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1、12.9 The Method of Lagrange Multipliers Problem Introduction Free Extremum Problem Such as finding the minimum value of ,=2+22 ,subject to some constraint ,=0.Such as finding the minimum value of ,=2+22,where ,is only resctricted in the domain.Constrained Extremum Problem Problem IntroductionSolutio
2、n.The sum of the length,width and height of a rectangle is 18,what values do they take respectively such that the volume of the rectangle is the largest?)18(yxxy 2218xyyxxy 18,0,0 yxyxThe domain The volume Assume the length,width,height are,respectively,18,xyzVxyz18zxy Problem Introduction The stati
3、onary point =216.6,xyz21820 xVyxyy21820yVxxxy(6,6)Since the volume has the only stationary point in the domain ,and must have maximum value,then 2218Vxyx yxy Problem Introductionvalue of subject to the condition +=18,objective function The above example can be considered as finding the extreme the c
4、onstraintHence,the example is called a constrained extremum problem.Vxyz18xyz Introduction In some case,the constrained extremum problem can be transformed into the free constrained extremum problem through The following we introduce the general solution to the constrained extremum problemLagranges
5、Method.But in general,it is difficult or impossible to do so.substituting the constraint into the objective function.Lagranges Method(ThA textbook in vector)Now we find the necessary conditions for the objective function subject to the constraint to obtain the extremum value.If the function has the
6、extremum value at(1),(00yxthen(1)(,)zf x y(2)(,)0 x y(3)00(,)0 xyBy the constraintifyis an implicit function of a variable xthe objective functionThat is,has extreme value at Lagranges Method(ThA textbook in vector)(,()zf x y x(,)0,x y()yy x0.xx By the necessary conditions for obtaining the extreme
7、value of univariate differentiable function,we have(4)Substituting into（4）to obtain 0ddxxxy),(),(0000yxyxyx (5)Lagranges Method(ThA textbook in vector)000000|0|x xx xx xx xy yy ydzffdydxxydxNote that=()is implicitly defined by(,)=0,so The equations(3)(5)and00(,)0,x y are the necessary conditions of
8、obtaining the extremum value for(1)(2).the function under the constraint Let0000(,)(,)yyfxyxy then the above necessary conditions turn into Lagranges Method(ThA textbook in vector)(6)The left of the first two equations of(6)are the two first-order partial derivatives of the function 00(,).xyat Lagra
9、nges Method(ThA textbook in vector)0000(,)(,)0 xxfxyxy0000(,)(,)0yyfxyxy00(,)0 xy(,)(,)(,)L x yf x yx y The function Lagranges Method(ThA textbook in vector)(,)(,)(,)L x yf x yx yis called Lagrange function.The corresponding is called a Lagrange multiplier,which is a unknown constant.Lagranges Metho
10、dTheorem A To maximize or minimize subject to the constraint ,solve the system of equations and for and .Each such point is a critical point for the constrained extremum problem,and the corresponding is called a Lagrange multiplier.()()fpgp()0gp()fp()0g p pp Example 1 Find second the four partial de
11、rivatives of Find the shortest distance from the point to the surface .,x y zf x y zxyz2221(,)(1)(1)()2 Objective function To simplify the calculation,let Let is a point on the surface,the distance between and 22zxy()()()2221112dxyz11,1,2 Example 1 The problem is to find the minimum for the function
12、 2221(,)(1)(1)()2f x y zxyz22.zxy222221(,)(1)(1)()()2L x y zxyzzxy02)1(2 xxLx 02)1(2 yyLy 0212 zLz22yxz (1)(2)(3)(4)Let subject to the constraint Example 1 the only stationary point is 3164,3164,342.So,substituting it into(4),then=1=1=1 2,By(1),(2)and(3),we have 33316164,442xyz02)1(2 xxLx 02)1(2 yyL
13、y 0212 zLz22yxz (1)(2)(3)(4)then=12,Example 1 Find second the four partial derivatives of Since the problem exists the minimum value really,222141412 33then at the point 3164,3164,342,has the minimum value:Example 2 Find second the four partial derivatives of 32,sinyxfx yxex yy000(,)P x y z,2|20axFP
14、x ,2|20byFPy 202|azFPz 000(,)P x y z0)()()(020020020 zzczyybyxxax We have the tangent plane of the ellipsoid in the first2222221xyzabc1),(222222 czbyaxzyxF octant such that the volume of the tetrahedron which is surrounded by the tangent plane and the three coordinate planes is minimum.Try to find t
15、he coordinate of the tangent point.The tangent plane which passes through the point thenis a point of ellipsoid,Example 2 Find second the four partial derivatives of Simplify it as1202020 czzbyyaxx,02xax ,02yby 02zcz xyzV61 0002226zyxcba Objective function 1220220220 czbyaxConstraint subject to the
16、constraint The volume of the tetrahdron is In each of the three axises,the intercepts of the tangent plane Example 2 Find second the four partial derivatives of So we get a constrained extremum problem.Let 000lnlnlnuxyz000000(,)lnlnlnL xyzxyz ,00 xL01220220220 czbyax,00 yL00 zL2220002221xyzabc Examp
17、le 2That is,021200 byy 021200 czz 01220220220 czbyax021200 axx 30ax 30by 30cz the tangent point (,).333abcmin3.2Vabc Similar to a function of single variable,we can find extremum to obtain minimum and maximum of the function over a closed and bounded set.Optimizing a Function over a Closed and Bound
18、ed Set The general method of finding the maximum or minimum maximum and minimum over (3)Evaluate the function value at obtained points to find (2)Find the points along the boundary that give the (1)Find all possible extreme points of the function on the interior of maximum or minimum.Example 3 Find
19、second the four partial derivatives of Find the minimum and maximum for the function on theround .22(2)(2)9xyThe stationary point in the interior of the round.(1)We first find the possible extreme point of the function 22zxy22zxy20 xzx20yzy (0,0).Example 3 222zxy2222(,)(2)(2)9L x yxyxy along the bou
20、ndary of the round.(2)Find the minimum or maximum for the functionThe lagrange function)(0)2(22axxLx )(0)2(22byyLy )(9)2()2(22cyx then225 yx22xyorExample 3 Find second the four partial derivatives of 5 2 5 222(0,0),2222zzz、22(2)(2)9,xy25z 0z minimum valueon the round maximum valueCompareing the valu
21、es Remarks Find second the four partial derivatives of How to determine the point obtained is the extreme point?Extension：The method of Lagrange multipliers can promote to the cases of more than two variables and several constraints.(1)In the practical problems,we can determine according to the natu
22、re of the problem.(2)For non-practical problems,we will not discuss further here SummaryLagranges Method To find the possible extreme point of the function(,)zf x ysubject to the constraint(,)0 x y first assume the function(,)(,)(,)L x yf x yx ywhere is a certain constant.Summarythen Finally,get the solution ,x y (,)0 x y(,)(,)0yyfx yx y(,)(,)0 xxfx yx ywhere is the coordinate of the possible extreme point.,x yThe Method of Lagrange Multipliers