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Mathematics LibreTexts

6: Exercises

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    17321
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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 \[\sum_{i=1}^{n}\alpha_{i}a_{ir}a_{is}= \begin{cases} 1 & \text{ if } r=s,\\0 & \text{ if }r\ne s. \end{cases}\]

    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

    \(L=\displaystyle{\frac{(x-1)^{2}+(y+2)^{2}+(z-3)^{2}}{2}-\lambda(2x+3y+z)}\) \[L_{x}=x-1-2\lambda,\quad L_{y}=y+2-3\lambda, \quad L_{z}=z-3-\lambda\] \[x_{0}=1+2\lambda, \quad y_{0}=-2+3\lambda, \quad z_{0}=3+\lambda \quad\] \[2(1+2\lambda)+3(-2+3\lambda)+(3+\lambda)=7, \quad \lambda=\displaystyle{\frac{4}{7}}\] \[x_{0}=\displaystyle{\frac{15}{7}}, \quad y_{0}=-\displaystyle{\frac{2}{7}},\quad z_{0}=\displaystyle{\frac{25}{7}}\] 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 \(i\).

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

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

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

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

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