lewis (scl876) – HW 14.8 – radin – (55315) 1This print-out should have 11 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsFinding the maximum value off(x, y) = 2x + y + 1subject to the constraintg(x, y) = 4x2+ 3y2− 3 = 0is equivalent to finding the height of the high-est point on the curve of intersection of thegraphs of f and g shown inyxzUse Lagrange multi pl iers to determine thismaximum value.1. max value = 62. max value = 53. max value = 74. max value = 3 correct5. max value = 4Explanation:The extreme values occur at solut ions of(∇f)(x, y) = λ(∇ g)(x, y) .Now(∇f)(x, y) = h2, 1 i,while(∇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= 3 , f−34, −12= −1 ,we thus see thatmax value = 3 .keywords:002 10.0 pointsUse Lagrange Multipliers to determine themaximum value off(x, y) = 6xysubject to the constraintg(x, y) =x29+y216− 1 = 0 .1. maximum = 36 correct2. maximum = 343. maximum = 37lewis (scl876) – HW 14.8 – radin – (55315) 24. maximum = 335. maximum = 35Explanation:By the Method of Lagrange multipliers, theextreme values of f occur at the commonsolutions of(∇f)(x, y) = λ(∇g)(x, y), g( x, y) = 0 .Now(∇f)(x, y) = h6y, 6x i,while(∇g)(x, y) =D29x,18yE.But then by the conditi on on ∇f and ∇g,6y =29λx , 6x =18λy ,which after simplification givesλ =27yx=48xy, i.e., y = ±43x .Thus by the constraint equat ion,g(x, ±43x) =x29+x29− 1 = 0 ,i.e., x = ±3√22. Consequently, the extremepoints are3√22, 2√2,3√22, −2√2,and−3√22, −2√2,−3√22, −2√2.Sincef3√22, 2√2= f−3√22, −2√2= 36 ,whilef3√22, −2√2= f−3√22, 2√2= −36 ,we thus see that f hasmax value = 36subject to the constraint g(x, y) = 0.keywords:003 10.0 pointsThe temperature T at a p oint (x, y, z) onthe surfacex2+ y2+ z2= 48is g iven byT (x, y, z) = x + y + zin degrees centigrade. Find the maximumtemperature on this surface.Correct answer: 12.Explanation:We have to maximize the function T sub-ject to the constraintx2+ y2+ z2= 48.The corresponding Lagrange function isF (x, y, z, λ) = x+y +z +λ(x2+y2+z2−48).At the critical point of F , therefore,Fx= 1 + 2λx = 0,Fy= 1 + 2λy = 0,Fz= 1 + 2λz = 0,Fλ= x2+ y2+ z2− 48 = 0.Thusx = y = z =p48/3 = 4.Substituting in the original definition of T wesee thatTmax= 12◦C.lewis (scl876) – HW 14.8 – radin – (55315) 3004 10.0 pointsA rectangular box with edges parallel to theaxes is inscribed in the ellipsoid6x2+ 2y2+ 4z2= 1similar to the one shown inUse Lagrange multipliers t o determine themaximum volume of this box.Note: all 8 vertices of the box will lie on theellipsoid when the volume is maximized.1. volume =16cu. units2. volume =29cu. units correct3. volume =23cu. units4. volume =13cu. units5. volume =19cu. unitsExplanation:The rectangula r box will be centered atthe origin, so if the corner lying in the firstoctant is P (x, y, z) , then the box will havesidelengths 2x, 2y and 2z. Thus the box willhave volumeV = 8xyz .But P li es on the ellipsoid6x2+ 2y2+ 4z2= 1 ,so we have to ma ximize V subject to this lastrestriction on x, y, and z. To use Lagrangemultipliers, setF (x, y, z, λ) = 8xyz−λ(6x2+2y2+4z2−1) .Then F will have a critical point when8yz = 12λx , 8zx = 4λy , 8xy = 8λz ,and6x2+ 2y2+ 4z2= 1 .From the first set of equations we see that8xyz = 12λx2, 8yzx = 4λy2, 8zxy = 8λz2,in whi ch case12x2= 4y2= 8z2.From the fact that P lies on the ellipsoid,therefore, it thus follows that F has a criticalpoint atx2=118, y2=16, z2=112.At this critical po int V will be maximized.Consequently, the box hasmaximum volume =29cu. units.005 10.0 pointsIf the method of Lagra nge multipliers isused to maximizef(x, y) = 3xysubject to the constraint x + y = 8, which ofthe following is true?(I) The maximum value of f(x, y) is lessthan or equal to 49.(II) The maximum value of f (x, y) occurswhen x = 8.(III) The maximum va lue of f(x, y) occurswhen y = 4.1. III onlylewis (scl876) – HW 14.8 – radin – (55315) 42. II and III only3. I and III only correct4. I only5. I, II and IIIExplanation:The met hod of Lagrange multipliers looksfor the critical points of the Lagrange functionF (x, y, λ) = 3xy − λ(x + y − 8 ).Now these critical points will be the commonsolutions of the equationsFx= 3y − λ = 0, Fy= 3x − λ = 0 ,andFλ= x + y − 8 = 0.Thus there is only one set of solutions andthese arex = y = 4 , λ = 3x = 12.Since f(4, 4) = 48, we see that I and III arethe correct statements.006 10.0 pointsDetermine the maximum value off(x, y) = 2x1/4y3/4subject to the constraintg(x, y) = x + 3y − 8 = 0 .1. maximum = 72. maximum = 33. maximum = 64. maximum = 55. maximum = 4 correctExplanation:By the Method of Lagrange multipliers, theextreme values occur at the common soluti onsof(∇f)(x, y) = λ(∇g)(x, y) , g(x, y) = 0 .Now∂f∂x=12x−3/4y3/4,∂f∂y=12x−3/4y3/4.Thus(∇f)(x, y) =D12x−3/4y3/4,32x1/4y−1/4E.On the other hand,(∇g)(x, y) = h1, 3 i.But then by the condition on ∇f and ∇g,12x−3/4y3/4= λ ,32x1/4y−1/4= 3λ ,which after simplification givesλ =12x−3/4y3/4=12x1/4y−1/4,i.e., y = x. Thus by the constraint equat ion,g (x, x) = x + 3x − 8 = 0 ,i.e., x = 2. Consequently,( 2, 2 ) ,is a critical po int at whichf(2, 2) = 2(2)1/4(2)3/4= 4 .And so f hasmaximum value = 4subject to the constraint g(x, y) = 0.keywords:007 10.0 pointslewis (scl876) – HW 14.8 – radin – (55315) 5Determine the minimum value off(x, y, z) = 2x2+ y2+ 2z2+ 1subject to the constraint3x + 3 y + 2z = 4.1. min value =6331correct2. min value =61313. min value =60314. min value =65315. min value =6731Explanation:By the met hod of Lagrange Multipliers theminimum val ue off(x, y, z) = 2x2+ y2+ 2z2+ 1subject to the constraint(†) 3x + 3 y + 2z = 4will occur at a critical point of the LagrangefunctionF (x, y, z, λ) = 2x2+ y2+ 2z2+ 1+ λ(3x + 3y + 2z − 4).This critical point will be the common solu-tion (x0, y0, z0, λ0) of the equations∂F∂x= 4x + 3λ = 0,∂F∂y= 2y + 3λ = 0,∂F∂z= 4z + 2λ = 0,and∂F∂λ= 3x + 3y + 2z − 4 = 0.Solving, we see thatx0= −34λ0, y0= −32λ0, z0= −12λ0,so314λ0+ 4 = 0, i.e., λ0= −1631.Thus the minimum value of f subject to theconstraint (†) occurs