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# 6: Exercises

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## 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 $f(x,y,z)=2x+3y+z\text{\quad subject to\quad} x^{2}+2y^{2}+3z^{2}=1.$

[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 $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1,$ 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 $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1.$ 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 $A= \sqrt{s(s-x)(s-y)(s-z)}.$

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

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

[exer:17] Let $$\mathbf{X}_{i}=(x_{i},y_{i},z_{i})$$, $$1 \le i \le n$$. Find the point in the plane $ax+by+cz=\sigma$ 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 $f(x,y,z)=2xy+2xz+2yz\text{\quad subject to\quad} x^{2}+y^{2}+z^{2}=1.$

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

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

[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 $4x^{2}+9y^{2}-36z^{2}=36.$

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

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

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

[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 $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}=d.$

[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 $f(x,y,z,w)=x^{2}+2y^{2}+z^{2}+w^2$ subject to \begin{aligned} x+y+\phantom{2}z+3w&=&1\\ x+y+2z+\phantom{3}w&=&2.\end{aligned}

[exer:33] Find the minimum value of $f(x,y,z)= \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}$ 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 $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1,$ 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 $3x^{2}+9y^{2}+6z^{2}=10$ and the plane $3x+3y+6z=70.$

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

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

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

[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 $ax^{p}+by^{q}+cz^{r}=1 \text{\; and\;\;} x,y,z>0 .$

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

[exer:44] Let $$a$$, $$b$$, $$c$$,and $$d$$ be positive. Find the extreme values of $f(x,y,z,w)=xw-yz$ subject to $ax^{2}+by^{2}=1, \quad cz^{2}+dw^{2}=1,$ 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 $a_{1}x+a_{2}y+a_{3}z=c\text{\; and\;\;} b_{1}x+b_{2}y+b_{3}z=d.$ Assume that $\alpha,\beta,\gamma>0,\quad a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\ne0, \text{\; and\;\;} b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\ne 0.$ 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 $\displaystyle{\sum_{i=1}^{n}a_{i}x_{i}=c} \text{\; and\;\;} \displaystyle{\sum_{i=1}^{n}b_{i}x_{i}=d},$ where $\sum_{i=1}^{n}a_{i}^{2}=\sum_{i=1}^{n}b_{i}^{2}=1 \text{\; and\;\;} \sum_{i=1}^{n}a_{i}b_{i}=0.$

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

[exer:48] Let $$p_{1}$$, $$p_{2}$$, …, $$p_{n}$$ and $$s$$ be positive numbers. Maximize $f({\bf X})= (s-x_{1})^{p_{1}}(s-x_{2})^{p_{2}}\cdots(s-x_{n})^{p_{n}}$ 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 $\sum_{i=1}^{n}\frac{x_{i}}{\sigma_{i}} = S,$ where $$p_{1}$$, $$p_{2}$$,…, $$p_{n}$$, $$\sigma_{1}$$, $$\sigma_{2}$$, …, $$\sigma_{n}$$, and $$V$$ are given positive numbers.

[exer:50] Maximize $f({\bf X})=\sum_{i=1}^{n}\frac{x_{i}}{\sigma_{i}}$ subject to $$x_{i}>0$$, $$1\le i\le n$$, and $x_{1}^{p_{1}}x_{2}^{p_{2}}\cdots x_{n}^{p_{n}}=V,$ 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 $Q({\bf X})=\sum_{i=1}^{n}\frac{(x_{i}-c_{i})^{2}}{\alpha_{i}}$ subject to $a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}=d.$

[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 $|a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}|\le (a_{1}^{2}+a_{2}^{2}+ \cdots+ a_{n}^{2})^{1/2} (x_{1}^{2}+x_{2}^{2}+ \cdots+ x_{n}^{2})^{1/2}.$ 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 $r_{1}+r_{2}+\cdots+r_{m}=r.$ Show that $\left(x_{1}^{r_{1}}x_{2}^{r_{2}}\cdots x_{m}^{r_{m}}\right)^{1/r} \le \frac{r_{1}x_{1}+r_{2}x_{2}+\cdots r_{m}x_{m}}{r},$ 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 $\sum_{j=1}^{m}\frac{1}{p_{j}}=1,$ and define $\sigma_{i}=\sum_{j=1}^{n}|a_{ij}|^{p_{i}}, \quad 1 \le i \le m.$ Use Exercise [exer:53] to show that $\left|\sum_{j=1}^{n}a_{ij}a_{2j}\cdots a_{mj}\right| \le \sigma_{1}^{1/p_{1}}\sigma_{2}^{1/p_{2}}\cdots \sigma_{m}^{1/p_{m}}.$ (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 $Q({\bf X})=\sum_{r=0}^{n}x_{r}^{2}$ subject to $\sum_{r=0}^{n}x_{r}r^{s}=c_{s},\quad 0\le s \le m,$ is attained when $x_{r}=\sum_{s=0}^{m}\lambda_{s}r^{s},\quad 0\le r\le n,$ where $\sum_{\ell=0}^{m}\sigma_{s+\ell}\lambda_{\ell}=c_{s} \text{\; and\;\;} \sigma_{s}= \sum_{r=0}^{n}r^{s},\quad 0\le s\le m.$ Show that if $$\{x_{r}\}_{r=0}^{n}$$ satisfies the constraints and $$x_{r}\ne x_{r0}$$ for some $$r$$, then $\sum_{r=0}^{n}x_{r}^{2}>\sum_{r=0}^{n}x_{r0}^{2}.$

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

[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 $\sum_{i=0}^{n}w_{i}P(r-i)=P(r+1)$ whenever $$r$$ is an integer and $$P$$ is a polynomial of degree $$\le k$$, is attained with $w_{i0}=\sum_{r=0}^{k}\lambda_{r}i^{r},\quad 0\le i\le n,$ where $\sum_{r=0}^{k}\sigma_{r+s}\lambda_{r}=(-1)^{s},\quad 0\le s \le k, \text{\quad and\quad } \sigma_{\ell}=\sum_{i=0}^{n}i^{\ell},\quad 0\le \ell\le 2k.$ Show that if $\sum_{i=0}^{n}u_{i}P(r-i)=P(r+1)$ whenever $$r$$ is an integer and $$P$$ is a polynomial of degree $$\le k$$, and $$u_{i}\ne w_{i0}$$ for some $$i$$, then $\sum_{i=0}^{n}u_{i}^{2}>\sum_{i=0}^{n}w_{i0}^{2}.$

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

[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

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

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

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

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

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

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

$L_{x}(x,y)=(x-2)-\lambda x,\quad L_{y}(x,y)=(y-4)-\lambda y$ $\frac{x_{0}-2}{x_{0}}=\frac{y_{0}-4}{y_{0}}=\lambda, \text{\; so\;\;} y_{0}=2x_{0}.$ 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.

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

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

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

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

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

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

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

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

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

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

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

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

We must maximize $$V=8xyz$$ subject to $$\displaystyle{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}}=1.$$ $L=xyz-\frac{\lambda}{2}\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}\right)$ $L_{x}=yz-\frac{\lambda x}{a^{2}}, \quad L_{y}=xz-\frac{\lambda y}{b^{2}}, \quad L_{z}=xy-\frac{\lambda z}{c^{2}}$ $y_{0}z_{0}=\frac{\lambda x_{0}}{a^{2}}, \quad x_{0}z_{0}=\frac{\lambda y_{0}}{b^{2}}, \quad x_{0}y_{0}=\frac{\lambda z}{c^{2}}\quad$ $\displaystyle{\frac{x_{0}^{2}}{a^{2}}}=\displaystyle{\frac{y_{0}^{2}}{b^{2}}}=\displaystyle{\frac{z_{0}^{2}}{c^{2}}} =\lambda x_{0}y_{0}z_{0}$ 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 $\tag{A} ax_{0}+by_{0}+cz_{0}=\sigma.$ $L=\displaystyle{\frac{(x-x_{1})^{2}+(y-y_{1})^{2}+(z-z_{1})^{2}}{2}}-\lambda(ax+by+cz)$ $L_{x}=(x-x_{1})-\lambda a,\quad L_{y}=(y-y_{1})-\lambda b,\quad L_{z}=(z-z_{1})-\lambda c$ $x_{0}=x_{1}+\lambda a,\quad y_{0}=y_{1}+\lambda b,\text{\quad and\quad} z_{0}=z_{1}+\lambda c, \tag{B}$ $\tag{C} d^{2}=\lambda^{2}(a^{2}+b^{2}+c^{2})$ (A) and (B) imply that $ax_{1}+by_{1}+cz_{1}+\lambda(a^{2}+b^{2}+c^{2})=\sigma,$ so $\lambda=\frac{\sigma-ax_{1}-by_{1}-cz_{1}}{a^{2}+b^{2}+c^{2}},$ and (C) implies that $d=\frac{|\sigma-ax_{1}-by_{1}-cz_{1}|}{\sqrt{a^{2}+b^{2}+c^{2}}}.$

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

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

$$\displaystyle{\sum_{i=1}^{n}x_{i0}^{2}=(1-\lambda)^{-2}\sum_{j=1}^{n}c_{j}^{2}}=1$$, 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].

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

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

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

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

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

$\left[\begin{array}{ccccccc} 3&0&1\\0&2&0\\1&0&3 \end{array}\right] \left[\begin{array}{ccccccc} x_{0}\\y_{0}\\z_{0} \end{array}\right]=\lambda \left[\begin{array}{ccccccc} x_{0}\\y_{0}\\z_{0} \end{array}\right]$ 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} $\left[\begin{array}{ccccccc} 1-\lambda & 4-\lambda \\ 4-\lambda &4(1-\lambda) \end{array}\right] \left[\begin{array}{ccccccc} x_{0}\\ y_{0} \end{array}\right]= \left[\begin{array}{ccccccc} 0\\0 \end{array}\right],$ so $4(\lambda-1)^{2}-(\lambda-4)^{2}=3(\lambda-2)(\lambda+2)=0.$ 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 $$f(x_{0},y_{0})=-\frac{2}{3}$$.

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

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

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

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

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

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

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

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

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 $2x_{0}=\lambda y_{0},\quad 2y_{0}=\lambda x_{0},\quad 2z_{0}=\lambda w_{0},\quad 2w_{0} =\lambda z_{0},$ and $2x_{0}=\lambda y_{0} =\frac{\lambda}{2}(2y_{0})=\frac{\lambda^{2}}{2}x_{0} \text{\; and \;\;}2z_{0}=\lambda w_{0}=\frac{\lambda}{2}(2w_{0})= \frac{\lambda^{2}}{2}z_{0}.$ If $$\lambda\ne2$$, then $$x_{0}=y_{0}=z_{0}=w_{0}$$, which does not satisfy the constraint. If $$\lambda=2$$, then $x_{0}=y_{0}=z_{0}=w_{0}=\pm\frac{1}{2}\text{\; and\;\;} (x_{0}+z_{0})(y_{0}+w_{0})=1.$ If $$\lambda=-2$$, then $x_{0}=-y_{0}=z_{0}=-w_{0}=\pm\frac{1}{2} \text{\; and\;\;} (x_{0}+z_{0})(y_{0}+w_{0})=-1.$ 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 $$\pm\left(\frac{1}{2},-\frac{1}{2},\frac{1}{2},-\frac{1}{2}\right)$$.

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

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

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

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

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

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

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

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 $$\pm\left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2},\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right)$$,

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

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

(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 $\frac{x_{10}}{y_{10}}=\frac{x_{20}}{y_{20}}=\frac{y_{20}}{x_{20}}.$ Therefore, $$x_{20}=y_{20}=1$$ and $$x_{10}=y_{10}=\frac{1}{\sqrt2}$$, so $(x_{10}-x_{20})^{2}+(y_{10}-y_{20})^{2}= 2\left(1-\frac{1}{\sqrt 2}\right)^{2}$ and the distance between the curves is $$\sqrt2-1$$.

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

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

$$\displaystyle{L(x_{1},x_{2},\dots,x_{n})=\frac{(x_{1}-c_{1})^{2}+(x_{2}-c_{2})^2+ \cdots+(x_{n}-c_{n})^2}{2}}$$ $-\lambda(a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n})$ $$L_{x_{i}}=x_{i}-c_{i}-\lambda a_{i}$$, $$1\le i\le n$$. 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} which holds if and only if $\lambda=\frac{d-a_{1}c_{1}-a_{2}c_{2}-\cdots-a_{n}c_{n}} {a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}}.$ Therefore, $x_{i0}=c_{i}+ \frac{(d-a_{1}c_{1}-a_{2}c_{2}-\cdots-a_{n}c_{n})a_{i}} {a_{1}^{2}+a_{2}^{2}+\cdots a_{n}^{2}},\quad 1\le i\le n,$ and the distance from $$(x_{10},x_{10},\dots,x_{n0})$$ to $$(c_{1},c_{2},\dots,c_{n})$$ is $\frac{|(d-a_{1}c_{1}-a_{2}c_{2}-\cdots-a_{n}c_{n})a_{i}|} {\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots a_{n}^{2}}}.$

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

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

[exer:31].

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Let $$(x_{1},y_{1},z_{1})$$ be on the ellipsoid and $$(x_{2},y_{2},z_{2})$$ be on the plane. $L= \frac{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}}{2} -\frac{\lambda}{2}(3x_{1}^{2}+9y_{1}^{2}+6z_{1}^{2}) -\mu(x_{2}+y_{2}+2z_{2}).$ $L_{x_{1}}=x_{1}-x_{2}-3\lambda x_{1}=0,\quad L_{y_{1}}=y_{1}-y_{2}-9\lambda y_{1}=0,\quad L_{z_{1}}=z_{1}-z_{2}-6\lambda z_{1}$ $L_{x_{2}}=x_{2}-x_{1}-\mu,\quad Ly_{2}=y_{2}-y_{1}-\mu,\quad L_{z_{2}}=z_{2}-z_{1}-2\mu$ \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} 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 $\text{\; (a)\;\;}(x_{10},y_{10},z_{10})=\left(1,\frac{1}{3},1\right) \text{\; or \quad (b)\;\;} (x_{10},y_{10},z_{10})=\left(-1,-\frac{1}{3},-1\right).$ 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 $\mu=\frac{70-3x_{10}-3y_{10}-6z_{10}}{18},$

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 $$\frac{10\sqrt{6}}{3}$$.

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

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

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

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

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

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

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

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

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

$L_{x}=w-\lambda x,\quad L_{y}=-z-2\lambda y,\quad L_{z}=-y-2\mu z,\quad L_{w}=x-\mu w$ $w_{0}=\lambda x_{0},\quad z_{0}=-2\lambda y_{0},\quad y_{0}=-2\mu z_{0},\quad x_{0}=\mu w_{0}$ 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 $f(2,0,0,3)=f(-2,0,0,-3)=6 \text{\; and\;\;} f(-2,0,0,3)=f(2,0,0,-3)=-6.$

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 $f\left(0,\sqrt{2},\frac{3}{\sqrt{2}},0\right)= f\left(0,-\sqrt{2},-\frac{3}{\sqrt{2}},0\right)=3$ and $f\left(0,\sqrt{2},-\frac{3}{\sqrt{2}},0\right)= f\left(0,-\sqrt{2},\frac{3}{\sqrt{2}},0\right)= -3.$ Hence the constrained maximum and minimum values are $$3$$ and $$-3$$.

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

$L_{x}=w-a\lambda x,\quad L_{y}=-z-b\lambda y,$ $L_{z}=-y-c\mu z=0,\quad L_{w}=x-d\mu w=0$ $x_{0}=\mu dw_{0},\quad y_{0}=-c\mu z_{0},\quad z_{0}=-b\lambda y_{0}, \text{\; and\;\;} w_{0}=\lambda a x_{0}.$ This implies that $x_{0}w_{0}-y_{0}z_{0}=\lambda (ax_{0}^{2}+by_{0}^{2})=\lambda \text{\; and\;\;} x_{0}w_{0}-y_{0}z_{0}=\mu(cz_{0}^{2}+dw_{0}^{2}) =\mu,$ so $$\lambda =\mu$$. Therefore, $x_{0}=\lambda dw_{0},\quad y_{0}=-c\lambda z_{0},\quad z_{0}=-b\lambda y_{0}, \text{\; and\;\;} w_{0}=\lambda a x_{0},$ 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 $\displaystyle{\frac{1}{\sqrt{ad}}},\text{\; attained at\;\;} \pm \displaystyle{\left(\frac{1}{\sqrt{a}},0,0,\frac{1}{\sqrt{d}}\right)}$ and the constrained minimum is $-\displaystyle{\frac{1}{\sqrt{ad}}},\text{\; attained at\;\;} \pm \displaystyle{\left(-\frac{1}{\sqrt{a}},0,0,\frac{1}{\sqrt{d}}\right)}.$ 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 $\displaystyle{\frac{1}{\sqrt{bc}}}, \text{\; attained at\;\;} \pm \displaystyle{\left(0,\frac{1}{\sqrt{b}},-\frac{1}{\sqrt{c}},0\right)},$ and the constrained minimum is $-\displaystyle{\frac{1}{\sqrt{bc}}}, \text{\; attained at\;\;} \pm \displaystyle{\left(0,\frac{1}{\sqrt{b}},\frac{1}{\sqrt{c}},0\right)}.$

Suppose that $$ad=bc$$. Since $$x_{0}=\lambda dw_{0}$$ and $$y_{0}=-c\lambda z_{0}$$, $1=ax_{0}^{2}+by_{0}^{2}=\lambda^{2}[(ad)dw_{0}^2+(bc)cz_{0}^{2}]= \lambda^{2}ad(cz_{0}^{2}+d(w_{0})^{2}=\lambda^{2}ad,$ 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, $$ax_{0}^{2}+by_{0}^{2}=1$$.

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

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

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

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

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

Assume that ${\bf u}=\frac{a_{1}}{\sqrt{\alpha}}{\bf i}+ \frac{a_{2}}{\sqrt{\beta}}{\bf j}+ \frac{a_{3}}{\sqrt{\gamma}}{\bf k} \text{\; and\;\;} {\bf v}=\frac{b_{1}}{\sqrt{\alpha}}{\bf i}+ \frac{b_{2}}{\sqrt{\beta}}{\bf j}+ \frac{b_{3}}{\sqrt{\gamma}}{\bf k}$ 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 $\lambda=\frac{c|{\bf v}|^{2}-d({\bf u}\cdot{\bf v})}{\Delta},\quad \mu=\frac{d|{\bf u}|^{2}-c({\bf u}\cdot{\bf v})}{\Delta}.$ 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}

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

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

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

$\lambda=c-\sum_{i=1}^{n}a_{i}\alpha_{i},\quad \mu=d-\sum_{i=1}^{n}b_{i}\alpha_{i}$ \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}

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

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

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

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

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

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

[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 $x_{i0}=\displaystyle{\frac{p_{i}\sigma_{i}S}{p_{1}+p_{2}+\cdots+p_{n}}}.$ and $x_{10}^{p_{1}}x_{20}^{p_{2}}\cdots x_{n0}^{p_{n}}= \left(\frac{S}{p_{1}+p_{2}+\cdots+ p_{n}}\right)^{p_{1}+p_{2}+\cdots+p_{n}} (p_{1}\sigma_{1})^{p_{1}} (p_{2}\sigma_{2})^{p_{2}} \cdots (p_{n}\sigma_{n})^{p_{n}}$

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

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

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

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

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

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

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

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 $S_{\sigma}=\left\{(x_{1},x_{2}, \dots, x_{m})\, \big|\, x_{i}>0, \,1 \le i \le m,\, r_{1}x_{1}+r_{2}x_{2}+\cdots+ r_{m}x_{m}=\sigma\right\}.$ $L=x_{1}^{r_{1}}x_{2}^{r_{2}}\cdots x_{m}^{r_{m}} -\lambda\sum_{i=1}^{m}r_{i}x_{i},\quad L_{x_{i}}=r_{i}\left(\frac{x_{1}^{r_{1}}x_{2}^{r_{2}}\cdots x_{m}^{r_{m}}}{x_{i}}-\lambda\right), \quad 1 \le i \le m.$ 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 $\left(x_{1}^{r_{1}}x_{2}^{r_{2}}\cdots x_{m}^{r_{m}}\right)^{1/r} \le \frac{\sigma}{r}=\frac{r_{1}{x_{1}+r_{2}x_{2}+\cdots+r_{k}x_{k}}}{r}$ 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 $\frac{|a_{1j}||a_{2j}|\cdots|a_{mj}|} {\sigma_{1}^{1/p_{1}}\sigma_{2}^{1/p_{2}}\cdots\sigma_{m}^{1/p_{m}}} \le \sum_{i=1}^{m} \frac{|a_{ij}|^{p_{i}}}{p_{i}\sigma_{i}}.$ Summing both sides from $$j=1$$ to $$n$$ yields the stated conclusion.

$$\displaystyle{L =\frac{1}{2}\sum_{r=0}^{n}x_{r}^{2}-\sum_{s=0}^{m} \lambda_{s}\sum_{r=0}^{n}x_{r}r^{s}}$$, $$L_{x_{r}}=x_{r}-\displaystyle{\sum_{s=0}^{m}\lambda_{s}r^{s}}$$ $x_{r0}=\sum_{s=0}^{m}\lambda_{s}r^{s},\quad 0\le r\le n.$ $\sum_{r=0}^{n}x_{r0}r^{s}=\sum_{r=0}^{n}\sum_{\ell=0}^{m}\lambda_{\ell}r^{\ell+s} =\sum_{\ell=0}^{m}\lambda_{\ell}\sum_{r=0}^{n}r^{\ell+s}= \sum_{\ell=0}^{m}\sigma_{s+\ell}\lambda_{\ell}=c_{s}, \quad 0\le s \le m,$ 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, $\sum_{r=0}^{n}y_{r}r^{s}=c_{s},\quad 0\le s \le m.$ Then $\sum_{r=0}^{n}(y_{r}-x_{r0})x_{r0}=\sum_{r=0}^{n}(y_{r}-x_{r0}) \sum_{s=0}^{m}\lambda_{s}r^{s}=\sum_{s=0}^{m}\lambda_{s}\sum_{r=0}^{n} (y_{r}-x_{r0})r^{s}=0,$ 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}

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 $\sum_{i=-n}^{n}w_{i}P(r-i)=P(r)$ whenever $$r$$ is an integer and $$P$$ is a polynomial of degree $$\le 2k$$. Therefore, $L=\frac{1}{2}\sum_{i=-n}^{n}w_{i}^{2}-\sum_{r=0}^{2k}\lambda_{r} \sum_{i=-n}^{n}w_{i}i^{r},$ $L_{w_{i}}=w_{i}-\sum_{r=0}^{2k}\lambda_{r}i^{r},\quad w_{i0}=\sum_{r=0}^{2k}\lambda_{r}i^{r}, \quad -n\le i\le n,$ and $\sum_{i=-n}^{n}w_{i0}i^{s}= \sum_{i=-n}^{n} \left(\sum_{r=0}^{2k} \lambda_{r}i^{r}\right)i^{s} =\sum_{r=0}^{2k}\lambda_{r}\sigma_{r+s}\text{\; where\;\;} \sigma_{m}=\sum_{i=-n}^{n}i^{m}.$

If $$\{w_{i}\}_{i=-n}^{n}$$ also satisfies the constraint, then $\sum_{i=-n}^{n}(w_{i}-w_{i0})w_{i0}= \sum_{i=-n}^{n}(w_{i}-w_{i0})\sum_{r=0}^{2k}\lambda_{r}i^{r}=0.$ 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} if $$w_{i}\ne w_{i0}$$ for some $$i$$.

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

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

If $$\{w_{i}\}_{i=0}^{n}$$ also satisfies the constraint, then $\sum_{i=0}^{n}(w_{i}-w_{i0})w_{i0}= \sum_{i=n}^{n}(w_{i}-w_{i0})\sum_{r=0}^{2k}\lambda_{r}i^{r}=0.$ 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} if $$w_{i}\ne w_{i0}$$ for some $$i$$.

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

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

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

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

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

6: Exercises is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.