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6.9: Local Extrema. Maxima and Minima

( \newcommand{\kernel}{\mathrm{null}\,}\)

We say that f:EE1 has a local maximum (minimum) at pE iff f(p) is the largest (least) value of f on some globe G about p; more precisely, iff

(xG)Δf=f(x)f(p)<0(>0).

We speak of an improper extremum if we only have Δf0(0) on G. In any case, all depends on the sign of Δf.

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

Theorem 6.9.1

If f:EE1 has a local extremum at p then Duf(p)=0 for all u0 in E.

In the case E=En(Cn), this means that d1f(p;)=0 on E.

(Recall that d1f(p;t)=nk=1Dkf(p)tk. It vanishes if the Dkf(p) do.

Note 1. This condition is only necessary, not sufficient. For example, if f(x,y)=xy, then d1f(0;)=0; yet f has no extremum at 0. (Verify!)

Sufficient conditions were given in Theorem 2 of §5, for E=E1. We now take up E=E2.

Theorem 6.9.2

Let f:E2E1 be of class CD2 on a globe G=Gp(δ). Suppose d1f(p;)=0 on E2. Set A=D11f(p),B=D12f(p), and C=D22f(p).

Then the following statements are true.

(i) If AC>B2, has a maximum or minimum at p, according to whether

A<0 or A>0.

(ii) If AC<B2,f has no extremum at p.

The case AC=B is unresolved.

Proof

Let xG and u=xp0.

As d1f(p;)=0, Theorem 2 in §5, yields

Δf=f(x)f(p)=R1=12d2f(s;u),

with sL(p,x)G (see Corollary 1 of §5). As fCD2, we have D12f=D21f on G (Theorem 1 in §5). Thus by formula (4) in §5,

Δf=12d2f(s;u)=12[D11f(s)u21+2D12f(s)u1u2+D22f(s)u22].

Now, as the partials involved are continuous, we can choose G=Gp(δ) so small that the sign of expression (1) will not change if s is replaced by p. Then the crucial sign of Δf on G coincides with that of

D=Au21+2Bu1u2+Cu22

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

From (2) we obtain, by elementary algebra,

AD=(Au1+Bu2)2+(ACB2)u22,CD=(Cu1+Bu2)2+(ACB2)u22.

Clearly, if AC>B2, the right-side expression in (3) is >0; so AD>0, i.e., D has the same sign as A.

Hence if A<0, we also have Δf<0 on G, and f has a maximum at p. If A>0, then Δf>0, and f has a minimum at p.

Now let AC<B2. We claim that no matter how small G=Gp(δ),Δf changes sign as x varies in G, and so f has no extremum at p.

Indeed, we have x=p+u,u=(u1,u2)0. If u2=0, (3) shows that D and Δf have the same sign as A(A0).

But if u20 and u1=Bu2/A (assuming A0), then D and Δf have the sign opposite to that of A; and x is still in G if u2 is small enough (how small?).

One proceeds similarly if C0 (interchange A and C, and use (3').

Finally, if A=C=0, then by (2), D=2Bu1u2 and B0 (since AC<B2). Again D and Δf change sign as u1u2 does; so f has no extremum at p. Thus all is proved.

Briefly, the proof utilizes the fact that the trinomial (2) is sign-changing iff its discriminant B2AC is positive, i.e., |ABBC|<0.

Note 2. Functions f:CE1 (of one complex variable) are likewise covered by Theorem 2 if one treats them as functions on E2 (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:EnE1 of the form

P(u)=nj=1ni=1aijuiuj,

where u=(ui,,un)En and aij=ajiE1.

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:EnE1 be a symmetric quadratic form,

P(u)=nj=1ni=1aijuiuj.

(i) P>0 on all of En{0} iff the following n determinants Ak are positive:

Ak=|a11a12a1ka21a22a2ka2kak1ak2akk|,k=1,2,,n.

(ii) We have P<0 on En{0} iff (1)kAk>0 for k=1,2,,n.

Now we can extend Theorem 2 to the case f:EnE1. (This will also cover f:CnE1, treated as f:E2nE1.) The proof resembles that of Theorem 2.

Theorem 6.9.3

Let f:EnE1 be of class CD2 on some G=Gp(δ). Suppose df(p;)=0 on En. Define the Ak as in (4), with aij=Dijf(p),i,j,kn Then the following statements hold.

(i) f has a local minimum at p if Ak>0 for k=1,2,,n.

(ii) f has a local maximum at p if (1)kAk>0 for k=1,,n.

(iii) f has no extremum at p if the expression

P(u)=nj=1ni=1aijuiuj

is >0 for some uEn and <0 for others (i.e., P changes sign on En).

Proof

Let again xG,u=xp0, and use Taylor's theorem to obtain

Δf=f(x)f(p)=R1=12d2f(s;u)=nj=1ni=1Dijf(s)uiuj,

with sL(x,p).

As fCD2, the partials Dijf are continuous on G. Thus we can make G so small that the sign of the last double sum does not change if s is replaced by p. Hence the sign of Δf on G is the same as that of P(u)=nj=1ni=1aijuiuj, with the aij as stated in the theorem.

The quadratic form P is symmetric since aij=aji 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(u)>0>P(v), i.e.,

nj=1ni=1aijuiuj>0>nj=1ni=1aijvivj for some u,vEn{0}.

If here u and v are replaced by tu and tv(t0), then uiuj and vivj turn into t2uiuj and t2vivj, respectively. Hence

P(tu)=t2P(u)>0>t2P(v)=P(tv).

Now, for any t(0,δ/|u|), the point x=p+tu lies on the u-directed line through p, inside G=Gp(δ). (Why?) Similarly for the point x=p+tv.

Hence for such x and x, Taylor's theorem again yields formulas analogous to (5) for some sL(p,x) and sL(p,x) lying on the same two lines. It again follows that for small δ,

f(x)f(p)>0>f(x)f(p),

just as P(u)>0>P(v).

Thus Δf changes sign on Gp(δ), and (iii) is proved.

Note 3. Still unresolved are cases in which P(u) vanishes for some u0, without changing its sign; e.g., P(u)=(u1+u2+u3)2=0 for u=(1,1,2). Then the answer depends on higher-order terms of the Taylor formula. In particular, if d1f(p;)=d2f(p;)=0 on En, then Δf=R2=16d3f(p;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 pA, and then fails to be among the local extrema (where, by definition, a globe GpA 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 p, i.e., those at which all directional derivatives vanish (or fail to exist, in which case Duf(p)=0 by convention).

In practice, to find such points in En(Cn), one equates the partials Dkf (kn) 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)=sinx+sinysin(x+y)

on the set AE2 bounded by the lines x=0,y=0 and x+y=2π.

We have

D1f(x,y)=cosxcos(x+y) and D2f(x,y)=cosycos(x+y).

Inside the triangle A, both partials vanish only at the point (2π3,2π3) at which f=323. On the boundary of A (i.e., on the lines x=0,y=0 and x+y=2π),f=0. Thus even without using Theorem 2, it is evident that f attains its largest value,

f(2π3,2π3)=323,

at this unique critical point.

(B) Find the largest and the least value of

f(x,y,z)=a2x2+b2y2+c2z2(ax2+by2+cz2)2,

on the condition that x2+y2+z2=1 and a>b>c>0.

As z2=1x2y2, we can eliminate z from f(x,y,z) and replace fby F:E2E1:

F(x,y)=(a2c2)x2+(b2c2)y2+c2[(ac)x2+(bc)y2+c]2.

(Explain!) For F, we seek the extrema on the disc ¯G=¯G0(1)E2, where x2+y21 (so as not to violate the condition x2+y2+z2=1).

Equating to 0 the two partials

D1F(x,y)=2x(ac){(a+c)2[(ac)x2+(bc)y2+c]2}=0,D2F(x,y)=2y(bc){(b+c)2[(ac)x2+(bc)y2+c]2}=0

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

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

(2) x=0,y=±212(F=14(bc)2); and

(3) x=±212,y=0(F=14(ac)2).

(Verify!)

Now, for the boundary of ¯G, i.e., the circle x2+y2=1, repeat this process: substitute y2=1x2 in the formula for F(x,y), thus reducing it to

h(x)=(a2b2)x2+b2+[(ab)x2+b]2,h:E1E1,

on the interval [1,1]E1. 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.


This page titled 6.9: Local Extrema. Maxima and Minima is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform.

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