6.10.E: Further Problems on Maxima and Minima
- Page ID
- 24102
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Fill in all details in Examples 1 and 2 and the proofs of all theorems in this section.
Redo Example (B) in §9 by Lagrange's method.
[Hint: Set \(F(x, y, z)=f(x, y, z)-r\left(x^{2}+y^{2}+z^{2}\right), g(x, y, z)=x^{2}+y^{2}+z^{2}-1\). Compare the values of \(f\) at all critical points.]
An ellipsoid
\[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1\]
is cut by a plane \(u x+v y+w z=0.\) Find the semiaxes of the section-ellipse, i.e., the extrema of
\[\rho^{2}=[f(x, y, z)]^{2}=x^{2}+y^{2}+z^{2}\]
under the constraints \(g=\left(g_{1}, g_{2}\right)=\overrightarrow{0},\) where
\[g_{1}(x, y, z)=u x+v y+w z \text { and } g_{2}(x, y, z)=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}-1.\]
Assume that \(a>b>c>0\) and that not all \(u, v, w=0\).
[Outline: By Note 2, explore the rank of the matrix
\[\left(\begin{array}{ccc}{x / a^{2}} & {y / b^{2}} & {z / c^{2}} \\ {u} & {v} & {z}\end{array}\right).\]
(Why this particular matrix?)
Seeking a contradiction, suppose all its \(2 \times 2\) determinants vanish at all points of the section-ellipse. Then the upper and lower entries in (14) are proportional (why?); so \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=0\) (a contradiction!).
Next, set
\[F(x, y, z)=x^{2}+y^{2}+z^{2}+r\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}\right)+2 s(u x+v y+w z).\]
Equate \(dF\) to \(0:\)
\[x+\frac{r x}{a^{2}}+s u=0, \quad y+\frac{r y}{b^{2}}+s v=0, \quad z+\frac{r z}{c^{2}}+s w=0.\]
Multiplying by \(x, y, z,\) respectively, adding, and combining with \(g=\overrightarrow{0},\) obtain \(r=\) \(-\rho^{2};\) so, by (15), for \(a, b, c \neq \rho\),
\[x=\frac{-s u a^{2}}{a^{2}-\rho^{2}}, \quad y=\frac{-s v b^{2}}{b^{2}-\rho^{2}}, \quad z=\frac{-s w c^{2}}{c^{2}-\rho^{2}}.\]
Find \(s, x, y, z,\) then compare the \(\rho\)-values at critical points.]
Find the least and the largest values of the quadratic form
\[f(\vec{x})=\sum_{i, k=1}^{n} a_{i k} x_{i} x_{k} \quad\left(a_{i k}=a_{k i}\right)\]
on the condition that \(g(\vec{x})=|\vec{x}|^{2}-1=0\left(f, g : E^{n} \rightarrow E^{1}\right)\).
[Outline: Let \(F(\vec{x})=f(\vec{x})-t\left(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}\right).\) Equating \(d F\) to \(0,\) obtain
\[\begin{array}{l}{\left(a_{11}-t\right) x_{1}+a_{12} x_{2}+\ldots+a_{1 n} x_{n}=0,} \\ {a_{21} x_{1}+\left(a_{22}-t\right) x_{2}+\ldots+a_{2 n} x_{n}=0,} \\ {\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots} \\ {a_{n 1} x_{1}+a_{n 2} x_{2}+\ldots+\left(a_{n n}-t\right) x_{n}=0.}\end{array}\]
Using Theorem 1(iv) in §6, derive the so-called characteristic equation of \(f\),
\[\left|\begin{array}{cccc}{a_{11}-t} & {a_{12}} & {\dots} & {a_{1 n}} \\ {a_{21}} & {a_{22}-t} & {\dots} & {a_{2 n}} \\ {\dots} & {\dots} & {\dots} & {\dots} \\ {a_{n 1}} & {a_{2 n}} & {\dots} & {a_{n n}-t}\end{array}\right|=0,\]
of degree \(n\) in \(t.\) If \(t\) is one of its \(n\) roots (known to be real), then equations (16) admit a nonzero solution for \(\vec{x}=\left(x_{1}, \ldots, x_{n}\right);\) by replacing \(\vec{x}\) by \(\vec{x} /|\vec{x}|\) if necessary, \(\vec{x}\) satisfies also the constraint equation \(g(\vec{x})=|\vec{x}|^{2}-1=0.\) (Explain!) Thus each root \(t\) of (17) yields a critical point \(\vec{x}_{t}=\left(x_{1}, \ldots, x_{n}\right).\)
Now, to find \(f\left(\vec{x}_{t}\right),\) multiply the \(k\)th equation in (16) by \(x_{k}, k=1, \ldots, n,\) and add to get
\[0=\sum_{i, k=1}^{n} a_{i k} x_{i} x_{k}-t \sum_{k=1}^{n} x_{k}^{2}=f\left(\vec{x}_{t}\right)-t.\]
Hence \(f\left(\vec{x}_{t}\right)=t\).
Thus the values of \(f\) at the critical points \(\vec{x}_{t}\) are simply the roots of (17). The largest (smallest) root is also the largest (least) value of \(f\) on \(S=\left\{\vec{x} \in E^{n}| | \vec{x} |=1\right\}\) (Explain!)]
Use the method of Problem 4 to find the semiaxes of
(i) the quadric curve in \(E^{2},\) centered at \(\overrightarrow{0},\) given by \(\sum_{i, k=1}^{2} a_{i k} x_{i} x_{k}=1;\) and
(ii) the quadric surface \(\sum_{i, k=1}^{3} a_{i k} x_{i} x_{k}=1\) in \(E^{3},\) centered at (\overrightarrow{0}\).
Assume \(a_{i k}=a_{k I}\).
[Hint: Explore the extrema of \(f(\vec{x})=|\vec{x}|^{2}\) on the condition that
\[g(\vec{x})=\sum_{i, k} a_{i k} x_{i} x_{k}-1=0.]\]
Using Lagrange's method, redo Problems 4, 5, 6, 7, 11, 12, and 13 of §9.
In \(E^{2},\) find the shortest distance from \(\overrightarrow{0}\) to the parabola \(y^{2}=2(x+a)\).
In \(E^{3}\), find the shortest distance from \(\overrightarrow{0}\) to the intersection line of two planes given by the formulas \(\vec{u} \cdot \vec{x}=a\) and \(\vec{v} \cdot \vec{x}=b\) with \(\vec{u}\) and \(\vec{v}\) different from \(\overrightarrow{0}.\) (Rewrite all in coordinate form!)
In \(E^{n},\) find the largest value of \(|\vec{a} \cdot \vec{x}|\) if \(|\vec{x}|=1.\) Use Lagrange's method.
(Hadamard's theorem.) If \(A=\operatorname{det}\left(x_{i k}\right)(i, k \leq n),\) then
\[|A| \leq \prod_{i=1}^{n}\left|\vec{x}_{i}\right|,\]
where \(\vec{x}_{i}=\left(x_{i 1}, x_{i 2}, \ldots, x_{i n}\right)\).
[Hints: Set \(a_{i}=\left|\vec{x}_{i}\right|.\) Treat \(A\) as a function of \(n^{2}\) variables. Using Lagrange's method, prove that, under the \(n\) constraints \(\left|\vec{x}_{i}\right|^{2}-a_{i}^{2}=0, A\) cannot have an extremum unless \(A^{2}=\operatorname{det}\left(y_{i k}\right),\) with \(y_{I k}=0\) (if \(i \neq k\)) and \(y_{ii}=a_{i}^{2}.]