6.9.E: Problems on Maxima and Minima
Verify Note 1.
Complete the missing details in the proof of Theorems 2 and 3.
Verify Examples (A) and (B). Supplement Example (A) by applying Theorem 2.
Test \(f\) for extrema in \(E^{2}\) if \(f(x, y)\) is
(i) \(\frac{x^{2}}{2 p}+\frac{y^{2}}{2 q}(p>0, q>0)\);
(ii) \(\frac{x^{2}}{2 p}-\frac{y^{2}}{2 q}(p>0, q>0)\);
(iii) \(y^{2}+x^{4}\);
(iv) \(y^{2}+x^{3}\).
(i) Find the maximum volume of an interval \(A \subset E^{3}\) (see Chapter 3, §7) whose edge lengths \(x, y, z\) have a prescribed sum: \(x+y+z=a\).
(ii) Do the same in \(E^{4}\) and in \(E^{n};\) show that \(A\) is a cube.
(iii) Hence deduce that
\[\sqrt[n]{x_{1} x_{2} \cdots x_{n}} \leq \frac{1}{n} \sum_{1}^{n} x_{k} \quad\left(x_{k} \geq 0\right),\]
i.e., the geometric mean of \(n\) nonnegative numbers is \(\leq\) their arithmetic mean.
Find the minimum value for the sum \(f(x, y, z, t)=x+y+z+t\) of four positive numbers on the condition that \(x y z t=c^{4}\) (constant).
[Answer: \(x=y=z=t=c; f_{\max }=4c\).]
Among all triangles inscribed in a circle of radius \(R,\) find the one of maximum area.
[Hint: Connect the vertices with the center. Let \(x, y, z\) be the angles at the center. Show that the area of the triangle \(=\frac{1}{2} R^{2}(\sin x+\sin y+\sin z),\) with \(z=2 \pi-(x+y)\).]
Among all intervals \(A \subset E^{3}\) inscribed in the ellipsoid
\[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1\]
find the one of largest volume.
[Answer: the edge lengths are \(\frac{2 a}{\sqrt{3}}, \frac{2 b}{\sqrt{3}}, \frac{2 c}{\sqrt{3}}\).]
Let \(P_{i}=\left(a_{i} \cdot b_{i}\right), i=1,2,3,\) be 3 points in \(E^{2}\) forming a triangle in which one angle (say, \(\,easureangle P_{1})\) is \(\geq 2 \pi / 3\).
Find a point \(P=(x, y)\) for which the sum of the distances,
\[P P_{1}+P P_{2}+P P_{3}=\sum_{i=1}^{3} \sqrt{\left(x-a_{i}\right)^{2}+\left(y-b_{i}\right)^{2}},\]
is the least possible.
[Outline: Let \(f(x, y)=\sum_{i=1}^{3} \sqrt{\left(x-a_{i}\right)^{2}+\left(y-b_{i}\right)^{2}}\).
Show that \(f\) has no partial derivatives at \(P_{1}, P_{2},\) or \(P_{3}\) (and so \(P_{1}, P_{2},\) and \(P_{3}\) are critical points at which an extremum may occur), while at other points \(P,\) partials do exist but never vanish simultaneously, so that there are no other critical points.
Indeed, prove that \(D_{1} f(P)=0=D_{2} f(P)\) would imply that
\[\sum_{i=1}^{3} \cos \theta_{i}=0=\sum_{1}^{3} \sin \theta_{I},\]
where \(\theta_{i}\) is the angle between \(\overline{P P_{i}}\) and the \(x\)-axis; hence
\[\sin \left(\theta_{1}-\theta_{2}\right)=\sin \left(\theta_{2}-\theta_{3}\right)=\sin \left(\theta_{3}-\theta_{1}\right) \quad\text {(why?),}\]
and so \(\theta_{1}-\theta_{2}=\theta_{2}-\theta_{3}=\theta_{3}-\theta_{1}=2 \pi / 3,\) contrary to \(\angle P_{1} \geq 2 \pi / 3.\) (Why?)
From geometric considerations, conclude that \(f\) has an absolute minimum at \(P_{1}\).
(This shows that one cannot disregard points at which \(f\) has no partials.)]
Continuing Problem 8, show that if none of \(\angle P_{1}, \angle P_{2},\) and \(\angle P_{3}\) is \(\geq\) \(2 \pi / 3,\) then \(f\) attains its least value at some \(P\) (inside the triangle) such that \(\angle P_{1} P P_{2}=\angle P_{2} P P_{3}=\angle P_{3} P P_{1}=2 \pi / 3\).
[Hint: Verify that \(D_{1} f=0=D_{2} f\) at \(P\).
Use the law of cosines to show that \(P_{1} P_{2}>P P_{2}+\frac{1}{2} P P_{1}\) and \(P_{1} P_{3}>P P_{3}+\frac{1}{2} P P_{1}\).
Adding, obtain \(P_{1} P_{3}+P_{1} P_{2}>P P_{1}+P P_{2}+P P_{3},\) i.e., \(f\left(P_{1}\right)>f(P).\) Similarly, \(f\left(P_{2}\right)>f(P)\) and \(f\left(P_{3}\right)>f(P).\)
Combining with Problem 8, obtain the result.]
In a circle of radius \(R\) inscribe a polygon with \(n+1\) sides of maximum area.
[Outline: Let \(x_{1}, x_{2}, \ldots, x_{n+1}\) be the central angles subtended by the sides of the polygon. Then its area \(A\) is
\[\frac{1}{2} R^{2} \sum_{k=1}^{n+1} \sin x_{k},\]
with \(x_{n+1}=2 \pi-\sum_{k=1}^{n} x_{k}.\) (Why?) Thus all reduces to maximizing
\[f\left(x_{1}, \ldots, x_{n}\right)=\sum_{k=1}^{n} \sin x_{k}+\sin \left(2 \pi-\sum_{k=1}^{r_{k}} x_{k}\right),\]
on the condition that \(0 \leq x_{k}\) and \(\sum_{k=1}^{n} x_{k} \leq 2 \pi.\) (Why?)
These inequalities define a bounded set \(D \subset E^{n}\) (called a simplex). Equating all partials of \(f\) to \(0,\) show that the only critical point interior to \(D\) is \(\vec{x}=\left(x_{1}, \ldots, x_{n}\right),\) with \(x_{k}=\frac{2 \pi}{n+1}, k \leq n\) (implying that \(x_{n+1}=\frac{2 \pi}{n+1},\) too). For that \(\vec{x},\) we get
\[f(\vec{x})=(n+1) \sin [2 \pi /(n+1)].\]
This value must be compared with the "boundary" values of \(f,\) on the "faces" of the simplex D (see Note 4).
Do this by induction. For \(n=2,\) Problem 6 shows that \(f(\vec{x})\) is indeed the largest when all \(x_{k}\) equal \(\frac{2 \pi}{n+1}.\) Now let \(D_{n}\) be the "face" of \(D,\) where \(x_{n}=0.\) On that face, treat \(f\) as a function of only \(n-1\) variables, \(x_{1}, \ldots, x_{n-1}\).
By the inductive hypothesis, the largest value of \(f\) on \(D_{n}\) is \(n \sin (2 \pi / n).\) Similarly for the other "faces." As \(n \sin (2 \pi / n)<(n+1) \sin 2 \pi /(n+1),\) the induction is complete.
Thus, the area \(A\) is the largest when the polygon is regular, for which
\[\left.A=\frac{1}{2} R^{2}(n+1) \sin \frac{2 \pi}{n+1}.\right]\]
Among all triangles of a prescribed perimeter \(2p,\) find the one of maximum area.
[Hint: Maximize \(p(p-x)(p-y)(p-z)\) on the condition that \(x+y+z=2 p\).]
Among all triangles of area \(A,\) find the one of smallest perimeter.
Find the shortest distance from a given point \(\vec{p} \in E^{n}\) to a given plane \(\vec{u} \cdot \vec{x}=c\) (Chapter 3, §§4-6). Answer:
\[\pm \frac{\vec{u} \cdot \vec{p}-c}{|\vec{u}|}.\]
[Hint: First do it in \(E^{3},\) writing \((x, y, z)\) for \(\vec{x}\).]