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# Multivariate Calculus MA 26100

Purdue

GPA 3.97

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This 5 page Class Notes was uploaded by Dorothea Bode on Saturday September 19, 2015. The Class Notes belongs to MA 26100 at Purdue University taught by David Drasin in Fall. Since its upload, it has received 6 views. For similar materials see /class/208106/ma-26100-purdue-university in Mathematics (M) at Purdue University.

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Date Created: 09/19/15

MA 261 On LAGRANGE MULTIPLIERS based on Stewart s text We assume that you are familiar with the standard tests for eztrema when we have a function f of more than one independent variable we usually study but the same ideas work for fzyz etc e important thing is being clear on the di erence between independent and nonindependent variables In this introduction we and examples we write the variable as zy or z y 2 but use vector notation in describing the general princip er 1 Independent Variables Suppose we want to nd the extrema of fzy where zy ranges over some domain D for convenience we always assume that f has continuous partials The standard method is a to identify the possible interior extrema by nding points z y where Vfz y 0 This is where the issue of the variables being independent arises since when they are independent we may leave I y in any direction and check whether f increases or decreases in that direction that is why directional derivatives are introduced we introduced directional derivatives in the previous lecturel Only if Vfzy 0 will the directional derivative of f in any direction from I y be zero which means that there is no obvious direction in which to go from zy where f would increase or decrease and so zy has to be considered a point at which f might attain an extremuml Step a is followed by b identifying the possible extrema taken on the bound ary of D the situation discussed here 2 Dependent variables The situation is different when we look for points on the boundary of D at which the extrema occur this is analogous to checking at the endpoints7 when f is de ned on ab and f might have an extremum at either I az b without f a or fb vanishing In two variables this usually that means that a given portion of 8D the boundary of D is represented by one of the forms y 91 z gy or zy zt yt so that t is a parameter It is important to see in that situation we have lost an independent variable In case we are saying that only I is independent so that once I is given we know y 7 that canlt occur were the variables inependentl ln y is the only independent variable and in only t is independent So for example in we have Fz in case fzy Fy and in fzy Ft I am using the letter F for a different function in each case When zy 6 8D the variables are not independent so we are not able to move I y in any direction we choose and remain on the boundary That is the reason f might have an extremum at z y without Vfz y being zero When we consider a function f of n variables de ned in a region D there will in general be n 7 1 independent variables on 8D In our earlier homework when doing these problems without Lagrange multi pliers we would simply study f on this portion of 8D by going back to onevariable maximumminimum techniques from calculus and so identify all possible extrema on the boundary This can lead to many cases when the boundary has pieces which are of various types There will be examples given 3 Na39l39ve way to handle boundary situation The Lagrange method is a more insightful 7 and simple 7 way of handling the situation that the number of independent variables is reduced 7 that is what we mean by there being a constraint As we have mentioned the constraint is that either y gzz hy or z zty yt etc which as we have observed reduces the number of independent variables 4 Lagrange s insight Lagrange had an insight that uses geometry and some elementary vector analysis The exposition I am giving here covers l4i8 in my opinion a simpler way and always leads to one less equation77 in a system of simultaneous equationsi No matter how we approach these problems solving these systems requires methods improvised for each problem 7 we do not use any general method Our problem is to extremize subject to the constraint gx 0 here x 1112 i i i zk is a vector It is governed by an elementary fact about vectorsi Principle Let x 112Hi1k and y yl yg i i i yk be be vectors our context usually h 2 or 3 Then x y if and only if all ratios l 2 i i i h are the same the understanding is that if we have zj 0 for some j then the corresponding entry yj must also be zero You should check with a few examples for what A is 14 9 72 78 A How would your answer change if you replace only 4 with 6 Or 9 with 0 Let us apply this to the situation of extremizing f subject to our constraint 9 0 and let x ab be a point we are testing as a possible extremal Then Lagrange in order that f have an extrema at x it is necessary that Vfx H V9XA 1 NOTE The word multiplier arises from the usual formulation of this principle it asserts that at a potential extrema we have the equation VfX AV9X7 where A is a scalar You should check that these two ways of expressing the Lagrange principle are the same the slight advantage in the formulation I prefer is that the scalar A usually has no physical signi cance and in practice that is homework problems means introducing A simply gives is an extra equation to consider with no extra information Once we identify the points where 1 holds we have to compute f at each and then the maximum of f among points which satisfy the constraint will be where f takes on the largest value among these points a similar remark will lead to the minimum 0 i 5 Some examples Lets see how this works with some of the homework and examples in the text We should always identify f and g We then use the Lagrange principle and what will come out is that the coordinates at a possible extremum must satisfy a certain relation In other to nd the exact values of the coordinates at these possible extrema however we have to return to the constraint 9 0 since if our constraint were 9 c c a constant with c y 0 the Lagrange equation would be the same and so would be a relation between the variables Our rst example is an exception to this principle because the relation 1 is more subtle since the coordinates at the extrema will be zero I start with a complicated example which was done earlier by other methods P 962 N0 39 Find a point on the surface 22 my 1 closest to the origin Step I Find f g Well it is f that we wish to minimize so f 1702y7022702 22y222 where z y 2 are restricted to the surface thus 9Iyyyz 22 i any the constraint is g 1 Step II Write the Lagrange equation for this case we can cancel common factors This leads to Vf H V91 17972 H 97 1722 2 Note that without the constraint of g 1 there is a trivial solution f has an absolute minimum w en 1 y 0 but the points 000 does not satisfy the constraint 22 1 Step III Using our Principle manipulate 2 to get a relation that must be satis ed at a possible extremumi I always assume rst that none of the coordinates are zero So if these vectors are parallel then looking at the rst two coordinates only we see that zy yz so I iy which means that zy ill But then that common ratio i1 of the rst two coordinates in 2 would have to be the same for the ratio of third coordinates and so the same as 222 however unless 2 0 222 can t be ill In other words we are at a dead end This means the extrema will occur when 2 0 or when we recall the case excluded in 96 that 1 0 if 1 0 then by our principle concerning parallel vectors weld have y 0 tool This means the only possible points to check are 0 0 i1 the i1 is the value of 2 when 1 y 0 and in addition returning to the case that 2 0 we would have as possibilities the points 1 71 0 and 71 l 0 At these points we compute f z 12 y2 22 and see that the extrema occur at 0 0il the closest points on the surface are distance 1 from the origin P 966 E1ample 1 Here f 1y2 and g 212 2y2 1y so the constraint is that g 12 Step II We see that Vf Vg when y2 12 1y 22 y 2y 1 21 2y When we divide it is simplest to put the entries of the lefthand vector in the denominators since the algebra is simpler lndeed if1y or 2 is zero the box has no volume so we may assume that 1y2 0 Step III Looking at the ratio of rst two coordinates and crossmultiplying we nd that 21221y2 2y221y2 or that 1 yr Then ifwe look at the ratio of rst and third coordinates in the same way we nd that 21y2 1y2 21y22y22 or more simply 1 22 In summary 1 y 22 and since 212 2y2 1y 12 we nd that 1 2 y2 1 Notice that you get a different answer if 212 2y2 1y 100 we would have a box of different size but as we mentioned at the beginning of the the same relations between the variables 1 y 2 would remain P 968 E1ample 5 Now f 12 2y2g 12 y2 7 1 when we look inside the circle of radius one 1 and y are independent variables and the Lagrange method is not relevant there Step II Vf Vg means that we have 17 ll Ivy So now we see that the extrema occue when either 1 or y is zero otherwise weld have nonsense thinking the vectors parallel 2 f 1 So we need consider f only at 0i1i10i P 971 N0 9 This is a problem in three variables Maxtimize f1 y 2 1y2 g z 12 2y2 322 6 So Vf Vg occurs when yzyrzyzy ll 729732 lf1y2 0 then f 0 So if1y2 0 we may put the entries of the second vector in the denominators and see that an extemum will occur when 122 2y22 12y 322yi That gives the relation that z iyz i zi We now use that g 6 note again that if the constraint were that g 1007 we would have the same relation between I y and 2 but the speci c number would be different This produces eight points where extrema may occur7 dependingon the choice of sign for each of the coordinates A minimum will occur when exactly one or three of z y 2 is negative and the other positive7 and a maximum occurs when either all three variables are positive or exactly two are negative

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