1、Section 9.81Gauss,Karl Friedrich Unrestricted Extreme Valuesis defined in00(,).U xyDefinition Suppose the that the function(,)f x ysuch that0,If 000000(,)(,)(,)(,),(,)(,),)f x yf xyf x yf xyx yUxy then we say the function f has an unrestricted local maximum(minimum),or f has a maximum(minimum)00(,)f
2、 xyat the point 00(,)xy00(,)xyis called a maximal(minimal)point or for short,and the point Maximum and minimum values are called by the jointextreme point.name extreme value.attains a minimum at the point22zxyFor example,the function(0,0)attains a maximum at the point(0,0).and 221zxy3The Necessary C
3、ondition for An Extreme ValueTheorem Suppose that both partial derivatives of the function(,)f x yis an extreme point of the function00(,)xyexist at the point,and f.Then0000(,)(,)(,)0.xyxyf xyff00(,)xy00(,)xyis called a stationary point(驻点)A point 00(,)0f xy,which satisfies However,just as for funct
4、ions of one variable,a stationary of the function f.point is not necessarily an extreme point.,the point(0,0)is a stationary22(,)f x yxy For example,for the function point but not a extreme point.4Critical Point1.Interior points where0.xyffdo not exist.yf2.Interior points where one or both of xfandT
5、he theorem of necessary condition for a extreme value says that thecan ever have an extreme value areplaces a function(,)f x y(,)f x ywhere Definition An interior point of the domain of a function both fx and fy are zero or where one or both fx and fy do notexist is a critical point of f.5Saddle Poi
6、nt can assume extreme(,)f x yThus,the only points where a function As with differentiablevalues are critical points and boundary points.functions of a single variable,not every critical point gives rise to a A differentiable function of a single variable mightlocal extremum.A differentiable function
7、 of two variables have a point of inflection.might have a saddle point(鞍点鞍点).has a saddle point at a(,)f x yDefinition A differentiable functionstationary point(a,b)if in every open disk centered at (a,b)there areand domain points(x,y)domain points (x,y)where(,)(,)f x yf a b(,)(,).f x yf a b The cor
8、responding point(a,b,f(a,b)on thewhere is called a saddle point of the surface.(,)zf x y surface 6Sufficient Condition for Extreme Valueshas continuous second(,)zf x y Theorem Suppose that function order partial derivatives in a neighbourhood of the point 000(,),P xyLet0Pand is a stationary point of
9、 the function f.000(),(),().xxxyyyAfPBfPCfPThen 4.Can not be determined if 20.ACBand0()f P1.0A is a minimum value of the function f if 20;ACBand0()f P2.0A is a maximum value of the function f if 20;ACB0()f P3.is not an extreme value of the function f if 20;ACB7Finding Local Extreme ValuesExample Fin
10、d the local extreme values of 22(,)224.f x yxyxyxyThe function is defined and differentiable for all x and y and Solution The function therefore has extremeits domain has no boundary points.These values only at the points where fx and fy are simultaneously zero.leads to 2.xy or220,220,xyfyxfxyTheref
11、ore,the point(-2,-2)is the only point f may take on an extremevalue.8Finding Local Extreme ValuesSolution(continued)To see if it does so,we calculate 2,2,1.xxyyxyfff The discriminant of f at(-2,-2)is22(2)(2)(1)413.xxyyxyfff The combination 20 xxyyxyfffand0 xyf The value of f at this pointtell us tha
12、t f has a local maximum at(-2,-2).is(2,2)8.f Example Find the local extreme values of 22(,)224.f x yxyxyxy9Searching for Local Extreme ValuesExample Find the local extreme values of(,).f x yxy Since f is differentiable everywhere,it can assume extremeSolution values only where0.yfxand0 xfyThus,the o
13、rigin is the only point where f might have an extreme value.To see what happens there,we calculate0,0,1.xxyyxyfffThe discriminant,21,xxyyxyfff is negative.Therefore,the function has a saddle point at(0,0).we has no local extreme values.(,)f x yxy conclude that10Global Maxima and Minima(最值最值)Suppose
14、that the function f is continuous on a closed bounded region Then the function must attain its maximum and minimum in.xyand theythe maximum or minimum must be an extreme valueIf the maximum or minimum is attained in the interior of the region,zmust be stationary point of f if the partial derivatives
15、 of f existat the extreme point.11Finding Global ExtremaExample Find the global maximum and minimum values of 22(,)222f x yxyxyon the triangular plate in the first quadrant bounded by the line 0,x 0,9.yyxthe only places where f can assume these values are pointsand point on the boundary.0 xyffinside
16、 the triangle where Solution Since f is differentiable,xOy0y 0 x (0,9)B(9,0)A9yx12Finding Global ExtremaSolution(continued)xOy0y 0 x (0,9)B(9,0)A9yx(1,1)Boundary PointsFor these we have 220,yfy220,xfxThe valueyielding the single point(1,1).of f there is(1,1)4.f Interior PointsThe function 0.y 1.On t
17、he segment OA,2(,)(,0)22f x yf xxxmay now be regarded as a function of x defined on the closed interval09.x13Finding Global ExtremaxOy0y 0 x (0,9)B(9,0)A9yx9 9,2 2(1,1)2(,)(,0)22f x yf xxxIts extreme values may occur at the endpoints(0,0)2f where 0 x (9,0)61f where 9x and at the interior points wher
18、e(,0)220.fxx where(,0)0fx The only interior point where1,x is(,0)(1,0)3.f xf14Finding Global ExtremaxOy0y 0 x (0,9)B(9,0)A9yx(1,1)2.On the segment OB,The function 0.x 2(,)(0,)22.f x yfyyyWe know from the symmetry of f inin x and y and from the analysis wejust carried out that the candidateson the se
19、gment are(0,0)2,(0,9)61,(0,1)3.fff 222(,)22xyf x yyx 15Finding Global ExtremaxOy0y 0 x (0,9)B(9,0)A9yx9 9,2 2(1,1)3.On the segment AB,we have 9,yx(,)f x ySetting At this value of x,189.42x 261182.xx 22222(9)(9)xxxx(,9)1840fxxx gives 9 941(,),.2 22f x yf and99922y 222(,)22xyf x yyx 16Finding Global E
20、xtremaSummary We list all the candidates:2361 361(,)f x y4412(0,0)(1,0)(9,0)(0,1)(0,9)(,)x y(1,1)9 9,2 2The minimum is-61,The maximum is 4,which f assumes at(1,1).which f assumes at(0,9)and(9,0).17The Method of Least SquaresSuppose we find a sequence of observations relating an independentvariable x
21、 and a dependent variable y,say1122(,),(,),(,).nnx yxyxy12,nyyyare the closet together12,nx xxand the observed values in some sense.at the observation points We wish to find a suitable function()yf x If we can accomplish this we can use the function()yf x as an approximate representation of thefunct
22、ional relation between the variables x and y.18The Method of Least SquaresThis method is called the method of least squares(最小二乘法最小二乘法).A method which is often used to determine such that the sum of the()f xthe function()(1,2,.,)iiirf xyinare minimized.squares of the errors Suppose we find a sequenc
23、e of observations relating an independentvariable x and a dependent variable y,say1122(,),(,),(,).nnx yxyxy19Application:Spring ConstantsHooks law states that the force applied to a spring is proportional toThus,if F is the force appliedthe distance that the spring is stretched.and x is the distance
24、 that the spring has been stretched,then.Fkx The proportionality constant k is called the spring constant.If we apply forces of 3,5 and 8 ponds,which have the effect of stretching the spring 4,7 and 11 inches,respectively,derive the following system of equations:118k 43k 75k The system is clearly in
25、consistent since each equation yields a differentvalue of k.Using Hookes law,they20Application:Spring ConstantsRather than use any one of these values,we use the least squares solution to the system.is theykx By Hooks law,we suppose thatThen,the errors arerelation between the force and the distance.
26、,(1,2,3).iiirykxiThus,332211()().iiiiiQ krykxBy the necessary condition for the extreme value,we need to solve theequation31()2()0.iiiidQ kxykxdk Then 1 122332221230.726.x yx yx ykxxx21Constrained Extreme ValuesxOyzconstraint onx and yconstrained maximumFreemaximumFrom now on,we explore a powerful m
27、ethod for finding extreme values of constrainedfunctions.22Finding a Minimum with Constraint closest to the origin on the plane(,)P x y zExample Find the point 250.xyzThe problem asks us to find the minimum value of theSolution function222|OPxyz|OP has a minimum value wherever the function Sincehas
28、a minimum value,we may solve the problem by finding the minimum250.xyzsubject to the constraint that(,)f x y zvalue of 222(,)f x y zxyz23Finding a Minimum with ConstraintIf we regard x and y as the independentSolution(continued)variables in this equation and write z as25,zxy at which the(,)x your pr
29、oblem reduces to one of finding the points function(,)(,25)h x yf x yxy222(25)xyxySince the domain of h is the entirehas its minimum value or values.xy plane,then any minima that h might have must occur at the pointswhere22(25)0,yhyxy22(25)(2)0,xhxxy24Finding a Minimum with ConstraintThis leads to55
30、525.366z and the solution 10420,4410,xyxyBy the geometric argument,it is easy to see that these values minimizeThe z coordinate of the corresponding point on the planeh.25zxyis55,.36xyTherefore,the point we seek is5 55,3 66P the distance is5/62.04.25Finding a Minimum with ConstraintAlthough the doma
31、in of h is the entire xy plane,the domain from which it does not include the band between the lines and 1.x 1x We can avoid this problem if we treat y and z as independent variablesand express x in the terms of y and z.Solution(continued)we can select the first two coordinates of the points(,)x y zo
32、n thecylinder is restricted to the“shadow”of the cylinder on the xy plane;closest to the origin on theExample Find the point hyperbolic cylinder(,)P x y z2210.xz26The Method of Lagrange MultipliersWe solved the last example by the methodof Lagrange multipliers.0g In general terms,(,)0g x y z are(,)f
33、 x y zwhose variables are at the pointsto be found on the surface fg for some scalar(called a Lagrange multiplier).the method says that the extreme values of a functionsubject to a constraintwhere Lagrange,Joseph(1736-1813)0000,0pppfgg The extreme point may be P0:27The Method of Lagrange Multipliers
34、Example Find the greatest and smallest values that the function 22(,),f x yxytakes on the circle22(2)1.xy28The Method of Lagrange Multiplierslocal maximum and minimum values of f subject to the constraint To find theSuppose that(,)f x y zandare differentiable.(,)g x y zthe equations(,)0.g x y fg and
35、find the values of x,y,z and that simultaneously satisfy(,)0,g x y z For functions of two independent variables,the appropriate equationsare(,)0.g x y z fg andNote(,)(,)(,)L x y zf x y zg x y zwhich is called If we introduce Lagrange function,then the method of Lagrange Multipliers can also be wrote
36、 into a system of equations in terms of partial derivatives of L.0fgfg 29Using the Method of Lagrange MultipliersExample Find the greatest and smallest values that the function(,),f x yxy takes on the ellipse221.82xySolution We want the extreme value of(,)f x yxy subject to theconstraint 22(,)10.82x
37、yg x y To do so,we first give the Lagrange function 22(,)(,)(,)(1).82xyL x yf x yg x yxySet the three partial derivatives of function L equal to zero,we obtain30Using the Method of Lagrange MultipliersThen the required points are(2,1).Solution(continued)22(,)(,)(,)(1).82xyL x yf x yg x yxy22(,)0,4(,)0,(,)10.82xyxLx yyLx yxyxyLx y 2;(,)(2,1)(,)(2,1)(,)(2,1)(,)(2,1)x yor x yor x yor x y (x,y)f(x,y)-22-22(2,1)(2,1)(2,1)(2,1)And the conditional maximum is 2,the conditional minimum is-2.
