lewis (scl876) – HW 14.7 – radin – (55315) 1This print-out should have 17 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsSuppose (1, 1) is a critical point of a func-tion f having continuous second derivativessuch thatfxx(1, 1) = −5, fxy(1, 1) = 2,fyy(1, 1) = 5 .Which of the following properties does f haveat (1, 1)?1. a saddle po int correct2. a local maximum3. a local minimumExplanation:Since (1, 1) is a criti cal point, the SecondDerivative test ensures that f w ill have(i) a local minimum at (1, 1) iffxx(1, 1) > 0, fyy(1, 1) > 0,fxx(1, 1)fyy(1, 1) > ( fxy(1, 1))2,(ii) a local maximum at (1, 1) iffxx(1, 1) < 0, fyy(1, 1) < 0,fxx(1, 1)fyy(1, 1) > ( fxy(1, 1))2,(iii) a saddle point at (1, 1) iffxx(1, 1)fyy(1, 1) < ( fxy(1, 1))2.From the given values of the second deriva-tives of f at (1, 1), it thus follows that f hasa saddle pointat (1, 1).002 10.0 pointsIn the contour map below identify thepoints P, Q, and R as local minima, localmaxima, or neither.3210-1-20-3-2-1012QPRA. local minimum at R,B. local maximum at P ,C. neither local max nor local min at Q.1. A only2. B only3. C only correct4. A and C only5. A and B only6. all of them7. B and C only8. none of themExplanation:A. FALSE: the contours near R are closedcurves enclosing R and the contours increasein value as we approch R. So the surface hasa local maxi mum a t R, not a local minimum.lewis (scl876) – HW 14.7 – radin – (55315) 2B. FALSE: the contours near P are closedcurves enclosing P and the contours decreasein value as we approch P . So the surface hasa local minimum at P , not a local ma ximum.C. TRUE: the point Q lies on the 0-contourand in this contour divides the region near Qinto two regions. In one region the contourshave values increasing to 0, while in the otherthe contours have values decreasing to 0. Sothe surface has neither a local maximum nora local minimum at Q.keywords: contour map, local extrema,True/False,003 (part 1 of 3) 10.0 pointsIf f is defined byf(x, y) =83x3+ x2+ 4xy + 2y2+ y ,locate the critical points of f.1.14, 0,12,342.14, 0,−12,343.14, 0,12, −344.−14, 0,12,345.−14, 0,12, −34correct6.−14, 0,−12,34Explanation:After differentiationfx= 8x2+ 2x + 4y,whilefy= 4x + 4y + 1 .Thus the critical poi nts of f are the commonsolutions of the equations8x2+ 2x + 4y = 04x + 4y + 1 = 0 .Eliminati ng y we see that8x2− 2x − 1= 8x +14x −12= 0 ,the soluti ons o f which are x = −14and12.Consequently, the critical points of f occur at−14, 0,12, −34.004 (part 2 of 3) 10.0 pointsWhich of the following most accurately de-scribes the behaviour of f.1. local minimum at−14, 02. local maximum at14, 03. saddle-point at−14, 0correct4. saddle-point at14, 05. local minimum at14, 06. local maximum at−14, 0Explanation:After differentiation once again we see thatfxx= 2(8x + 1), fyy= 4, fxy= 4,and so at the critical point−14, 0,A = fxx−14, 0= −2 ,lewis (scl876) – HW 14.7 – radin – (55315) 3andC = fyy−14, 0= 4 > 0 ,whileB = fxy−14, 0= 4 .But then,AC − B2= −24 < 0 .Hence by the second derivative test, f has asaddle-point at−14, 0.005 (part 3 of 3) 10.0 pointsWhich of the following most accurately de-scribes the behaviour of f.1. saddle-point at12, −342. local ma ximum at12,343. local minimum at12, −34correct4. local minimum at−12,345. local ma ximum at12, −346. saddle-point at−12,347. local minimum at12,348. local ma ximum at−12,349. saddle-point at12,34Explanation:After differentiation once again we see thatfxx= 2 (8x + 1 ), fyy= 4, fxy= 4 ,and so at the critical point12, −34,A = fxx12, −34= 10 ,andC = fyy12, −34= 4 > 0 ,whileB = fxy12, −34= 4 .But then,AC − B2= 24 > 0 .Hence by the second derivative test, f has alocal minimum at12, −34.006 10.0 pointsLocate and classify the local extremum off whenf(x, y) = 3x +y3+1xy+ 2, (x, y > 0).1. local min at (3, 3)2. local min at13, 3correct3. saddle at (3, 3)4. local max at (3, 3)5. saddle at13, 36. local max at13, 3Explanation:lewis (scl876) – HW 14.7 – radin – (55315) 4Differentiating once we see thatfx= 3 −1x2y, fy=13−1xy2.At a local extremum these first partial deriva-tives are zero. Thus f has a local extremumat13, 3.To classify the local extremum we use thesecond derivative test. Nowfxx=2x3y, fxy=1x2y2, fyy=2xy3.But thenA = fxx13, 3= 18 > 0,C = fyy13, 3=29> 0,andB = fxy13, 3= 1.ThusAC − B2= 3 > 0.Hence by the second derivative test, f has alocal min at13, 3.007 10.0 pointsLocate and classify all the local extrema off(x, y) = 2x3+ 2y3+ 6xy −1.1. local max at (−1, −1), saddle at (0, 0)correct2. local min at (−1, −1), saddle at (0, 0)3. saddle at (−1, −1), local max at (0, 0)4. local min at (0, 0), local max at (−1, −1)5. local max at (1, 1), saddle at (0, 0)Explanation:Local extrema occur at the cri tical pointsof f. Now after differentiation of f we obtainfx= 6(x2+ y), fy= 6(y2+ x).The critical points of f are thus the commonsolutions of the equationsx2+ y = 0, y2+ x = 0.This yields only the two extremum points(−1, −1) and (0, 0). B ut after differentiatingagain we see t hatfxx= 12x, fxy= 6, fyy= 12y;consequently,fxxfyy− (fxy)2= (12)2xy − 36.Hence by the second derivative test there arelocal max at (−1, −1), saddle at (0, 0).008 (part 1 of 2) 10.0 pointsWhen f is the function defined for x, y > 0byf(x, y) = ln(xy) + 8 x2− x − xy + 4 ,locate its critical point.1. critical po int at16,142. critical po int at4, 163. critical po int at116, 44. critical po int at14,1165. critical po int at116, 16correctExplanation:lewis (scl876) – HW 14.7 – radin – (55315) 5The critical point of f is the common solu-tion of the equatio ns∂f∂x=1x+ 16x − 1 − y = 0,and∂f∂y=1y− x = 0.Thus f has acritical point at116, 16.009 (part 2 of 2) 10.0 pointsWhich of the following statements most ac-curately describes f at this criti cal point?1. local minimum value of127322. local minimum value of95323. saddle-point correct4. local ma ximum value of95325. local ma ximum value