ha (lvh262) – Homework 14.7 – karakurt – (56295) 1This print-out should have 26 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsDetermine the absolute maximum value off(x, y) = −2 cos x cos yon the square 0 ≤ x, y ≤ π.1. absolute max = 2π2. absolute max = 03. absolute max = 2 correct4. absolute max = −25. absolute max = −2πExplanation:The absolute extrema of f(x, y) occur ei-ther at a crit ical point inside the square, or onthe boundary of the square as shown inedge 1edge 2edge 3edge 4(π, π)xyNow∂f∂x= 2 sin x cos y ,∂f∂y= 2 cos x sin y .Thus the critical points are at the solutions o f2 sin x cos y = 0 , 2 cos x sin y = 0 .Since0 < x, y < π =⇒ sin x, sin y > 0 ,the only critical points inside the square occurwhencos x = 0, cos y = 0 ,i.e., at (π/ 2, π/2), and at this critical pointfπ2,π2= 0 .We look at the edges separately:(i) on Edge 1, f (x, 0) = −2 cos x somax0≤x≤πf(x, 0) = 2 .(i) on Edge 2, f (0, y) = −2 cos y somax0≤y≤πf(0, y) = 2 .(i) on Edge 3, f (x, π) = 2 cos x somax0≤x≤πf(x, π) = 2 .(i) on Edge 4, f (π, y) = 2 cos x somax0≤y≤πf(π, y) = 2 .Consequently, on t he square, f hasabsolute maximum = 2 .keywords:002 10.0 pointsDetermine the absolute maximum off(x, y) = x2+ y2− x −y + 1on the unit diskD = {(x, y) : x2+ y2≤ 1 }.1. absolute max = 2 −√22. absolute max = 33. absolute max = 2 +√2 correctha (lvh262) – Homework 1 4 .7 – kara kurt – ( 56295) 24. absolute max =125. absolute max = 2Explanation:The Absolute Extrema occur either at acritical point inside the disk, or on the b ound-ary o f the disk.1. Inside D: whenf(x, y) = x2+ y2− x − y + 1 ,the critical points occur at the solutions to∂f∂x= 2x − 1 ,∂f∂y= 2y − 1 = 0 .So (1/2, 1/2) is the only critical point, andthis lies inside the disk. At t his critical pointf12,12=12.2. On the boundary of D: the boundary ofD is the unit circle which can be parametrizedbyr(t) = cos t i + sin t j , 0 ≤ t ≤ 2π .Thus the restriction of f to the boundary isthe functionf(r(t)) = 2 − cos t − sin t .Finding the absolute extrema of f ( r(t)) on[0, 2π] is a single vari able problem: they willoccur either at a critical point in (0, 2π), orat the endpoints t = 0, 2π. Butdfdt(r(t)) = sin t − cos t ,so the crit ical points of f (r(t)) on (0, 2π)occur when tan t = 1, i.e. at t = π/4, 5π/4.Butfrπ4= 2 −√2 , fr5π4= 2 +√2 ,whilef(r(0)) = 1 = f(r(2π)) .Consequently, on D, f(x, y) hasAbsolute max = 2 +√2 .keywords:003 10.0 pointsSuppose (1, 1) is a critical point of a func-tion f having continuous second derivativessuch thatfxx(1, 1) = 4, fxy(1, 1) = 2,fyy(1, 1) = 5 .Which of the following properties does f haveat (1, 1)?1. a local max imum2. a local minimum correct3. a saddle pointExplanation:Since (1, 1) is a critical point, the SecondDerivative test ensures that f will 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 local minimumha (lvh262) – Homework 1 4 .7 – kara kurt – ( 56295) 3at (1, 1).004 10.0 pointsIn the contour map below identify thepoints P, Q , and R as local minima, lo calmaxima, or neither.3210-1-20-3-2-1012QPRA. local minimum at P ,B. local maximum at R,C. local maximum at Q.1. A and C only2. A only3. none of them4. all of them5. A and B only correct6. C onl y7. B and C only8. B onlyExplanation:A. TRUE: the contours near P are closedcurves enclosing P and the contours decreasein va lue as we approch P . So the surface hasa local minimum at P .B. TRUE: the contours near R are closedcurves enclosing R and the contours increasein va lue as we approch R. So the surface hasa local maximum at R.C. FALSE: the point Q lies on the 0-contour and this contour divides the regionnear Q into two regions. In one region thecontours have values i ncreasing to 0, while inthe other the contours have va lues decreasingto 0 . So the surface does not have a localminimum at Q.keywords: contour map, local extrema,True/False,005 10.0 pointsLocate and classify t he local extremum ofthe functionf(x, y) = x2+ y2− 8x − 6y + 2.1. saddle-point at (4, 3)2. local maximum at (4, 3)3. saddle-point at (4, 4)4. local minimum at ( 4, 3) correct5. local maximum at (3, 3)Explanation:Differentiating t he function with respect tox and y we see thatfx= 2x − 8, fy= 2y − 6.As the local extremum is a critical point,therefore, t he local extremum will occur at(4, 3).To classify the critical point we use thesecond derivative test. Now fxx= fyy= 2,while fxy= 0. Thusfxx= fyy> 0, fxxfyy− (fxy)2> 0.Hence f has alocal minimum at (4, 3) .ha (lvh262) – Homework 1 4 .7 – kara kurt – ( 56295) 4006 10.0 pointsLocate and classify t he local extremum ofthe functionf(x, y) = 4 + 8x + 2y − x2− 2y2.1. saddle-point at (1, 2)2. local max at (1, 2)3. saddle-point at4,124. local min at4,125. local max at4,12correct6. local min at (1, 2)Explanation:Differentiating once we see thatfx= 8 − 2x , fy= 2 − 4y.At a local extremum the first partial deriva-tives are zero. T hus8 − 2x = 0 , 2 − 4y = 0,so f has a local extremum at4,12.To classify this local extremum we use thesecond derivative test withfxx= −2, fxy= 0, fyy= −4 .For thenfxx< 0 , fyy< 0 , fxxfyy− f2xy> 0 .Consequently, by the second derivative t est,f has alocal max at4,12.007 10.0 pointsLocate and classify all the local extrema o ff(x, y) = 2x3+ 2y3+ 6xy − 4.1. local max at (1, 1), saddle at ( 0, 0)2. saddle at (−1, −1), l ocal max at (0, 0)3. local max at (−1, −1), saddle at (0, 0)correct4. local min a t ( 0, 0), local max at (−1, −1)5. local min at (−1, −1), saddle at (0, 0)Explanation:Local extrema occur at the critical pointsof f . Now after differentiation of f we obtainfx= 6(x2+ y), fy= 6(y2+ x) .The crit ical points of f are t hus the commonsolutions of the equationsx2+ y = 0 , y2+ x = 0.This yields only the two extremum po ints(−1, −1) and (0, 0). But after di fferentiatingagain we see thatfxx= 12 x , fxy= 6, fyy= 12 y;consequently,fxxfyy− (fxy)2= ( 12)2xy − …