Unit #23 - Lagrange Multipliers Lagrange Multipliers

1 Unit #23 - Lagrange MultipliersSome problems and solutions selected or adapted from Hughes-Hallett MultipliersIn Problems 1 4, use Lagrange Multipliers tofind the maximum and minimum values offsubject to the given constraint, if such valuesexist. Make an argument supporting the classi-fication of your minima and (x,y) =x+y,x2+y2= 1We use the constraint to build the constraint function,g(x,y) =x2+y2. We then take all the partial deriva-tives which will be needed for the Lagrange multiplierequations:fx= 1gx= 2xfy= 1gy= 2ySetting up the Lagrange multiplier equations:fx= gx 1 = 2x(1)fy= gy 1 = 2y(2)constraint: x2+y2= 1(3)Taking (1) / (2), (assuming 6= 0)11= 2x 2y=xysoy=xSub into (3) to find2x2= 1 x= 1/2 Combining withy=x, we get the solutions (x,y) =( 1/2, 1/2) and ( 1/2, 1/2).Since our constraint is closed and bounded (only pointson the circlex2+y2= 1 are allowed), we can simplycompare the value offat these two points to determinethe maximum and minimum values offsubject to ( 1/2, 1/2) = 2 1/2f( 1/2, 1/2) = 2 1/2 From this, the maximum of f onx2+y2=1 is at ( 1/2, 1/2) and the minimum is at( 1/2, 1/2) (x,y) =xy, 4x2+y2= 8fx=ygx= 8xfy=xgy= 2ySet up the Lagrange multiplier equations:fx= gx y= 8x(4)fy= gy x= 2y(5)constraint: 4x2+y2= 8(6)Taking (4) / (5), (assuming 6= 0)yx= 8x 2y=8x2ysoy2= 4x2ory= 2xSub into (6) to find4x2+ 4x2= 8 x= 1 Combining withy= 2x, we get the solutions (x,y) =(1,2),(1, 2),( 1,2) and ( 1, 2).

2 Since our constraint is closed and bounded, we can sim-ply compare the value offat these four points to deter-mine the maximum and minimum values offsubjectto the (1,2) = 2f(1, 2) = 2f( 1,2) = 2f( 1, 2) = 2 From this, the maximum offon the constraint 4x2+y2= 8is at two points, (1, 2) and (-1, -2); thefvaluethere is +2. The minimum offoccurs at (1, -2) and (-1, 2);thefvalue there is (x,y) =x2+y,x2 y2= 1fx= 2xgx= 2xfy= 1gy= 2y1 Set up the Lagrange multiplier equations:fx= gx 2x= (2x)(7)fy= gy 1 = ( 2y)(8)constraint: x2 y2= 1(9)From (7), we must have = 1 orx= 0 If = 1, then (8) gives 1 = (1)( 2y), ory= 12, and from (9)x2 ( 12)2= 1, sox= 1 +14= 54 Ifx= 0, then (9) gives 02 y2= 1, but thishas no solution! In other words, no point withx= 0 belongs to the constraint, so we won t getany candidate points from this solutions to the Lagrange Multiplier equations aretherefore (x,y) = ( 54, 12), and ( 54, 12).

3 The associated function values at these points are: f( 54, 12)=x2+y=54+ 12=34 f( 54, 12)=x2+y=54+ 12=34 Since the constraint isnotbounded, it is not as easyto demonstrate that these values are minimums offon the constraint. However, with a little mathematicalinsight it can be done in just a few steps:f(x,y) =x2+y,but we are limited to the constraintx2 y2= 1,orx2=y2+ 1 Substituting this intof, we getf(x,y) = (y2+ 1) +y=y2+y+ 1 on the constraintCompleting the square givesf(x,y) =(y+12)2+34 Since squared values are always positive, we can say thatf(x,y) =(y+12)2+34 34on the constraint curveTherefore, the values we found(f=34)are minimumsoffon the constraint.[On a test or exam, this kind of check would not beexpected without some prompting steps.] (x,y) =x2+ 2y2,x2+y2 4 Note that we are dealing with an inequality for theconstraint. We can consider any pointin or ontheboundary of a circle with radius 2.

4 To lookontheboundary, we use Lagrange Multipliers . To look at theinterior, we identify the critical points off(x,y).We ll start with the Lagrange Multipliers :fx= 2xgx= 2xfy= 4ygy= 2ySet up the Lagrange multiplier equations:fx= gx 2x= 2x(10)fy= gy 4y= 2y(11)constraint: x2+y2= 4(12)From (10), eitherx= 0 or = 1. Ifx= 0, then (12)saysy= 2. Alternatively, if = 1, then (11) meansy= 0, sox= 2. Our solutions are(x,y) = (0,2),(0, 2),(2,0) and ( 2,0)At these points,f(0,2) = 8f(0, 2) = 8f(2,0) = 4f( 2,0) = 4 Before we can say these are global max or mins, weneed to look for critical points in the interior of thecirclex2+y2 0 2x= 0andfy= 0 4y= 0 The only critical points is (0, 0), and this is in theinterior of the circle. The value off(0,0) = the results on the boundary with the onlycritical point we see: f(0,2) andf(0, 2) are global maxes with valuesoff= 8 f(0,0) is the global min on the region, withf= contour diagram showing the region and contours offis included below to illustrate the (a) Draw contours off(x,y) = 2x+yforz= 7, 5, 3, 1,1,3,5,7.

5 (b) On the same axes, graph the constraintx2+y2= 5.(c) Use the graph to approximate the pointsat whichfhas a maximum or a minimumvalue subject to the constraintx2+y2= 5.(d) Use Lagrange Multipliers to find the max-imum and minimum values off(x,y) =2x+ysubject tox2+y2= 5.(a) The contours offare straight lines with slope 2(inxyterms), as shown below.(b) Overlaying the constraint, we are allowed to moveon a circle of radius 5.(c) From the graph, the maximum values occurswhere the constraint circle just touches thef= 5contour line, at (x,y) = (2,1). The minimumvalue isf= 5, which occurs on the oppositeside of the circle, at ( 2, 1).(d) Computing the constrained optimum locations us-ing Lagrange Multipliers ,fx= 2gx= 2xfy= 1gy= 2ySet up the Lagrange multiplier equations:fx= gx 2 = 2x(13)fy= gy 1 = 2y(14)constraint: x2+y2= 5(15)Taking (13) / (14), (assuming 6= 0)21= 2x 2y=xyso 2y=xSub into (15) to find4y2+y2= 5 y= 1 Combining with 2y=x, we get the solutions(x,y) = (2,1) and ( 2, 1).

6 These are the samepoints we found in (c), and knowing theirzval-ues, we know thatf(2,1) is a maximum whilef( 2, 1) is a minimum on the A company manufacturesxunits of one itemandyunits of another. The total cost in dol-lars,C, of producing these two items is approx-imated by the functionC= 5x2+ 2xy+ 3y2+ 800(a) If the production quota for the total num-ber of items (both types combined) is 39,find the minimum production cost.(b) Estimate the additional production cost orsavings if the production quota is raised to40 or lowered to 38.(a) If the total production is 39, thenx+y g(x,y)= 39 kThis is our constraint in the formg(x,y) =kSetting up the Lagrange multiplier equations,10x+ 2y Cx= 1 gx(16)2x+ 6y Cy= 1 gy(17)x+y= 39(18)3 Setting (16) equal to (17),10x+ 2y= 2x+ 6y8x= 4yy= 2xSub that into (18),x+ (2x) = 39x= 13and soy= 2x= 26 The optimal production levels arex= 13 units andy= 26 units, giving a total production of 39 units.

7 (b) We are asked to evaluate the impact on the cost ofadding one or removing one item from the Lagrange multiplier value gives us the approx-imate effect on the cost of adding one unit to theconstraint valuek, which in this caseisthe changein the quota. Usingx= 12 andy= 26, (16) givesus = 10(13) + 2(26) = 182so adding one unit to the total production (or pro-ducing 40 units) will increase the cost by $ , by removing one unit from the quota (orproducing 38 units), the production cost will dropby $ A firm manufactures a commodity at two differ-ent factories. The total cost of manufacturingdepends on the quantities,q1andq2, suppliedby each factory, and is expressed by the jointcost function,C=f(q1,q2) = 2q21+q1q2+q22+ 500 The company s objective is to produce 200units, while minimizing production costs. Howmany units should be supplied by each factory?We want to minimizeC=f(q1,q2) = 2q21+q1q2+q22+ 500subject to the constraintq1+q2= 200 (sog(q1,q2) =q1+q2).

8 Since f= (4q1+q2,2q2+q1) and g= (1,1), setting f= ggives4q1+q2= 1q1+ 2q2= 1 Solving, we get4q1+q2=q1+ 2q2so3q1= wantq1+q2= 200q1+ 3q1= 4q1= 200 Therefore,q1= 50andq2= 150 From the problem statement, we can conclude that thisproduction level will minimize the total manufacturingcost, given the desired size of production Each person tries to balance his or her time be-tween leisure and work. The trade-off is thatas you work less your income falls. Thereforeeach person has indifference curves which con-nect the number of hours of leisure,l, and in-come,s. If, for example, you are indifferentbetween 0 hours of leisure and an income of$1125 a week on the one hand, and 10 hoursof leisure and an income of $750 a week on theother hand, then the pointsl= 0,s= 1125, andl= 10,s= 750 both lie on the same indiffer-ence curve. The table below gives informationon three indifference curves, I, II, and incomeWeekly leisure hoursIIIIIIIIIIII11251250137502040750875 1000103050500625750204060375500625305070 250375500507090(a) Graph the three indifference curves.

9 (b) You have 100 hours a week available forwork and leisure combined, and you earn$10/ hour. Write an equation in terms oflandswhich represents this constraint.(c) On the same axes, graph this constraint.(d) Estimate from the graph what combina-tion of leisure hours and income you wouldchoose under these circumstances. Givethe corresponding number of hours perweek you would work.(a) The graphs are shown, along with the constraintfrom part (c), (b) Since you re earning $10 per hour, andsis yourincome,s/10 is the number of hours worked. Tolimit yourself to 100 hours per week, you mustsatisfyl+s/10 = 100(c) See the graph from (a)(d) Since the constraint line just touches the indif-ference curve II att= 50,s= 500, we can tachieve a higher level of satisfaction than level achieve that level of satisfaction, we shouldsplit our time intot= 50 hours of leisure, ands/10 = 500/10 = 50 hours of The director of a neighborhood health clinic hasan annual budget of $600,000.

10 He wants to al-locate his budget so as to maximize the numberof patient visits,V, which is given as a functionof the number of doctors,D, and the number ofnurses,N, byV= doctor s salary is $40,000; nurses get $10,000.(a) Set up the director s constrained optimiza-tion problem.(b) Describe, in words, the conditions whichmust be satisfied by V Dand V NforVto have an optimum value.(c) Solve the problem formulated in part (a)(d) Find the value of the Lagrange multiplierand interpret its meaning in this problem.(e) At the optimum point, what is themarginal cost of a patient visit (that is,the cost of an additional visit)?(a) The problem is to maximizeV= to the budget constraint that40,000D+ 10,000N 600,000 This will be easier to deal with if we divide by10,000, so4D+N 60 Since there our function,V, always grows largerwith largerNandD, there is no point in us-ing less than our budget, so we want to find themaximum value ofVon the line 4D+N= this problem, then, our constraint function isg(D,N) = 4D+N.