{4}{13},\, y_{0}=\frac{2}{13},\, z_{0}=\frac{53}{65},\, w_{0}=-\frac{6}{65},\,\ \min=\frac{689}{845} \nonumber$$

$\displaystyle{L=\frac{1}{2}\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}\right) -\lambda(p_{1}x+p_{2}y+p_{3}z)}$

$L_{x}=\displaystyle{\frac{x}{a^{2}}}-\lambda p_{1}$, $L_{y}=\displaystyle{\frac{y}{b^{2}}}-\lambda p_{2}$, $L_{z}=\displaystyle{\frac{z}{c^{2}}}-\lambda p_{3}$

$x_{0}=\lambda p_{1}a^{2}$, $y_{0}=\lambda p_{2}b^{2}$, $z_{0}=\lambda p_{3}c^{2}$

$p_{1}x_{0}+p_{2}y_{0}+p_{3}z_{0}= \lambda(p_{1}^{2}a^{2}+p_{2}^{2}b^{2}+p_{3}^{2}c^{2})=d$

$\lambda=\displaystyle{\frac{d}{p_{1}^{2}a^{2}+p_{2}^{2}b^{2}+p_{3}^{2}c^{2}}}$, $\displaystyle{\frac{x_{0}}{a}=\lambda p_{1}a}$, $\displaystyle{\frac{y_{0}}{b}=\lambda p_{2}b}$, $\displaystyle{\frac{z_{0}}{b}=\lambda p_{3}c}$

$$\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}+\frac{z_{0}^{2}}{c^{2}} =\lambda^{2}(p_{1}^{2}a^{2}+p_{2}^{2}b^{2}+p_{3}^{2}c^{2}) =\frac{d^{2}}{p_{1}^{2}a^{2}+p_{2}^{2}b^{2}+p_{3}^{2}c^{2}} \nonumber$$

$L=p_{1}x+p_{2}y+p_{3}z- \displaystyle{\frac{\lambda}{2}\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+ \frac{z^{2}}{c^{2}}\right)}$

$L_{x}=p_{1}-\lambda\displaystyle{\frac{x}{a^{2}}}$, $L_{y}=p_{2}-\lambda\displaystyle{\frac{y}{b^{2}}}$, $L_{z}=p_{3}-\lambda\displaystyle{\frac{z}{c^{2}}}$

$x_{0}=\displaystyle{\frac{p_{1}a^{2}}{\lambda}}$, $y_{0}=\displaystyle{\frac{p_{2}b^{2}}{\lambda}}$, $z_{0}=\displaystyle{\frac{p_{3}c^{2}}{\lambda}}$

$$\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}+\frac{z_{0}^{2}}{c^{2}} =\frac{p_{1}^{2}a^{2}+p_{2}^{2}b^{2}+p_{3}^{2}c^{2}}{\lambda^{2}}=1 \nonumber$$ $$\lambda=\pm(p_{1}^{2}a^{2}+p_{2}^{2}b^{2}+p_{3}^{2}c^{2})^{1/2} \nonumber$$ $$p_{1}x_{0}+p_{2}y_{0}+p_{3}z_{0}= \frac{p_{1}^{2}a^{2}+p_{2}^{2}b^{2}+p_{3}^{2}c^{2}}{\lambda} =\pm (p_{1}^{2}a^{2}+p_{2}^{2}b^{2}+p_{3}^{2}c^{2})^{1/2} \nonumber$$

$L=\displaystyle{\frac{(x+1)^{2}+(y-2)^{2}+(z-3)^{2}}{2}}-\lambda(x+2y-3z)-\mu(2x-y+2z)$

$$L_{x}=x+1-\lambda-2\mu,\quad L_{y}=y-2-2\lambda +\mu,\quad L_{z}=z-3+3\lambda-2\mu \nonumber$$ $$x_{0}=-1+\lambda+2\mu,\quad y_{0}=2+2\lambda-\mu, \quad z_{0}=3-3\lambda+2\mu \nonumber$$ $$x_{0}+2y_{0}-3z_{0}-4=-10+14\lambda-6\mu,\quad 2x_{0}-y_{0}+2z_{0}-5=-3-6\lambda+9\mu \nonumber$$ $$7\lambda-3\mu=5, \quad -2\lambda+3\mu=1,\quad \lambda=\frac{18}{15}, \quad \mu=\frac{17}{15} \nonumber$$ $$(x_{0},y_{0},z_{0})=\left(\frac{37}{15}, \frac{49}{15},\frac{25}{15}\right),\quad \nonumber$$ $$\sqrt{(x_{0}+1)^{2}+(y_{0}-2)^{2}+(z_{0}-3)^{2}} =\left[\left(\frac{52}{15}\right)+\left(\frac{19}{15}\right)^{2}+ \left(\frac{20}{15}\right)^{2}\right]^{1/2}=\sqrt{\frac{693}{45}} \nonumber$$

$L=2x+y+2z-\displaystyle{\frac{\lambda}{2}}(x^{2}+y^{2})-\mu(x+z)$

$$L_{x}=2-\lambda x-\mu,\quad L_{y}=1-\lambda y,\quad L_{z}=2-\mu \nonumber$$ $\mu=2$, so $\lambda x_{0}=0$. Since $\lambda y_{0}=1$, $\lambda\ne0$; hence, $x_{0}=0$. Since $x_{0}^{2}+y_{0}^{2}=4$, $y_{0}=\pm2$. Therefore, $(0,2,2)$ and $(0,-2,2)$, are constrained extreme points, and the constrained extreme values are $f(0,2,2)=6$ and $f(0,-2,2)=2$.

Let $(x_{1},y_{1})$ be on the parabola, $(x_{2},y_{2})$ on the line. $$L=\frac{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}{2} -\lambda(y_{1}-x_{1}^{2})-\mu(x_{2}+y_{2}). \nonumber$$ $$L_{x_{1}}=x_{1}-x_{2}+2\lambda x_{1},\, L_{x_{2}}=x_{2}-x_{1}-\mu,\, \nonumber$$ $$L_{y_{1}}=y_{1}-y_{2}-\lambda,\, L_{y_{2}}=y_{2}-y_{1}-\mu \nonumber$$ $$\begin{aligned} x_{10}-x_{20}&=&-2\lambda x_{10}\\ x_{20}-x_{10}&=&\mu\\ y_{10}-y_{20}&=&\lambda\text{\quad (i)}\\ y_{20}-y_{10}&=&\mu\text{\quad (ii)}\end{aligned} \nonumber$$ From (i) and (ii), $\lambda=-\mu$, so $$\begin{aligned} x_{10}-x_{20}&=& 2\mu x_{10}\text{\quad (i)}\\ x_{20}-x_{10}&=&\mu\text{\quad\quad \;\, (ii)}\\ y_{20}-y_{10}&=&\mu\text{\quad\quad \;\, (iii)}\end{aligned} \nonumber$$ From (i) and (ii), $x_{10}=-1/2$, so $y_{10}=1+x_{10}^{2}=5/4$ and $$2\mu=x_{20}+y_{20}-x_{10}-y_{10}=-1+\frac{1}{2}-\frac{5}{4}=-\frac{7}{4}, \nonumber$$ since $x_{20}+y_{20}=-1$ (constraint). Therefore, $\mu=-7/8$ so (ii) and (iii) imply that $$x_{20}=x_{10}=\mu=-\frac{1}{2}-\frac{7}{8}=-\frac{11}{8} \text{\; and\;\;} y_{20}=y_{10}-\frac{7}{8}=\frac{5}{4}-\frac{7}{8}=\frac{3}{8}. \nonumber$$ The distance between the line and the parabola is $$\sqrt{(x_{10}-x_{20})^{2}+(y_{10}-y_{20})^{2}}=\frac{7}{4\sqrt{2}}. \nonumber$$

Let $(x_{1},y_{1},z_{1})$ be on the ellipsoid and $(x_{2},y_{2},z_{2})$ be on the plane. $$L= \frac{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}}{2} -\frac{\lambda}{2}(3x_{1}^{2}+9y_{1}^{2}+6z_{1}^{2}) -\mu(x_{2}+y_{2}+2z_{2}). \nonumber$$ $$L_{x_{1}}=x_{1}-x_{2}-3\lambda x_{1}=0,\quad L_{y_{1}}=y_{1}-y_{2}-9\lambda y_{1}=0,\quad L_{z_{1}}=z_{1}-z_{2}-6\lambda z_{1} \nonumber$$ $$L_{x_{2}}=x_{2}-x_{1}-\mu,\quad Ly_{2}=y_{2}-y_{1}-\mu,\quad L_{z_{2}}=z_{2}-z_{1}-2\mu \nonumber$$ $$\begin{aligned} x_{10}-x_{20}&=& 3\lambda x_{10}\\ y_{10}-y_{20}&=& 9\lambda y_{10}\\ z_{10}-z_{20}&=& 6\lambda z_{10}\\ x_{20}-x_{10}&=& \mu\\ y_{20}-y_{10}&=& \mu \\ z_{20}-z_{10}&=& 2\mu\end{aligned} \nonumber$$ Therefore, $3\lambda x_{10}=-\mu$, $9\lambda y_{10}=-\mu$, and $3\lambda z_{10}=-\mu$, so $y_{10}=x_{1}/3$ and $z_{10}=x_{10}$. Since $(x_{10},x_{10}/3,x_{10})$ is on the ellipsoid if and only if $x_{10}=\pm1$, either $$\text{\; (a)\;\;}(x_{10},y_{10},z_{10})=\left(1,\frac{1}{3},1\right) \text{\; or \quad (b)\;\;} (x_{10},y_{10},z_{10})=\left(-1,-\frac{1}{3},-1\right). \nonumber$$ Since $$\tag{A} x_{2}=x_{1}+\mu,\quad y_{2}=y_{1}+\mu,\quad z_{2}=z_{1}+2\mu,$$ $$\tag{B} (x_{10}-x_{20})^{2}+(y_{10}-y_{20})^{2}+(z_{10}-z_{20})=6\mu^{2}, \text{\; so\;\;} d=\mu.\sqrt{6}.$$ Since $3x_{20}+3y_{20}+6z_{20}=70$, (A) implies that $$\mu=\frac{70-3x_{10}-3y_{10}-6z_{10}}{18}, \nonumber$$

In Case (a) $\mu=\frac{10}{3}$ so (A) implies that $d=\frac{10\sqrt{6}}{3}$ In case (b) $\mu=\frac{40}{9}>\frac{10}{3}$, so the distance between the plane and the ellipsoid is $\frac{10\sqrt{6}}{3}$.

$L=xy+yz+zx-\displaystyle{\frac{\lambda}{2} \left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}\right)}$ $$L_{x}=y+z-\lambda\frac{x}{a^{2}},\quad L_{y}=z+x-\lambda\frac{y}{b^{2}},\quad L_{z}=x+y-\lambda\frac{z}{c^{2}} \nonumber$$ $$y_{0}+z_{0}=\lambda\frac{x_{0}}{a^{2}},\quad z_{0}+x_{0}=\lambda\frac{y_{0}}{b^{2}},\quad x_{0}+y_{0}-\lambda\frac{z_{0}}{c^{2}} \nonumber$$ $$\left[\begin{array}{ccccccc} 0&a^{2}&a^{2}\\ b^{2}&0&b^{2}\\c^{2}&c^{2}&0 \end{array}\right] \left[\begin{array}{ccccccc} x_{0}\\y_{0}\\z_{0} \end{array}\right]=\lambda \left[\begin{array}{ccccccc} x_{0}\\y_{0}\\z_{0} \end{array}\right] \nonumber$$ $$x_{0}(y_{0}+z_{0})=\lambda\frac{x_{0}^{2}}{a^{2}},\quad y_{0}(z_{0}+x_{0})=\lambda\frac{y_{0}^{2}}{b^{2}},\quad z_{0}(x_{0}+y_{0})=\lambda\frac{z_{0}^{2}}{c^{2}}, \nonumber$$

$$x_{0}(y_{0}+z_{0})+ y_{0}(z_{0}+x_{0})+ z_{0}(x_{0}+y_{0})= \lambda\left(\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}+\frac{z_{0}^{2}}{c^{2}}\right) =\lambda \nonumber$$

$L=xy+2yz+2zx-\lambda \displaystyle{\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}\right)}$ $$L_{x}=y+2z-2\lambda\frac{x}{a^{2}},\quad L_{y}=x+2z-2\lambda\frac{y}{b^{2}},\quad L_{z}=2x+2y-2\lambda\frac{z}{c^{2}} \nonumber$$ $$y_{0}+2z_{0}=2\lambda\frac{x_{0}}{a^{2}},\quad x_{0}+2z_{0}=2\lambda\frac{y_{0}}{b^{2}},\quad 2x_{0}+2y_{0}-2\lambda\frac{z_{0}}{c^{2}} \nonumber$$

$$\left[\begin{array}{ccccccc} 0&a^{2}/2&a^{2}\\ b^{2}/2&0&b^{2}\\c^{2}&c^{2}&0 \end{array}\right] \left[\begin{array}{ccccccc} x_{0}\\y_{0}\\z_{0} \end{array}\right]=\lambda \left[\begin{array}{ccccccc} x_{0}\\y_{0}\\z_{0} \end{array}\right]. \nonumber$$

$$x_{0}(y_{0}+2z_{0})=2\lambda\frac{x_{0}^{2}}{a^{2}},\quad y_{0}(x_{0}+2z_{0})=2\lambda\frac{y_{0}^{2}}{b^{2}};\quad z_{0}(2x_{0}+2y_{0})=2\lambda\frac{z_{0}^{2}}{c^{2}}, \nonumber$$ $$\frac{x_{0}(y_{0}+2z_{0})+ y_{0}(x_{0}+2z_{0})+ z_{0}(2x_{0}+2y_{0})}{2}= \lambda\left(\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}+\frac{z_{0}^{2}}{c^{2}}\right) =\lambda, \nonumber$$

$L=xz+yz-\displaystyle{\frac{\lambda}{2} \left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}\right)}$ $$L_{x}=z-\lambda\frac{x}{a^{2}},\quad L_{y}=z-\lambda\frac{y}{b^{2}},\quad L_{z}=x+y-\lambda\frac{z}{c^{2}} \nonumber$$ $$z_{0}=\lambda\frac{x_{0}}{a^{2}},\quad z_{0}=\lambda\frac{y_{0}}{b^{2}},\quad x_{0}+y_{0}=\lambda\frac{z_{0}}{c^{2}},\text{\; so\;\;} \frac{a^{2}}{\lambda}+\frac{b^{2}}{\lambda}=\frac{\lambda}{c^{2}}. \nonumber$$ Therefore, $\lambda=\pm |c|\sqrt{a^{2}+b^{2}}$. To determine $z_{0}$, note that $x_{0}=\displaystyle{\frac{a^{2}z_{0}}{\lambda}}$ and $y_{0}=\displaystyle{\frac{b^{2}z_{0}}{\lambda}}$. Therefore, $$1=\frac{x_{0}^{2}}{a^{2}} +\frac{y_{0}^{2}}{b^{2}}+\frac{z_{0}^{2}}{c^{2}} = \left(\frac{a^{2}+b^{2}}{\lambda^{2}}+\frac{1}{c^{2}}\right)z_{0}^{2}= \frac{2z_{0}^{2}}{c^{2}}, \nonumber$$ so $$z_{0}=\pm\frac{|c|}{\sqrt2}\text{\; and\;\;} (x_{0},y_{0},z_{0})=\pm \left(\frac{a^{2}}{\sqrt{2(a^{2}+b^{2})}}, \frac{b^{2}}{\sqrt{2(a^{2}+b^{2})}}, \displaystyle{\frac{|c|}{\sqrt{2}}} \right) \nonumber$$ $$(x_{0}+y_{0})z_{0}=\displaystyle{\frac{\lambda z_{0}^{2}}{c^{2}}}=\pm\frac{\lambda}{2}=\pm \frac{|c|\sqrt{a^{2}+b^{2}}}{2}. \nonumber$$

$L=x^{\alpha}y^{\beta}z^{\gamma}-\lambda(ax^{p}+by^{q}+cz^{r})$

$$L_{x}=\alpha x^{\alpha-1}y^{\beta}z^{\gamma}-\lambda pax^{p-1},\quad L_{y}=\beta x^{\alpha}y^{\beta-1}z^{\gamma}-\lambda qby^{q-1} \nonumber$$ $$L_{z}=\gamma x^{\alpha}y^{\beta}z^{\gamma-1}-\lambda rcz^{r-1} \nonumber$$ $$\displaystyle{\frac{p}{\alpha}ax_{0}^{p}=\frac{q}{\beta}by_{0}^{q}=\frac{r}{\gamma}cz_{0}^{q}=C} \nonumber$$ where $C$ is to be determined as follows: $$\displaystyle{ax_{0}^{p}=\frac{C\alpha}{p},\quad by_{0}^{q}=\frac{C\beta}{q},\quad cz_{0}^{q}=\frac{C\gamma}{r}} \nonumber$$ From the constraint, $$ax_{0}^p+by_{0}^{p}+cz_{0}^{r}=1, \nonumber$$ so $$C=\displaystyle{\left(\frac{\alpha}{p}+\frac{\beta}{q}+\frac{\gamma}{r}\right)^{-1}} \text{\; and\;\;} \displaystyle{x_{0}^{p}y_{0}^{q}z_{0}^{r}=\frac{\alpha\beta\gamma}{pqr} \left(\frac{\alpha}{p}+\frac{\beta}{q}+\frac{\gamma}{r}\right)^{-3}}. \nonumber$$

$L=xw-yz-\displaystyle{\frac{\lambda (x^{2}+2y^{2})}{2}-\frac{\mu(2z^{2}+w^{2})}{2}}$

$$L_{x}=w-\lambda x,\quad L_{y}=-z-2\lambda y,\quad L_{z}=-y-2\mu z,\quad L_{w}=x-\mu w \nonumber$$ $$w_{0}=\lambda x_{0},\quad z_{0}=-2\lambda y_{0},\quad y_{0}=-2\mu z_{0},\quad x_{0}=\mu w_{0} \nonumber$$ The first and last equality imply that $w_{0}=\lambda\mu w_{0}$ and $z_{0}=4\lambda\mu z_{0}$. Since
$2z_{0}^{2}+w_{0}^{2}=9$, $w_{0}$ and $z_{0}$ cannot both be zero, so either $\lambda\mu=1$ or $4\lambda\mu=1$.

If $\lambda\mu=1$, $z_{0}=y_{0}=0$, $x_{0}^{2}=4$, and $w_{0}^{2}=9$, so the constrained critical values are $$f(2,0,0,3)=f(-2,0,0,-3)=6 \text{\; and\;\;} f(-2,0,0,3)=f(2,0,0,-3)=-6. \nonumber$$

If $4\lambda\mu=1$, then $x_{0}=w_{0}=0$, $y_{0}^{2}=2$ and $z_{0}^{2}=9/2$, so the constrained critical values are $$f\left(0,\sqrt{2},\frac{3}{\sqrt{2}},0\right)= f\left(0,-\sqrt{2},-\frac{3}{\sqrt{2}},0\right)=3 \nonumber$$ and $$f\left(0,\sqrt{2},-\frac{3}{\sqrt{2}},0\right)= f\left(0,-\sqrt{2},\frac{3}{\sqrt{2}},0\right)= -3. \nonumber$$ Hence the constrained maximum and minimum values are $3$ and $-3$.

$L=xw-yz-\displaystyle{ \frac{\lambda}{2}(ax^{2}+by^{2})-\frac{\mu}{2}(cz^{2}+dw^{2})}$

$$L_{x}=w-a\lambda x,\quad L_{y}=-z-b\lambda y, \nonumber$$ $$L_{z}=-y-c\mu z=0,\quad L_{w}=x-d\mu w=0 \nonumber$$ $$x_{0}=\mu dw_{0},\quad y_{0}=-c\mu z_{0},\quad z_{0}=-b\lambda y_{0}, \text{\; and\;\;} w_{0}=\lambda a x_{0}. \nonumber$$ This implies that $$x_{0}w_{0}-y_{0}z_{0}=\lambda (ax_{0}^{2}+by_{0}^{2})=\lambda \text{\; and\;\;} x_{0}w_{0}-y_{0}z_{0}=\mu(cz_{0}^{2}+dw_{0}^{2}) =\mu, \nonumber$$ so $\lambda =\mu$. Therefore, $$x_{0}=\lambda dw_{0},\quad y_{0}=-c\lambda z_{0},\quad z_{0}=-b\lambda y_{0}, \text{\; and\;\;} w_{0}=\lambda a x_{0}, \nonumber$$ so $z_{0}=bc\lambda^{2} z_{0}$ and $w_{0}=ad\lambda^{2}w_{0}$. Since $cz_{0}^{2}+dw_{0}^{2}=1$, $w_{0}$ and $z_{0}$ cannot both be zero; hence, either $ad\lambda^{2}=1$ or $bc\lambda^{2}=1$.

Suppose that $ad\ne bc$. If $\lambda^{2} ad=1$, then $\lambda^{2} bc\ne1$, so $z_{0}=y_{0}=0$, and the constraints imply that $x_{0}^{2}=1/a$, and $w_{0}^{2}=1/d$. Therefore, the constrained maximum is $$\displaystyle{\frac{1}{\sqrt{ad}}},\text{\; attained at\;\;} \pm \displaystyle{\left(\frac{1}{\sqrt{a}},0,0,\frac{1}{\sqrt{d}}\right)} \nonumber$$ and the constrained minimum is $$-\displaystyle{\frac{1}{\sqrt{ad}}},\text{\; attained at\;\;} \pm \displaystyle{\left(-\frac{1}{\sqrt{a}},0,0,\frac{1}{\sqrt{d}}\right)}. \nonumber$$ If $\lambda^{2} bc=1$, then $\lambda^{2} ad\ne1$, so $x_{0}=w_{0}=0$ and the constraints imply that $y_{0}^{2}=1/b$ and $z_{0}^{2}=1/c$. Therefore, the constrained maximum is $$\displaystyle{\frac{1}{\sqrt{bc}}}, \text{\; attained at\;\;} \pm \displaystyle{\left(0,\frac{1}{\sqrt{b}},-\frac{1}{\sqrt{c}},0\right)}, \nonumber$$ and the constrained minimum is $$-\displaystyle{\frac{1}{\sqrt{bc}}}, \text{\; attained at\;\;} \pm \displaystyle{\left(0,\frac{1}{\sqrt{b}},\frac{1}{\sqrt{c}},0\right)}. \nonumber$$

Suppose that $ad=bc$. Since $x_{0}=\lambda dw_{0}$ and $y_{0}=-c\lambda z_{0}$, $$1=ax_{0}^{2}+by_{0}^{2}=\lambda^{2}[(ad)dw_{0}^2+(bc)cz_{0}^{2}]= \lambda^{2}ad(cz_{0}^{2}+d(w_{0})^{2}=\lambda^{2}ad, \nonumber$$ so $\lambda=\pm \displaystyle{\frac{1}{\sqrt {ad}}}=\pm\frac{1}{\sqrt{bc}}$. Therefore, the constrained maximum value of $f$ is $\displaystyle{\frac{1}{\sqrt {ad}}}=\frac{1}{\sqrt{bc}}$, is attained at all points of the form $\displaystyle{\left(w_{0}\sqrt{\frac{d}{a}},-z_{0}\sqrt{\frac{c}{b}},z_{0},w_{0}\right)}$ and the constrained minimum value of $f$ is $-\displaystyle{\frac{1}{\sqrt {ad}}}=-\frac{1}{\sqrt{bc}}$, attained at all points of the form $\displaystyle{\left(-w_{0}\sqrt{\frac{d}{a}},z_{0}\sqrt{\frac{c}{b}},z_{0},w_{0}\right)}$ where, in both cases, $cz_{0}^2+dw_{0}^{2}=1$. Alternatively, all the constrained maximum points are of the form $\displaystyle{\left(x_{0},y_{0},-y_{0}\sqrt{\frac{b}{c}},x_{0}\sqrt{\frac{a}{d}}\right)}$ and all the constrained minimum points are of the form $\displaystyle{\left(x_{0},y_{0},y_{0}\sqrt{\frac{b}{c}},-x_{0}\sqrt{\frac{a}{d}}\right)}$ where, in both cases, $ax_{0}^{2}+by_{0}^{2}=1$.

$L=\displaystyle{\frac{\alpha x^{2}+\beta y^{2}+\gamma z^{2}}{2}} -\lambda(a_{1}x+a_{2}y+a_{3}z)-\mu(b_{1}x+b_{2}y+b_{3}z)$

$$L_{x}=\alpha x-\lambda a_{1}-\mu b_{1},\quad L_{y}=\beta y-\lambda a_{2}-\mu b_{2}, \quad L_{z}=\gamma y-\lambda a_{3}-\mu b_{3} \nonumber$$

$$\tag{A} x_{0}=\frac{\lambda a_{1}+\mu b_{1}}{\alpha},\quad y_{0}=\frac{\lambda a_{2}+\mu b_{2}}{\beta},\quad z_{0}=\frac{\lambda a_{3}+\mu b_{3}}{\gamma}.$$

$$\tag{B} \frac{a_{1}(\lambda a_{1}+\mu b_{1})}{\alpha}+ \frac{a_{2}(\lambda a_{2}+\mu b_{2})}{\beta}+ \frac{a_{3}(\lambda a_{3}+\mu b_{3})}{\gamma}=c.$$

$$\tag{C} \frac{b_{1}(\lambda a_{1}+\mu b_{1})}{\alpha}+ \frac{b_{2}(\lambda a_{2}+\mu b_{2})}{\beta}+ \frac{b_{3}(\lambda a_{3}+\mu b_{3})}{\gamma}=d.$$

Assume that $${\bf u}=\frac{a_{1}}{\sqrt{\alpha}}{\bf i}+ \frac{a_{2}}{\sqrt{\beta}}{\bf j}+ \frac{a_{3}}{\sqrt{\gamma}}{\bf k} \text{\; and\;\;} {\bf v}=\frac{b_{1}}{\sqrt{\alpha}}{\bf i}+ \frac{b_{2}}{\sqrt{\beta}}{\bf j}+ \frac{b_{3}}{\sqrt{\gamma}}{\bf k} \nonumber$$ are linearly independent. Then (B) and (C) can be written as $$\tag{D} |{\bf u}|^{2}\lambda+({\bf u}\cdot{\bf v})\mu=c,\quad ({\bf u}\cdot{\bf v})\lambda+|{\bf v}|^2\mu=d.$$ Since ${\bf u}$ and ${\bf v}$ are linearly independent, $\Delta=_\text{def}|{\bf u}|^{2}|{\bf v}|^{2}-({\bf u}\cdot{\bf v})^{2}\ne0$. Therfore the solution of (D) is $$\lambda=\frac{c|{\bf v}|^{2}-d({\bf u}\cdot{\bf v})}{\Delta},\quad \mu=\frac{d|{\bf u}|^{2}-c({\bf u}\cdot{\bf v})}{\Delta}. \nonumber$$ From (A), $$\begin{aligned} \alpha x_{0}^2+\beta y_{0}^{2}+\gamma z_{0}^{2} &=& (\lambda a_{1}+\mu b_{1})^{2}+ (\lambda a_{2}+\mu b_{2})^{2}+ (\lambda a_{3}+\mu b_{3})^{2}\\ &=& \lambda^{2} (a_{1}^{2}+a_{2}^{2}+a_{3}^{2})+ \mu^{2} (b_{1}^{2}+b_{2}^{2}+b_{3}^{2})\\ &&+ 2\lambda\mu(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}).\end{aligned} \nonumber$$

$L=\displaystyle{\frac{1}{2}\sum_{i=1}^{n}(x_{i}-\alpha_{i})^{2}- \lambda\sum_{i=1}^{n}a_{i}x_{i}-\mu\sum_{i=1}^{n}b_{i}x_{i}}$

$$L_{x_{i}}=x_{i}-\alpha_{i}-\lambda a_{i}-\mu_{i} b_{i},\quad x_{i0}=\alpha_{i}+\lambda a_{i}+\mu b_{i} \nonumber$$ $$c=\sum_{i=1}^{n}a_{i}x_{i0}=\sum_{i=1}^{n}a_{i}\alpha_{i} +\lambda \sum_{i=1}^{n}a_{i}^{2}+\mu\sum_{i=1}^{n}a_{i}b_{i} = \sum_{i=1}^{n}a_{i}\alpha_{i} +\lambda \nonumber$$

$$d=\sum_{i=1}^{n}b_{i}x_{i0}=\sum_{i=1}^{n}b_{i}\alpha_{i} +\lambda \sum_{i=1}^{n}a_{i}b_{i}+\mu\sum_{i=1}^{n}b_{i}^{2} = \sum_{i=1}^{n}b_{i}\alpha_{i} +\mu \nonumber$$

$$\lambda=c-\sum_{i=1}^{n}a_{i}\alpha_{i},\quad \mu=d-\sum_{i=1}^{n}b_{i}\alpha_{i} \nonumber$$ $$\begin{aligned} \sum_{i=1}^{n}(x_{i0}-\alpha_{i})^{2}&=& \sum_{i=1}^{n}(\lambda a_{i}+\mu b_{i})^{2} =\lambda^{2}\sum_{i=1}^{n}a_{i}^{2}\\&+&2\lambda \mu\sum_{i=1}^{n}a_{i}b_{i} +\mu^{2}\sum_{i=1}^{n}b_{i}^{2}=\lambda^{2}+\mu^{2}\\ &=&\left(c-\sum_{i=1}^{n}a_{i}\alpha_{i}\right)^{2} +\left(d-\sum_{i=1}^{n}b_{i}\alpha_{i}\right)^{2}\end{aligned} \nonumber$$

$L=\displaystyle{\frac{1}{2}}\sum_{i=1}^{n}x_{i}^{2}- \lambda\sum_{i=1}^{n}x_{i}-\mu\sum_{i=1}^{n}jx_{i}$; $L_{x_{i}}=x_{i}-\lambda-\mu i$, so $x_{i0}=\lambda+\mu i$. To satisfy the constraints, $$\displaystyle{\sum_{i=1}^{n}(\lambda+ \mu i)=1} \text{\; and\;\;} \displaystyle{\sum_{i=1}^{n}i(\lambda+ \mu i)=0}. \tag{A}$$ Let $$s_{0}=n, \quad s_{1}=\sum_{j=1}^{n}i=\frac{n(n+1)}{2},\text{\; and\;\;} s_{2}=\sum_{i=1}^{n}i^{2}=\frac{n(n+1)(2n+1)}{6}. \nonumber$$ Then (A) is equvalent to, $$\left[\begin{array}{ccccccc} s_{0}&s_{1}\\ s_{1}&s_{2} \end{array}\right] \left[\begin{array}{ccccccc} \lambda\\\mu \end{array}\right] = \left[\begin{array}{ccccccc} 1\\0 \end{array}\right]. \nonumber$$

By Cramer’s rule, $$\lambda=\frac{s_{2}}{s_{0}s_{2}-s_{1}^{2}}=\frac{2(2n+1)}{n(n-1)} \text{\; and\;\;} \mu=-\frac{s_{1}}{s_{0}s_{2}-s_{1}^{2}}=-\frac{6}{n(n-1)}. \nonumber$$ Therefore, $$x_{i0}=\displaystyle{\frac{4n+2-6i}{n(n-1)}},\quad 1\le i\le n. \nonumber$$

If $$\sum_{i=1}^{n}y_{i}=1\text{\; and\;\;}\sum_{i=1}^{n}iy_{i}=0, \text{\; then\;\;}\sum_{i=1}^{n}(y_{i}-x_{i0})x_{i0}=0, \nonumber$$ so $$\begin{aligned} \sum_{i=1}^{n}y_{i}^{2}&=&\sum_{i=1}^{n}(y_{i}-x_{i0}+x_{i0})^{2} +\sum_{i=1}^{n}(y_{i}-x_{i0})^{2}+ 2\sum_{i=1}^{n}(y_{i}-x_{i0})x_{i0} +\sum_{i=1}^{n}x_{i0}^{2}\\ &=& \sum_{i=1}^{n}(y_{i}-x_{i0})^{2}+ \sum_{i=1}^{n}x_{i0}^{2}>\sum_{i=1}^{n}x_{i0}^{2}\end{aligned} \nonumber$$ if $y_{i}\ne x_{i0}$ for some $i\in\{1,2,\dots,n\}$.

$L=f({\bf X})- \lambda (x_{1}+x_{2}+\cdots+x_{n})$

$$L_{x_{i}}=-\frac{p_{i}f({\bf X})}{s-x_{i}}- \lambda,\text{\; so\;\;} \frac{s-x_{10}}{p_{1}}= \frac{s-x_{20}}{p_{2}}=\cdots= \frac{s-x_{n0}}{p_{n}}=_\text{ def}C. \nonumber$$ $x_{i0}=s-Cp_{i}$, $1\le i\le n$. Denote $P=p_{1}+p_{2}+\cdots +p_{n}$.

$$x_{1}+x_{2}+\cdots+x_{n}=ns-C(p_{1}+p_{2}+\cdots+p_{n})=ns-CP=s. \nonumber$$ $$\displaystyle{C=\frac{(n-1)s}{P}};\quad x_{i0}=\displaystyle{\frac{[P-(n-1)]sp_{i}}{P}}. \nonumber$$ $$f_\text{max}=C^{P}p_{1}^{p_{1}}p_{2}^{p_{2}}\cdots p_{n}^{p_{n}}= \left[\frac{(n-1)s}{P}\right]^{P}p_{1}^{p_{1}}p_{2}^{p_{2}}\cdots p_{n}^{p_{n}} \nonumber$$

[exer:49]. $L({\bf X})=\displaystyle{f({\bf X})-\lambda \sum_{i=1}^{n}\frac{x_{i}}{\sigma_{i}}}$, $L_{x_{i}}=\displaystyle{\frac{p_{i}f({\bf X})}{x_{i}}-\frac{\lambda}{\sigma_{i}}}$, so $\displaystyle{\frac{x_{i0}}{\sigma_{i}}}=Cp_{i}$. To satisfy the constraint, $C=(p_{1}+p_{2}\cdots+p_{n})^{-1}$, so $$x_{i0}=\displaystyle{\frac{p_{i}\sigma_{i}S}{p_{1}+p_{2}+\cdots+p_{n}}}. \nonumber$$ and $$x_{10}^{p_{1}}x_{20}^{p_{2}}\cdots x_{n0}^{p_{n}}= \left(\frac{S}{p_{1}+p_{2}+\cdots+ p_{n}}\right)^{p_{1}+p_{2}+\cdots+p_{n}} (p_{1}\sigma_{1})^{p_{1}} (p_{2}\sigma_{2})^{p_{2}} \cdots (p_{n}\sigma_{n})^{p_{n}} \nonumber$$

$\displaystyle{L=\sum_{i=1}^{n}\frac{x_{i}}{\sigma_{i}}- \lambda x_{1}^{p_{1}}x_{2}^{p_{2}}\cdots x_{n}^{p_{n}}}$, $L_{x_{i}}=\displaystyle{\frac{1}{\sigma_{i}}-\lambda\frac{p_{i}V}{x_{i}}}$, $\displaystyle{\frac{x_{i0}}{\sigma_{i}}}=Cp_{i}$, where $C$ must be chosen to satisfy the constraints.

$x_{i0}=C\sigma_{i}p_{i}$, $x_{i0}^{p_{i}}=(C\sigma_{i}p_{i})^{p_{i}}$, $V=(C\sigma_{1}p_{1})^{p_{1}} (C\sigma_{2}p_{2})^{p_{2}}\cdots (C\sigma_{n})^{p_{n}}$ $$C^{p_{1}+p_{2}+\cdots+p_{n}}=\frac{V}{(\sigma_{1}p_{1})^{p_{1}}(\sigma_{2}p_{2})^{p_{2}} \cdots (\sigma_{ n}p_{n})^{p_{n}}} \nonumber$$

$$C=\left(\frac{V}{(\sigma_{1}p_{1})^{p_{1}}(\sigma_{2}p_{2})^{p_{2}} \cdots (\sigma_{ n}p_{n})^{p_{n}}}\right)^{\frac{1}{p_{1}+p_{2}+\cdots+p_{n}}} \nonumber$$ $$\frac{x_{i0}}{\sigma_{i}}=p_{i} \left(\frac{V}{(\sigma_{1}p_{1})^{p_{1}}(\sigma_{2}p_{2})^{p_{2}} \cdots (\sigma_{ n}p_{n})^{p_{n}}}\right)^{\frac{1}{p_{1}+p_{2}+\cdots+p_{n}}} \nonumber$$

$$\sum_{i=1}^{n}\frac{x_{i0}}{\sigma_{i}}=(p_{1}+p_{2}+\cdots+p_{n}) \left(\frac{V}{(\sigma_{1}p_{1})^{p_{1}}(\sigma_{2}p_{2})^{p_{2}} \cdots (\sigma_{ n}p_{n})^{p_{n}}}\right)^{\frac{1}{p_{1}+p_{2}+\cdots+p_{n}}}. \nonumber$$

$L=\displaystyle{\frac{1}{2}\sum_{i=1}^{n}\frac{(x_{i}-c_{i})^{2}}{\alpha_{i}}} -\lambda (a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n})$

$$L_{x_{i}}=\displaystyle{\frac{x_{i}-c_{i}}{\alpha_{i}}}-\lambda a_{i},\quad x_{i0}=c_{i}+\lambda a_{i}\alpha_{i} \nonumber$$ $$\sum_{i=1}^{n}a_{i}x_{i0}=\sum_{i=1}^{n}a_{i}c_{i}+ \lambda\sum_{i=1}^{n}a_{i}^{2}\alpha_{i}=d,\quad \lambda=\frac{d-\sum_{i=1}^{n}a_{i}c_{i}}{\sum_{i=1}^{n}a_{i}^{2}\alpha_{i}} \nonumber$$ $$\sum_{i=1}^{n}\frac{(x_{i0}-c_{i})^{2}}{\alpha_{i}}=\lambda^{2} \sum_{i=1}^{n}a_{i}^{2}\alpha_{i}= \frac{(d-\sum_{i=1}^{n}a_{i}c_{i})^{2}}{\sum_{i=1}^{n}a_{i}^{2}\alpha_{i}} \nonumber$$

It suffices to extremize $\displaystyle{\sum_{i=1}^{n}a_{i}x_{i}}$ subject to $\sum_{i=1}^{n}x_{i}^{2}=\sigma^{2}$ for arbitrary $\sigma>0$. $$L=\displaystyle{\sum_{i=1}^{n}a_{i}x_{i}-\frac{\lambda}{2}\sum_{i=1}^{n}x_{i}^{2}}, \quad L_{y_{i}}=a_{i}-\lambda x_{i},\quad a_{i}=\lambda x_{i0}, \nonumber$$ $$\sum_{i=1}^{n}a_{i}^{2}=\lambda^{2}\sum_{i=1}^{n}x_{i0}^{2}=\lambda^{2}\sigma^{2} \nonumber$$ $$\sum_{i=1}^{n}a_{i}x_{i0}=\lambda \sum_{i=1}^{n}x_{i0}^{2}=\lambda\sigma^{2}=(\lambda\sigma)\sigma = \pm\left(\sum_{i=1}^{n}a_{i}^{2}\right)^{1/2} \left(\sum_{i=1}^{n}x_{i0}^{2}\right)^{1/2} \nonumber$$

For every $\sigma>0$, $f({\bf X})= x_{m})=x_{1}^{r_{1}}x_{2}^{r_{2}}\cdots x_{m}^{r_{m}}$ assumes a maximum value on the closed set $$S_{\sigma}=\left\{(x_{1},x_{2}, \dots, x_{m})\, \big|\, x_{i}>0, \,1 \le i \le m,\, r_{1}x_{1}+r_{2}x_{2}+\cdots+ r_{m}x_{m}=\sigma\right\}. \nonumber$$ $$L=x_{1}^{r_{1}}x_{2}^{r_{2}}\cdots x_{m}^{r_{m}} -\lambda\sum_{i=1}^{m}r_{i}x_{i},\quad L_{x_{i}}=r_{i}\left(\frac{x_{1}^{r_{1}}x_{2}^{r_{2}}\cdots x_{m}^{r_{m}}}{x_{i}}-\lambda\right), \quad 1 \le i \le m. \nonumber$$ Therefore, the constrained extremum is attained at $x_{1}=x_{2}=\cdots =x_{m}=\sigma/r$, and the value of the constrained extremum is $(\sigma/r)^{r}$, so $$\left(x_{1}^{r_{1}}x_{2}^{r_{2}}\cdots x_{m}^{r_{m}}\right)^{1/r} \le \frac{\sigma}{r}=\frac{r_{1}{x_{1}+r_{2}x_{2}+\cdots+r_{k}x_{k}}}{r} \nonumber$$ with equality if and only if $x_{1}=x_{2}=\cdots= x_{m}=\sigma/r$.

The statement is trivial if $\sigma_{i}=0$ for some $i$. If $\sigma_{i}\ne0$, $1 \le i \le m$, then Exercise [exer:53] with $r_{i}=\displaystyle{\frac{1}{p_{i}}}$ and $x_{i}=\displaystyle{\frac{|a_{ij}|^{p_{i}}}{\sigma_{i}}}$ implies that $$\frac{|a_{1j}||a_{2j}|\cdots|a_{mj}|} {\sigma_{1}^{1/p_{1}}\sigma_{2}^{1/p_{2}}\cdots\sigma_{m}^{1/p_{m}}} \le \sum_{i=1}^{m} \frac{|a_{ij}|^{p_{i}}}{p_{i}\sigma_{i}}. \nonumber$$ Summing both sides from $j=1$ to $n$ yields the stated conclusion.

$\displaystyle{L =\frac{1}{2}\sum_{r=0}^{n}x_{r}^{2}-\sum_{s=0}^{m} \lambda_{s}\sum_{r=0}^{n}x_{r}r^{s}}$, $L_{x_{r}}=x_{r}-\displaystyle{\sum_{s=0}^{m}\lambda_{s}r^{s}}$ $$x_{r0}=\sum_{s=0}^{m}\lambda_{s}r^{s},\quad 0\le r\le n. \nonumber$$ $$\sum_{r=0}^{n}x_{r0}r^{s}=\sum_{r=0}^{n}\sum_{\ell=0}^{m}\lambda_{\ell}r^{\ell+s} =\sum_{\ell=0}^{m}\lambda_{\ell}\sum_{r=0}^{n}r^{\ell+s}= \sum_{\ell=0}^{m}\sigma_{s+\ell}\lambda_{\ell}=c_{s}, \quad 0\le s \le m, \nonumber$$ so $(x_{10},x_{20},\dots,x_{n0})$ is a critical point of $L$. To see that it is constrained minimum point of $Q$, suppose that $(y_{0},y_{1},\dots,y_{n})$ also satisfies the constraints; thus, $$\sum_{r=0}^{n}y_{r}r^{s}=c_{s},\quad 0\le s \le m. \nonumber$$ Then $$\sum_{r=0}^{n}(y_{r}-x_{r0})x_{r0}=\sum_{r=0}^{n}(y_{r}-x_{r0}) \sum_{s=0}^{m}\lambda_{s}r^{s}=\sum_{s=0}^{m}\lambda_{s}\sum_{r=0}^{n} (y_{r}-x_{r0})r^{s}=0, \nonumber$$ so $$\begin{aligned} \sum_{r=0}^{n}y_{r}^{2}&=&\sum_{r=0}^{n}(y_{r}-x_{r0}+x_{r0})^{2} =\sum_{r=0}^{n}[(y_{r}-x_{r0})^{2}+2(y_{r}-x_{r0})x_{r0} +x_{r0}^{2}]\\ &=&\sum_{r=0}^{n}[(y_{r}-x_{r0})^{2} +\sum_{r=0}^{n}x_{r0}^{2}> \sum_{r=0}^{n}x_{r0}^{2}.\end{aligned} \nonumber$$

Imposing the constraint with $r=0$ and $P(x)=x^{s}$, $1\le s\le 2k$, yields the necessary condition $$\tag{A} \sum_{i=-n}^{n}w_{i}i^{s}= \begin{cases} 1& \text{if } s=0,\\ 0&\text{if }1\le s\le 2k. \end{cases}$$ If $P$ is an arbitrary polynomial of degree $\le 2k$ and $r$ is an arbitrary integer, then
$P(r-i)=P(r)+$ a linear combination of $i$, $i^{2}$, …, $i^{2k}$, so (A) implies that $$\sum_{i=-n}^{n}w_{i}P(r-i)=P(r) \nonumber$$ whenever $r$ is an integer and $P$ is a polynomial of degree $\le 2k$. Therefore, $$L=\frac{1}{2}\sum_{i=-n}^{n}w_{i}^{2}-\sum_{r=0}^{2k}\lambda_{r} \sum_{i=-n}^{n}w_{i}i^{r}, \nonumber$$ $$L_{w_{i}}=w_{i}-\sum_{r=0}^{2k}\lambda_{r}i^{r},\quad w_{i0}=\sum_{r=0}^{2k}\lambda_{r}i^{r}, \quad -n\le i\le n, \nonumber$$ and $$\sum_{i=-n}^{n}w_{i0}i^{s}= \sum_{i=-n}^{n} \left(\sum_{r=0}^{2k} \lambda_{r}i^{r}\right)i^{s} =\sum_{r=0}^{2k}\lambda_{r}\sigma_{r+s}\text{\; where\;\;} \sigma_{m}=\sum_{i=-n}^{n}i^{m}. \nonumber$$

If $\{w_{i}\}_{i=-n}^{n}$ also satisfies the constraint, then $$\sum_{i=-n}^{n}(w_{i}-w_{i0})w_{i0}= \sum_{i=-n}^{n}(w_{i}-w_{i0})\sum_{r=0}^{2k}\lambda_{r}i^{r}=0. \nonumber$$ Therefore, $$\begin{aligned} \sum_{i=-n}^{n}w_{i}^{2}&=&\sum_{i=-n}^{n}(w_{i0}+w_{i}-w_{i0})^{2}= \sum_{i=-n}^{n}\left(w_{i0}^{2}+2(w_{i}-w_{i0})w_{i0}+(w_{i}-w_{i0})^{2}\right)\\ &=&\sum_{i=-n}^{n}w_{i0}^{2}+\sum_{i=-n}^{n}(w_{i}-w_{i0})^{2} >\sum_{i=-n}^{n}w_{i0}^{2}\end{aligned} \nonumber$$ if $w_{i}\ne w_{i0}$ for some $i$.

The coefficients $w_{0}$, $w_{1}$, …, $w_{k}$ satisfy the constraint if and only if $$\sum_{i=0}^{n}w_{i}(r-i)^{j}=(r+1)^{j},\quad 0\le j\le k, \nonumber$$ for all integers $r$. This is equivalent to $$\sum_{i=0}^{n}w_{i}\sum_{s=0}^{j}(-1)^{s} \binom{j}{s}s^{j}r^{j-s} =\sum_{s=0}^{j}\binom{j}{s}r^{j-s},\quad 0\le j\le k, \nonumber$$ which is equivalent to $$\tag{A} \sum_{i=0}^{n}w_{i}i^{s}=(-1)^{s},\quad 0\le s\le k.$$ $$L=\frac{1}{2}\sum_{i=0}^{k}w_{i}^{2}-\sum_{r=0}^{k}\lambda_{r} \sum_{i=0}^{k}w_{i}i^{r};\quad L_{x_{i}}=w_{i}-\sum_{r=0}^{k}\lambda_{r} i^{r};\quad w_{i0}=\sum_{r=0}^{k}\lambda_{r}i^{r}. \nonumber$$ Now must choose $\lambda_{1}$, $\lambda_{2}$, …, $\lambda_{k}$ to satisfy (A).

$$\sum_{i=0}^{n}w_{i0}i^{s}\sum_{r=0}^{k}\lambda_{r}i^{r} =\sum_{r=0}^{k}\lambda_{r}\sum_{i=0}^{n}i^{r+s} = \sum_{r=0}^{k}\sigma_{r+s}\lambda_{r}=(-1)^{s}, \quad 0\le s\le k. \nonumber$$

If $\{w_{i}\}_{i=0}^{n}$ also satisfies the constraint, then $$\sum_{i=0}^{n}(w_{i}-w_{i0})w_{i0}= \sum_{i=n}^{n}(w_{i}-w_{i0})\sum_{r=0}^{2k}\lambda_{r}i^{r}=0. \nonumber$$ Therefore, $$\begin{aligned} \sum_{i=0}^{n}w_{i}^{2}&=&\sum_{i=0}^{n}(w_{i0}+w_{i}-w_{i0})^{2}= \sum_{i=0}^{n}\left(w_{i0}^{2}+2(w_{i}-w_{i0})w_{i0}+(w_{i}-w_{i0})^{2}\right)\\ &=&\sum_{i=0}^{n}w_{i0}^{2}+\sum_{i=-n}^{n}(w_{i}-w_{i0})^{2} >\sum_{i=0}^{n}w_{i0}^{2}\end{aligned} \nonumber$$ if $w_{i}\ne w_{i0}$ for some $i$.

$L=\displaystyle{\frac{1}{2}\sum_{i=1}^{n}\frac{(x_{i}-c_{i})^{2}}{\alpha_{i}} -\sum_{s=1}^{m}\lambda_{s}\sum_{i=1}^{n}a_{is}x_{i}}$

$$L_{x_{i}}=\frac{x_{i}-c_{i}}{\alpha_{i}}, \quad x_{i0}=c_{i}+\alpha_{i}\displaystyle{\sum_{s=1}^{m}\lambda_{s}a_{is}} \nonumber$$

$$\displaystyle{\sum_{i=1}^{n}a_{ir}x_{i0}}=\displaystyle{\sum_{i=1}^{n}a_{ir}c_{i}+ \sum_{s=1}^{m}\lambda_{s}\sum_{i=1}^{n}\alpha_{i}a_{ir}a_{is} = \sum_{i=1}^{n}a_{ir}c_{i}+\lambda_{r} =d_{r}} \nonumber$$

$$\lambda_{r}=d_{r}-\displaystyle{\sum_{i=1}^{n}a_{ir} c_{i}},\quad \displaystyle{\frac{(x_{i}-c_{i})^{2}}{\alpha_{i}}=\alpha_{i}\sum_{r,s=1}^{m} \lambda_{r}\lambda_{s}a_{ir}a_{is}} \nonumber$$

$$\sum_{i=1}^{n}\frac{(x_{i}-c_{i})^{2}}{\alpha_{i}}=\sum_{r,s=1}^{m} \lambda_{r}\lambda_{s}\sum_{i=1}^{n}\alpha_{i}a_{ir}a_{is} =\sum_{r=1}^{m}\lambda_{r}^{2} =\sum_{r=1}^{m} \left(d_{r}-\sum_{i=1}^{n}a_{ir}c_{i}\right)^{2} \nonumber$$