ha (lvh262) – Homework 14.8 – karakurt – (56295) 1This print-out should have 13 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsUse the method o f Lagrange multipliers tominimizef(x, y) = 2x2+ 6y2subject to the constraintx + y = 2 .1. min value = 6 correct2. min value =2743. min value =2344. no min value exists5. min value =2546. min value =132Explanation:Setg(x, y) = x + y −2 .Then by the method of Lagra nge multipliersthe extreme values of f under the constraintg = 0 occur at solutions of∇f = λ∇g , g(x, y) = 0 .But whenf(x, y) = 2x2+ 6y2we see that∇f = 4x i + 12x j .Since∇g = i + j ,the equation ∇f = λ∇g thus becomes4x i + 12y j = λ(i + j) .After comparing coefficients this reduces tothe pair of equationsλ = 4x , λ = 12y ,i.e., y =13x. But we still have the constraintequationg(x, y) = x + y −2 = 0 .Substituting y =13x givesx +13x − 2 =43x − 2 = 0 .Consequently, the only solution of∇f = λ∇g , g(x, y) = 0 ,occurs at(x, y) =32,12,and at t hi s pointf32,12= 6 .But is this a maximum or a minimum value?There are several ways of deciding this, alge-braically and graphically. Which to use de-pends on f and g.Notice first that i f we solve for y in theconstraint equation, then y = 2 − x, and soon g = 0, the function f becomesf(x, 2 −x) = 2x2+ 6(2 −x)2.Its graph is a parabo la opening upwards, andsuch a parabola has a minimum value. Ofcourse, we could have solved the problem byfinding this minimum value with single vari-able calculus techniques wi thout using La-grange multipliers, but the problem did askus to use Lagrange multipliers!!Graphically, we could draw the contourmap for f , superimpose the graph of g = 0,ha (lvh262) – Homework 14.8 – karakurt – (562 95) 2and then look for points where the the con-tour lines for f are tangential to the graph ofg = 0. Alternatively, we can note that thegraph ofz = f(x, y) = 2x2+ 6y2is a paraboloidzxywhile t he graph ofg(x, y) = x + y − 2 = 0in 3-space is a vertical plane. Minimizing fon g = 0 corresponds to finding the heightof the lowest point on the intersection of thisvertical plane with the paraboloid. Since theintersection will be a parabola opening up-wards, again we see that f has a minimum ong = 0 and thismin value = 6 .keywords: Lagrange multipliers, optimiza-tion, gradient, quadratic function, constraint,constrained optimization,002 10.0 pointsUse the method o f Lagrange multipliers tominimizef(x, y) =p3x2+ y2subject to the constraintx + y = 2 .1. min value = 22. min value = 2√33. no min value exists4. min value =√3 correct5. min value = 1Explanation:Setg(x, y) = x + y − 2 .Then by the method of Lagra nge multipliersthe extreme values of f under the constraintg = 0 occur at the solutio ns of∇f = λ∇g , g(x, y) = 0 .But whenf(x, y) =p3x2+ y2we see that∇f =3xp3x2+ y2i +yp3x2+ y2j .Since∇g = i + j ,the equation ∇f = λ∇g thus becomes1p3x2+ y23x i + y j= λ(i + j) .After comparing coefficients this reduces tothe pair of equat ionsλ =3xp3x2+ y2, λ =yp3x2+ y2,i.e., y = 3x . But we still have the constraintequationg(x, y) = x + y − 2 = 0 .Substituting y = 3x givesx + 3x −2 = 4x − 2 = 0 .ha (lvh262) – Homework 14.8 – karakurt – (562 95) 3Consequently, the only solution of∇f = λ∇g , g(x, y) = 0 ,occurs at(x, y) =12,32,and at t hi s pointf12,32=√3 .But is this a maximum or a minimum value?We can decide algebraicall y or graphically, thebest choice depending on f and g. Let’ s do itgraphically because the graphs i n 3-space off and g = 0 are easy to describe. Indeed, thegraph ofz = f(x, y) =p3x2+ y2is a conezxywhile t he graph ofg(x, y) = x + y − 2 = 0in 3-space is a vertical plane. Minimizing f ong = 0 corresponds to finding the height of thelowest point on the intersection of this verticalplane with the cone. Since the intersectionwill be half of a hyperbola opening upwards,we see t hat f has a minimum on g = 0 andthismin value =√3 .keywords: Lagrange multipliers, optimiza-tion, gradient, quadratic function, constraint,constrained optimization,003 10.0 pointsFinding the minimum value off(x, y) = 2x + y −1subject to the constraintg(x, y) = 4x2+ 3y2− 3 = 0is equivalent to finding the height of the l owestpoint on the curve of intersection of the graphsof f and g shown inyxzUse Lagrange multipliers to determine thisminimum value.1. min value = −52. min value = −43. min value = −3 correct4. min value = −25. min value = −6Explanation:The extreme values occur at solutio ns of(∇f)(x, y) = λ(∇g)(x, y) .Now(∇f)(x, y) = h2, 1 i,ha (lvh262) – Homework 14.8 – karakurt – (562 95) 4while(∇g)(x, y) = h8x, 6y i.Thus2 = 8λx , 1 = 6λy ,and soλ =14x=16y, i .e., x =32y .Butg32y, y= 12y2− 3 = 0 , i .e., y = ±12.Consequently, the extreme points are34,12,−34, −12.Sincef34,12= 1 , f−34, −12= −3 ,we thus see thatmin value = −3 .keywords:004 10.0 pointsDetermine the maximum value off(x, y) = 4x − 3y + 3subject to the constraintg(x, y) = x2+ y2− 1 = 0 .1. max value = 8 correct2. max value = 123. max value = 94. max value = 115. max va lue = 10Explanation:The extreme values of f subject to the con-straint g = 0 occur at solutions of(∇f)(x, y) = λ(∇g)(x, y) , g(x, y) = 0 .Now(∇f)(x, y) = h4, −3 i,while(∇g)(x, y) = h2x, 2y i.Thus4 = 2λx , −3 = 2λy ,and soλ =2x= −32y, i .e., y = −34x .Butgx, −34x= x2+916x2− 1 = 0 ,i.e., x = ±4/5. Consequently, the extremepoints are45, −35,−45,35.Sincef45, −35= 8 , f−45,35= −2 ,we thus see thatmax value = 8 .keywords:005 10.0 pointsIf the method of Lagra nge multipliers isused to maximizef(x, y) = 6xyha (lvh262) – Homework 14.8 – karakurt – (562 95) 5subject to the constraint x + y = 2, whi ch ofthe following is true?(I) The maximum value of f(x, y) is lessthan or equal to 7.(II) The maximum value of f ( x , y) occurswhen x = 2.(III) The maximum value of f(x, y) occurswhen y = 1.1. I, II and III2. I only3. II and III only4. III only5. I and III only correctExplanation:The method of Lagra nge multipliers looksfor the critical points of the Lagrange functionF (x, y, λ) = 6xy − λ(x + y − 2).Now these criti cal points wil l be the commonsolutions of t he equat ionsFx= 6y − λ = 0, Fy= 6x −λ = 0,andFλ= x + y −2 = 0.Thus there is only one set of solutions andthese arex = y = 1 , λ =