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1、外文原文EXTREME VALUES OF FUNCTIONS OF SEVERAL REAL VARIABLES1. Stationary PointsDefinition 1.1 Let and . The point a is said to be:nRD RDf:Da(1) a local maximum iffor all points sufficiently close to ;)()(afxfxa(2) a local minimum iffor all points sufficiently close to ;)()(afxfxa(3) a global (or absol
2、ute) maximum iffor all points ;)()(afxfDx(4) a global (or absolute) minimum iffor all points ;)()(afxfDx(5) a local or global extremum if it is a local or global maximum or minimum.Definition 1.2 Let and . The point a is said to be critical nRD RDf:Daor stationary point if and a singular point if do
3、es not exist at .0)(affaFact 1.3 Let and .If has a local or global extremum at the point nRD RDf:f, then must be either:Daa(1) a critical point of , orf(2) a singular point of , orf(3) a boundary point of .DFact 1.4 If is a continuous function on a closed bounded set then is bounded ffand attains it
4、s bounds.Definition 1.5 A critical point which is neither a local maximum nor minimum is acalled a saddle point.Fact 1.6 A critical point is a saddle point if and only if there are arbitrarily small avalues of for which takes both positive and negative values.h)()(afhafDefinition 1.7 If is a functio
5、n of two variables such that all second order RRf2:partial derivatives exist at the point , then the Hessian matrix of at is ),(baf),(bathe matrix yyyxxyxx ffffHwhere the derivatives are evaluated at.),(baIf is a function of three variables such that all second order partial RRf3:derivatives exist a
6、t the point , then the Hessian of f at is the matrix),(cba),(cbazzzyzxyzyyyxxzxyxxfffffffff Hwhere the derivatives are evaluated at.),(cbaDefinition 1.8 Let be an matrix and, for each ,let be the Annnr 1rArr matrix formed from the first rows and columns of .The determinants det(),rrArA,are called th
7、e leading minors of nr 1ATheorem 1.9(The Leading Minor Test). Suppose that is a sufficiently RRf2:smooth function of two variables with a critical point atand H the Hessian of ),(baat.If , then is:f),(ba0)det(H),(ba(1) a local maximum if 0det(H1) = fxx and 0det(H1), 0det(H3);(2) a local minimum if 0
8、det(H3);(3) a saddle point if neither of the above hold.where the partial derivatives are evaluated at.),(cbaIn each case, if det(H)= 0, then can be either a local extremum or a saddle ),(bapoint.Example. Find and classify the stationary points of the following functions:(1) ; 1),(2224xzzyyxxzyxf(2)
9、 ;) 1() 1(),(422xyxyyxfSolution. (1) ,so1),(2224xzzyyxxzyxfijk)24),(3zxyxyxf（)2(2yx )2(xz Critical points occur when ,i.e. when0f(1) zxyx2403(2) yx202(3) xz 20Key Points.A continuous function on a closed bounded set is bounded and achieves its bounds.To find the extreme values of a function on a clo
10、sed bounded set it is necessary to consider the value of the function at stationary points(), singular points 0f(does not exist) and boundary points(points on the edge of the set).fStationary points can be classified as local maxim , local minim or saddle points.aaIf The Leading Minor Test 1.9 is no
11、t applicable, the stationary point must be classified by directly applying Definition 1.1 and Fact 1.6. For example in the two variable case, if has a stationary point at ,we consider the sign off),(ba),(),(bafhbhaffor arbitrarily small, positive and negative values of and (that are not both zero).h
12、kUsing equations (2) and (3) to eliminate y and z from (1), we see thator ,giving , and .Hence we 021433xxx0) 16(2xx0x66x66xhave three stationary points: , and .）（0 , 0 , 0）（126,121,66）（126,121,66Since, and ,the Hessian yxfxx2122xfxy21xzf2yyf0yzf2zzfmatrix is 201022122122xxyx HAt ,）（126,121,66201023
13、/613/66/11Hwhich has leading minors 0,6110396 311 23/63/66/11det And det .By the Leading Minor Test, then, 042912 322His a local minimum.）（126,121,66At ,）（126,121,66 201023/613/66/11 Hwhich has leading minors 0,6110396 311 23/63/66/11det And det .By the Leading Minor Test, then, 042912 322His also a
14、 local minimum.）（126,121,66At , the Hessian is）（0 , 0 , 0 201020100HSince det, we can apply the leading minor test which tells us that this is a 2)(Hsaddle point since the first leading minor is 0. An alternative method is as follows. In this case we consider the value of the expression，hllkkhhlkhff
15、D22240 ,0 ,00 , 0 , 0）（）（for arbitrarily small values of h, k and l. But for very small h, k and l, cubic terms and above are negligible in comparison to quadratic and linear terms, so that.If h, k and l are all positive, . However, if and and hllkD220D0k0h,then .Hence close to ,both increases and decreases, so hl 00D）（0 , 0 , 0fis a saddle point.）（0 , 0 , 0(2) so422) 1() 1(),(xyxyyxfij.) 1(4) 1(2(),(3xyxyxf)1(2(2xyStationary points occ
