hw-14-sec-15.7-solns

hw-14-sec-15.7-solns - ‘— SECTION 15.7 MAXIMUM AND...

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Unformatted text preview: ‘— SECTION 15.7 MAXIMUM AND MINIMUM VALUES ET SECTION 14.7 C 6. f(:c, y) = $33,, +1237;2 — 83/ => fm = 3wa + 24x, f, = $3 — 8, fm 2 6x3; + 24, fly = 3332, fyy : 0. Then fy = 0 implies m = 2, and substitution into fx 2 0 gives 12y + 48 = 0 :> y = —4. Thus, the only critical point is (2, —4). D(2, *4) = (—24)(0) — 122 = —144 < 0, so (2, —4) is a saddle point. , yv‘ V I ' f” = 12x2, f“, = —4, fyy = 12342. Then fI 2 0 implies y : $3, ' puma" “VIN/’1‘, I " Weaw “Vt-241$ 1” 9 I i and substitution into fy = 0 => :10 : 3;3 gives x9 — w = 0 :> $(m8 — 1) = 0 => :0 = 0 orw = :tl. Thus the critical points are (0,0), (1, 1), and (-1, —1). Now D(0, 0) z 0 . 0 ~ (—4)2 = —16 < 0, 2 so (0, 0) is a saddle point. D(1, 1) = (12)(12) — (-4)2 > 0 and f”(1, 1) = 12 > 0, so f(1, 1) = 0 is alocal minimum. D(—1,—1) fm = (—1, —1) = 12 > 0, so f(—1, ~1) : 0 is alsoalocal minimum. 2 2 2 2 (m) = w ;~ 12 = —2ace4w w , fy : (4 _ 2y)e4y—~z2~y27 fix : (43:2 _ 2)€4y—12_y2, = (12)(12) * (—4)? > Oand fzy = —2z(4 — 2y)e4y~$2—y2, fyy = (43,,2 ~ 16y + 14)e4y*$2*y2. Then fw = 0 and fy = 0 implies a: = 0 and y z 2, so the only critical point is (0, 2). Now D(0, 2) = (—2e4)(—2e4) ~ 02 = 4.38 > 0 and fxz(0, 2) = —2e4 < 0, so f(0, 2) = e4 is a local maximum. (wyy)=(1+xy)(w+y) =m+y+rc2y+xy2 3» 5 fz=1+2xy+y2, fy=1+$2+2wy, “=29, fsy=2$+2y, o "WW" "I~'1~.¢.~‘~:t~szi« ‘Q r \ fyy = 236. Then fr 2 0 implies 1 + 2223/ + yr‘f: 0 and fy = 0 implies 2 cm” ' ' 1 + :c + 2mg 2 0. Subtracting the second equation from the first gives ‘W'?" H“! g y2ez2=0 :> y:i$,butify=xthen1+2my+y2=0 :s 2 “ALMA 1 + 3:132 2 0 which has no real solution. If y = —ac then 1+ 250.11 + 3/2 = 0 :> 1— $2 = 0 => 2: : il, so critical points are (1, —1) and (—1, l). D(1,‘1)=(—2)(2)— 0 < 0 and D(—1,1)=(2)(—2)— 0 < 0, so (—1,1) and (1, —1) are saddle points. 407 408 1: CHAPTER 15 PARTIAL DERIVATlVES ET CHAPTER 14 @ f(x,y) = 2:123 +5163;2 + 5.722 +112 :> fz = 6m2 +1;2 + 10x, fy = 2333/ + 2y, fm = 1223 + 10, fyy = 2a: + 2, fig 2 2y. Then fy = 0 implies y = 0 or x = ——1. Substituting into fx 2 0 gives the critical points (0, 0), (—g, 0), (—1, :12). Now D(0, 0) = 20 > 0 and fm(0, 0) = 10 > 0, so f(0, 0) = 0 is a local minimum. Also fut—3,0) < 0, D(—§,o) > 0, and D(—1,i2) < 0. Hence f (—g, 0) = $2275 is a local maximum while (—1, :12) are saddle points. 11. f 93,3; 2 $3 —— 12mg + 81/3 => f1 = 3372 - 12y, f = ~12m+ 24312, y 200 fm = 6m, fxy = —12, fyy = 48y. Then fx = 0 implies :52 = 4y‘anld' fy 7—— 0 implies cc = 2y2. Substituting the second equation into the first gives (2y2)2 = 43/ => 4y4 2 4y => 4y(y3 — 1): 0 => y 2 0 or —200 y = 1. Ify = 0 then at = 0 and ify = 1 then x = 2, so the critical points are (0,0) and (2,1). D(0,0) = (O)(0) — (—12)2 = —144 < 0, so (0, 0) is a saddle point. D(2, 1) = (12)(48) — (—12)2 = 432 > 0 and fm(2, 1) : 12 > 0 so f(2, 1) = —8 is alocal minimum. 1 1 1 1 2 => fEZyfifiafyz‘T—?’ 113:?) 2 . . 1 . . fzy = 1, fyy = Then ft = O Implies y = E and fy = O imphes 1 . . . . , a: = Substituting the first equatlon into the second gives 11:: 1 => w2m4 :> $(13—1)=0 => m:00rac=1. (1/932)2 f is not defined when x = O, and when ac = 1 we have y = 1, so the only critical point is (1, 1). D(1, 1) = (2)(2) — 12 2 3 > 0 and f”(1, 1) = 2 > 0, so f(1, 1) = 3 is a local minimum. 13. f(ac,y) 2 ex cosy => fa: 2 ex cos y, fy = flex siny. Now fr : 0 implies cos y = 0 or y = g + mr for n an integer. But sin(§ + mr) 7E 0, so there are no critical points. 416 El CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14 Sec; 67 32. fan (on, y) 2 4 — 2m and fy(ac, y) 2 6 — 2y, so the only critical point is (2, 3) (which is in D) where f(2, 3) 2 13. Along L1: y 2 0, so flat, 0) 2 4m — x2 2 —(a: — 2)2 + 4, 0 g as g 4, which has a maximum value when x 2 2 where f(2, 0) 2 4 and a minimum value both when a: 2 0 and as 2 4, where f(0, 0) 2 f(4, 0) 2 0. Along L2: as 2 4, so f(4, y) 2 6y — y2 2 —(y — 3)2 + 9, 0 g y g 5, which has a maximum value when y 2 3 where f(4, 3) 2 9 and a minimum value when y 2 0 where f(4, 0) 2 0. Along L3: y 2 5, so f(x, 5) 2 —z2 + 4x + 5 2 ~(a: — 2)2 + 9, 0 S as g 4, which has a maximum value when w 2 2 where f (2, 5) 2 9 and aminimum value both when a: 2 O and m 2 4, where f(0, 5) 2 f(4, 5) 2 5. AlongL42x20,sof(0,y) 26y—y2 = —(y—3)2+9,03y55, which has a maximum value when y 2 3 where f (0, 3) 2 9 and a minimum value when y 2 0 where f (0, 0) 2 0. Thus the absolute maximum is f (2, 3) 2 13 and the absolute minimum is attained at both (0, 0) and (4,0), where f(0,0) 2 f(4,0) 2 0. I (39 f (9:, y) 2 m4 + y4 — 4201/ + 2 is a polynomial and hence continuous on D, so it has an absolute maximum and minimum on D. In Exercise 7, we found the critical points off; only (1,1) with f(1, 1) 2 0 is inside D. On L1: 3/ 2 0, f (w, 0) 2 m4 + 2, 0 S m S 3, a polynomial in a: which attains its maximum at m 2 3, f(3, 0) 2 83, and its minimum at a: 2 0, f(0, 0) 2 2. On L221 2 3, f(3,y) 2 y4 — 12y + 83, 0 g y S 2, apolynomial iny which attains its minimum at y 2 {73, f(3, 2 83 — 9 \3/3 2 70.0, and its maximum at y 2 0, f(3, 0) 2 83. On L3: y 2 2, f(:r;, 2) 2 m4 ~ 813 + 18, 0 S as g 3, a polynomial in a: which attains its minimum at x 2 \3/2, f(\3/2,2) 21826922 10.4,anditsmaximumatw23,f(3,2) 275. OnL4: x20,f(0,y) 2y4+2,0 g y S 2,a polynomial in y which attains its maximum at y 2 2, f (0, 2) 2 18, and its minimum at y 2 0, f (O, 0) 2 2. Thus the absolute maximum off on D is f(3, 0) 2 83 and the absolute minimum is f(1, 1) 2 0. fan 2 3/2 and fy 2 2337;, and since f1 2 0 <2 y 2 0, there are no critical points in the interior of D. Along L1: y 2 0 and f(x, 0) 2 0. Along L2: :2: 2 0 and f(0,y) 2 0. Along L3: 3; 2 W, so let g(m) 2 finm) 2 335—133 forO g x S Then g'(m) 2 3 — 33:2 2 0 <=> ac 2 1. The maximum value is f(1, 2 2 and. the minimum occurs both at ac 2 0 and a: 2 \/3 where f (0, \/3 ) 2 f (V3, 0) 2 0. Thus the absolute maximum of f on D is f(1, 2 2, and the absolute minimum is 0 which occurs at all points along L1 and L2. 39. 418 I: CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14 38. f (w, y) = Bacey — m3 — 83y is differentiable everywhere, so the requirement for critical points is that fx : 3ey — 32:2 = 0 (1) and fy : 3mg — 3639 = 0 (2). From (1) we obtain ey 2 m2, and then (2) gives y 33:3 —3x6 = 0 : as = 1 or0,butonlyx = 1 isvalid,sinceac : 0 makes (1) impossible. So substituting a: = 1 into (1) gives y = 0, and the only critical point is (1,0). The Second Derivatives Test shows that this gives a local maximum, since D 1,0 = —6a:(3mey — 953'” — 36y)2 : 27 > Oand fu(1,0 = —693 = —6 < 0. But f(1,0 = 1 is not an (1,0) (1,0) absolute maximum because, for instance, f (—3, 0) = 17. This can also be seen from the graph. Let d be the distance from (2,1,—1) to any point (at, y, 2) on the plane a: + y — z = 1, so d : (cc — 2)2 + (y — l)2 + (z + 1)2 where z = as + y — 1, and we minimize d2 = may) = (x — 2)2 + (y — 1)2 + (ac + 21)? Then Mm) = 2(rc — 2) + 2(w + y) = 4a: + 2y — 4, fy(m, y) = 2(3) — 1) + 2(23 + y) = 227 + 4y — 2. Solving 4x + 2y ~ 4 = 0 and 2x + 4y 2 = 0 simultaneously gives a: = 1, y : 0. An absolute minimum exists (since there is a minimum distance fiom the point to the plane) and it must occur at a critical point, so the shortest distance occurs for ac : 1, y = 0 for which (1 = \/(1 — 2)2 + (0 — 1)2 + (0 + 1)2 = ere the distance d from a point on the plane to the point (1, 2, 3) is d = (a: — 1)2 + (y — 2)2 + (z — 3)2, 41. 42. wherez =4—m+y. We canminimize d2 =f(m,y) = (m—1)2+(y—2)2+(1—x+y)2, so f$(m,y)=2(x—1)+2(1—x+y)(—1)=4x—2y—4andfy(a:,y)=2(y—2)+2(1‘m+y)=4y—2zr—2. Wlb and y = g, so the only critical point is (g, ). Solving 4m — 2y — 4 = O and 4y ~ 236 — 2 = 0 simultaneously gives a: = g This point must correspond to the minimum distance, so the point on the plane closest to (1, 2, 3) is (g, g, Let d be the distance from the point (4,2,0) to any point (x, y, 2) on the cone, so (1 = W where z2 = x2 + if, and we minimize d2 = (1: ~ 4)2 + (y — 2)2 + m2 + y2 2 f (w, Then fac (90,31) 2 2 (m — 4) + 2x = 4m — 8, fy (x,y) = 2 (y — 2) + 2y : 4y — 4, and the critical points occur when fm = O :> at z 2, fy = 0 :> y = 1. Thus the only critical point is (2, 1). An absolute minimum exists (since there is a minimum distance from the cone to the point) which must occur at a critical point, so the points on the cone closest to (4, 2, 0) are (2, 1, The distance from the origin to a point (x, y, 2) on the surface is d = y m2 + y2 + Z2 where y2 = 9 + :vz, so we minimize d2 2332+9+acz+22 =f(ac,z).Thenfz =2x+z,fz 2912—1—22,andfm :0,fz :0 => x=0,z=0,sotheonly critical point is (0,0). D (0,0) = (2)(2) — 1 : 3 > 0 with f” (0,0) = 2 > 0, so this is a minimum. Thus y2 = 9 + 0 => y = i3 and the points on the surface closest to the origin are (0, i3, 0). 3 » x l ...
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This note was uploaded on 05/02/2011 for the course MATH 408D taught by Professor Chu during the Spring '09 term at University of Texas.

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hw-14-sec-15.7-solns - ‘— SECTION 15.7 MAXIMUM AND...

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