6.9: Local Extrema. Maxima and Minima

We say that $f : E^{\prime} \rightarrow E^{1}$ has a local maximum (minimum) at $\vec{p} \in E^{\prime}$ iff $f(\vec{p})$ is the largest (least) value of $f$ on some globe $G$ about $\vec{p};$ more precisely, iff

$$(\forall \vec{x} \in G) \quad \Delta f=f(\vec{x})-f(\vec{p})<0(>0).$$

We speak of an improper extremum if we only have $\Delta f \leq 0( \geq 0)$ on $G.$ In any case, all depends on the sign of $\Delta f.$

From Problem 6 in §1, recall the following necessary condition.

Theorem $\PageIndex{1}$

If $f : E^{\prime} \rightarrow E^{1}$ has a local extremum at $\vec{p}$ then $D_{\vec{u}} f(\vec{p})=0$ for all $\vec{u} \neq \overrightarrow{0}$ in $E^{\prime}.$

In the case $E^{\prime}=E^{n}\left(C^{n}\right),$ this means that $d^{1} f(\vec{p} ; \cdot)=0$ on $E^{\prime}$.

(Recall that $d^{1} f(\vec{p} ; \vec{t})=\sum_{k=1}^{n} D_{k} f(\vec{p}) t_{k}.$ It vanishes if the $D_{k} f(\vec{p})$ do.

Note 1. This condition is only necessary, not sufficient. For example, if $f(x, y)=x y,$ then $d^{1} f(\overrightarrow{0} ; \cdot)=0 ;$ yet $f$ has no extremum at $\overrightarrow{0}.$ (Verify!)

Sufficient conditions were given in Theorem 2 of §5, for $E^{\prime}=E^{1}.$ We now take up $E^{\prime}=E^{2}.$

Theorem $\PageIndex{2}$

Let $f : E^{2} \rightarrow E^{1}$ be of class $C D^{2}$ on a globe $G=G_{\vec{p}}(\delta).$ Suppose $d^{1} f(\vec{p} ; \cdot)=0$ on $E^{2}.$ Set $A=D_{11} f(\vec{p}), B=D_{12} f(\vec{p}),$ and $C=D_{22} f(\vec{p})$.

Then the following statements are true.

(i) If $A C>B^{2},$ has a maximum or minimum at $\vec{p},$ according to whether

$A<0$ or $A>0.$

(ii) If $A C<B^{2}, f$ has no extremum at $\vec{p}$.

The case $A C=B$ is unresolved.

Proof

Let $\vec{x} \in G$ and $\vec{u}=\vec{x}-\vec{p} \neq \overrightarrow{0}$.

As $d^{1} f(\vec{p} ; \cdot)=0,$ Theorem 2 in §5, yields

$$\Delta f=f(\vec{x})-f(\vec{p})=R_{1}=\frac{1}{2} d^{2} f(\vec{s} ; \vec{u}),$$

with $\vec{s} \in L(\vec{p}, \vec{x}) \subseteq G$ (see Corollary 1 of §5). As $f \in C D^{2},$ we have $D_{12} f=D_{21} f$ on $G$ (Theorem 1 in §5). Thus by formula (4) in §5,

$$\Delta f=\frac{1}{2} d^{2} f(\vec{s} ; \vec{u})=\frac{1}{2}\left[D_{11} f(\vec{s}) u_{1}^{2}+2 D_{12} f(\vec{s}) u_{1} u_{2}+D_{22} f(\vec{s}) u_{2}^{2}\right].$$

Now, as the partials involved are continuous, we can choose $G=G_{\vec{p}}(\delta)$ so small that the sign of expression (1) will not change if $\vec{s}$ is replaced by $\vec{p}$. Then the crucial sign of $\Delta f$ on $G$ coincides with that of

$$D=A u_{1}^{2}+2 B u_{1} u_{2}+C u_{2}^{2}$$

(with $A, B,$ and $C$ as stated in the theorem).

From (2) we obtain, by elementary algebra,

$$\begin{aligned} A D &=\left(A u_{1}+B u_{2}\right)^{2}+\left(A C-B^{2}\right) u_{2}^{2}, \\ C D &=\left(C u_{1}+B u_{2}\right)^{2}+\left(A C-B^{2}\right) u_{2}^{2}. \end{aligned}$$

Clearly, if $A C>B^{2},$ the right-side expression in (3) is $>0;$ so $A D>0$, i.e., $D$ has the same sign as $A.$

Hence if $A<0,$ we also have $\Delta f<0$ on $G,$ and $f$ has a maximum at $\vec{p}.$ If $A>0,$ then $\Delta f>0,$ and $f$ has a minimum at $\vec{p}$.

Now let $A C<B^{2}$. We claim that no matter how small $G=G_{\vec{p}}(\delta), \Delta f$ changes sign as $\vec{x}$ varies in $G,$ and so $f$ has no extremum at $\vec{p}$.

Indeed, we have $\vec{x}=\vec{p}+\vec{u}, \vec{u}=\left(u_{1}, u_{2}\right) \neq \overrightarrow{0}.$ If $u_{2}=0,$ (3) shows that $D$ and $\Delta f$ have the same sign as $A(A \neq 0).$

But if $u_{2} \neq 0$ and $u_{1}=-B u_{2} / A$ (assuming $A \neq 0),$ then $D$ and $\Delta f$ have the sign opposite to that of $A;$ and $\vec{x}$ is still in $G$ if $u_{2}$ is small enough (how small?).

One proceeds similarly if $C \neq 0$ (interchange $A$ and $C,$ and use (3').

Finally, if $A=C=0,$ then by (2), $D=2 B u_{1} u_{2}$ and $B \neq 0$ (since $A C<B^{2})$. Again $D$ and $\Delta f$ change sign as $u_{1} u_{2}$ does; so $f$ has no extremum at $\vec{p}.$ Thus all is proved.$\quad \square$

Briefly, the proof utilizes the fact that the trinomial (2) is sign-changing iff its discriminant $B^{2}-A C$ is positive, i.e., $\left|\begin{array}{cc}{A} & {B} \\ {B} & {C}\end{array}\right|<0$.

Note 2. Functions $f : C \rightarrow E^{1}$ (of one complex variable) are likewise covered by Theorem 2 if one treats them as functions on $E^{2}$ (of two real variables).

Functions of n variables. Here we must rely on the algebraic theory of so-called symmetric quadratic forms, i.e., polynomials $P : E^{n} \rightarrow E^{1}$ of the form

$$P(\vec{u})=\sum_{j=1}^{n} \sum_{i=1}^{n} a_{i j} u_{i} u_{j},$$

where $\vec{u}=\left(u_{i}, \ldots, u_{n}\right) \in E^{n}$ and $a_{i j}=a_{j i} \in E^{1}$.

We take for granted a theorem due to J. J. Sylvester (see S. Perlis, Theory of Matrices, 1952, p. 197), which may be stated as follows.

Let $P : E^{n} \rightarrow E^{1}$ be a symmetric quadratic form,

$$P(\vec{u})=\sum_{j=1}^{n} \sum_{i=1}^{n} a_{i j} u_{i} u_{j}.$$

(i) $P>0$ on all of $E^{n}-\{\overrightarrow{0}\}$ iff the following $n$ determinants $A_{k}$ are positive:

$$A_{k}=\left|\begin{array}{cccc}{a_{11}} & {a_{12}} & {\dots} & {a_{1 k}} \\ {a_{21}} & {a_{22}} & {\dots} & {a_{2 k}} \\ {\ldots} & {\ldots \ldots \ldots} & {\ldots} & {a_{2 k}} \\ {a_{k 1}} & {a_{k 2}} & {\dots} & {a_{k k}}\end{array}\right|, \quad k=1,2, \ldots, n.$$

(ii) We have $P<0$ on $E^{n}-\{\overrightarrow{0}\}$ iff $(-1)^{k} A_{k}>0$ for $k=1,2, \ldots, n$.

Now we can extend Theorem 2 to the case $f : E^{n} \rightarrow E^{1}$. (This will also cover $f : C^{n} \rightarrow E^{1},$ treated as $f : E^{2 n} \rightarrow E^{1}.)$ The proof resembles that of Theorem 2.

Theorem $\PageIndex{3}$

Let $f : E^{n} \rightarrow E^{1}$ be of class $C D^{2}$ on some $G=G_{\vec{p}}(\delta).$ Suppose $d f(\vec{p} ; \cdot)=0$ on $E^{n}.$ Define the $A_{k}$ as in (4), with $a_{i j}=D_{i j} f(\vec{p}), i, j, k \leq n$ Then the following statements hold.

(i) $f$ has a local minimum at $\vec{p}$ if $A_{k}>0$ for $k=1,2, \ldots, n$.

(ii) $f$ has a local maximum at $\vec{p}$ if $(-1)^{k} A_{k}>0$ for $k=1, \ldots, n$.

(iii) $f$ has no extremum at $\vec{p}$ if the expression

$$P(\vec{u})=\sum_{j=1}^{n} \sum_{i=1}^{n} a_{i j} u_{i} u_{j}$$

is $>0$ for some $\vec{u} \in E^{n}$ and $<0$ for others (i.e., $P$ changes sign on $E^{n})$.

Proof

Let again $\vec{x} \in G, \vec{u}=\vec{x}-\vec{p} \neq \overrightarrow{0},$ and use Taylor's theorem to obtain

$$\Delta f=f(\vec{x})-f(\vec{p})=R_{1}=\frac{1}{2} d^{2} f(\vec{s} ; \vec{u})=\sum_{j=1}^{n} \sum_{i=1}^{n} D_{i j} f(\vec{s}) u_{i} u_{j},$$

with $\vec{s} \in L(\vec{x}, \vec{p})$.

As $f \in C D^{2},$ the partials $D_{i j} f$ are continuous on $G.$ Thus we can make $G$ so small that the sign of the last double sum does not change if $\vec{s}$ is replaced by $\vec{p}$. Hence the sign of $\Delta f$ on $G$ is the same as that of $P(\vec{u})=\sum_{j=1}^{n} \sum_{i=1}^{n} a_{i j} u_{i} u_{j}$, with the $a_{i j}$ as stated in the theorem.

The quadratic form $P$ is symmetric since $a_{i j}=a_{j i}$ by Theorem 1 in §5. Thus by Sylvester's theorem stated above, one easily obtains our assertions (i) and (ii). Indeed, they are immediate from clauses (i) and (ii) of that theorem.

Now, for (iii), suppose $P(\vec{u})>0>P(\vec{v}),$ i.e.,

$$\sum_{j=1}^{n} \sum_{i=1}^{n} a_{i j} u_{i} u_{j}>0>\sum_{j=1}^{n} \sum_{i=1}^{n} a_{i j} v_{i} v_{j} \quad \text { for some } \vec{u}, \vec{v} \in E^{n}-\{\overrightarrow{0}\}.$$

If here $\vec{u}$ and $\vec{v}$ are replaced by $t \vec{u}$ and $t \vec{v}(t \neq 0),$ then $u_{i} u_{j}$ and $v_{i} v_{j}$ turn into $t^{2} u_{i} u_{j}$ and $t^{2} v_{i} v_{j},$ respectively. Hence

$$P(t \vec{u})=t^{2} P(\vec{u})>0>t^{2} P(\vec{v})=P(t \vec{v}).$$

Now, for any $t \in(0, \delta /|\vec{u}|),$ the point $\vec{x}=\vec{p}+t \vec{u}$ lies on the $\vec{u}$-directed line through $\vec{p},$ inside $G=G_{\vec{p}}(\delta).$ (Why?) Similarly for the point $\vec{x}^{\prime}=\vec{p}+t \vec{v}.$

Hence for such $\vec{x}$ and $\vec{x}^{\prime},$ Taylor's theorem again yields formulas analogous to (5) for some $\vec{s} \in L(\vec{p}, \vec{x})$ and $\vec{s}^{\prime} \in L\left(\vec{p}, \vec{x}^{\prime}\right)$ lying on the same two lines. It again follows that for small $\delta$,

$$f(\vec{x})-f(\vec{p})>0>f\left(\vec{x}^{\prime}\right)-f(\vec{p}),$$

just as $P(\vec{u})>0>P(\vec{v})$.

Thus $\Delta f$ changes sign on $G_{\vec{p}}(\delta),$ and (iii) is proved.$\quad \square$

Note 3. Still unresolved are cases in which $P(\vec{u})$ vanishes for some $\vec{u} \neq \overrightarrow{0},$ without changing its sign; e.g., $P(\vec{u})=\left(u_{1}+u_{2}+u_{3}\right)^{2}=0$ for $\vec{u}=(1,1,-2)$. Then the answer depends on higher-order terms of the Taylor formula. In particular, if $d^{1} f(\vec{p} ; \cdot)=d^{2} f(\vec{p} ; \cdot)=0$ on $E^{n},$ then $\Delta f=R_{2}=\frac{1}{6} d^{3} f(\vec{p} ; \vec{s}),$ etc.

Note 4. The largest or least value of $f$ on a set $A$ (sometimes called the absolute maximum or minimum) may occur at sominterior (e.g., boundary) point $\vec{p} \in A,$ and then fails to be among the local extrema (where, by definition, a globe $G_{\vec{p}} \subseteq A$ is presupposed). Thus to find absolute extrema, one must also explore the behaviour of $f$ at noninterior points of $A.$

By Theorem 1, local extrema can occur only at so-called critical points $\vec{p}$, i.e., those at which all directional derivatives vanish (or fail to exist, in which case $D_{\vec{u}} f(\vec{p})=0$ by convention).

In practice, to find such points in $E^{n}\left(C^{n}\right),$ one equates the partials $D_{k} f$ $(k \leq n)$ to $0.$ Then one uses Theorems 2 and 3 or other considerations to determine whether an extremum really exists.

Examples

(A) Find the largest value of

$$f(x, y)=\sin x+\sin y-\sin (x+y)$$

on the set $A \subseteq E^{2}$ bounded by the lines $x=0, y=0$ and $x+y=2 \pi$.

We have

$$D_{1} f(x, y)=\cos x-\cos (x+y) \text { and } D_{2} f(x, y)=\cos y-\cos (x+y).$$

Inside the triangle $A,$ both partials vanish only at the point $\left(\frac{2 \pi}{3}, \frac{2 \pi}{3}\right)$ at which $f=\frac{3}{2} \sqrt{3}.$ On the boundary of $A$ (i.e., on the lines $x=0, y=0$ and $x+y=2 \pi ), f=0.$ Thus even without using Theorem 2, it is evident that $f$ attains its largest value,

$$f\left(\frac{2 \pi}{3}, \frac{2 \pi}{3}\right)=\frac{3}{2} \sqrt{3},$$

at this unique critical point.

(B) Find the largest and the least value of

$$f(x, y, z)=a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}-\left(a x^{2}+b y^{2}+c z^{2}\right)^{2},$$

on the condition that $x^{2}+y^{2}+z^{2}=1$ and $a>b>c>0$.

As $z^{2}=1-x^{2}-y^{2},$ we can eliminate $z$ from $f(x, y, z)$ and replace $f$by $F : E^{2} \rightarrow E^{1}:$

$$F(x, y)=\left(a^{2}-c^{2}\right) x^{2}+\left(b^{2}-c^{2}\right) y^{2}+c^{2}-\left[(a-c) x^{2}+(b-c) y^{2}+c\right]^{2}.$$

(Explain!) For $F,$ we seek the extrema on the disc $\overline{G}=\overline{G}_{0}(1) \subset E^{2},$ where $x^{2}+y^{2} \leq 1$ (so as not to violate the condition $x^{2}+y^{2}+z^{2}=1)$.

Equating to 0 the two partials

$$\begin{array}{l}{D_{1} F(x, y)=2 x(a-c)\left\{(a+c)-2\left[(a-c) x^{2}+(b-c) y^{2}+c\right]^{2}\right\}=0}, \\ {D_{2} F(x, y)=2 y(b-c)\left\{(b+c)-2\left[(a-c) x^{2}+(b-c) y^{2}+c\right]^{2}\right\}=0}\end{array}$$

and solving this system of equations, we find these critical points inside $G:$

(1) $x=y=0$ ($F=0)$;

(2) $x=0, y=\pm 2^{-\frac{1}{2}}\left(F=\frac{1}{4}(b-c)^{2}\right);$ and

(3) $x=\pm 2^{-\frac{1}{2}}, y=0\left(F=\frac{1}{4}(a-c)^{2}\right)$.

(Verify!)

Now, for the boundary of $\overline{G},$ i.e., the circle $x^{2}+y^{2}=1,$ repeat this process: substitute $y^{2}=1-x^{2}$ in the formula for $F(x, y),$ thus reducing it to

$$h(x)=\left(a^{2}-b^{2}\right) x^{2}+b^{2}+\left[(a-b) x^{2}+b\right]^{2}, \quad h : E^{1} \rightarrow E^{1},$$

on the interval $[-1,1] \subset E^{1}.$ In $(-1,1)$ the derivative

$$h^{\prime}(x)=2(a-b) x\left(1-2 x^{2}\right)$$

vanishes only when

(4) $x=0$ ($h=0),$ and

(5) $x=\pm 2^{-\frac{1}{2}}\left(h=\frac{1}{4}(a-b)^{2}\right)$.

Finally, at the endpoints of $[-1,1],$ we have

(6) $x=\pm 1$ ($h=0)$.

Comparing the resulting function values in all six cases, we conclude that the least of them is $0,$ while the largest is $\frac{1}{4}(a-c)^{2}.$ These are the desired least and largest values of $f,$ subject to the conditions stated. They are attained, respectively, at the points

$$(0,0, \pm 1),(0, \pm 1,0),( \pm 1,0,0), \text { and }\left( \pm 2^{-\frac{1}{2}}, 0, \pm 2^{-\frac{1}{2}}\right).$$

Again, the use of Theorems 2 and 3 was redundant. However, we suggest as an exercise that the reader test the critical points of $F$ by using Theorem 2.

Caution. Theorems 1 to 3 apply to functions of independent variables only. In Example (B), $x, y, z$ were made interdependent by the imposed equation

$$x^{2}+y^{2}+z^{2}=1$$

(which geometrically limits all to the surface of $G_{\overrightarrow{0}}(1)$ in $E^{3}),$ so that one of them, $z,$ could be eliminated. Only then can Theorems 1 to 3 be used.