This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 : $3, ' puma" “VIN/’1‘, I
" Weaw
“Vt241$
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 ...
View
Full
Document
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.
 Spring '09
 Chu

Click to edit the document details