Absolute minimum change.doc

Absolute minimum change.doc - 18 The function f.1,y = 32(2...

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Unformatted text preview: 18. The function f(.1:,y) = 32(2+y)+y2+4 on the disk D = {32+(y-i-1)2 S 9/4} has critical points at (0,0). (2,—2) and (—2, —2). The following graph is of z = f(:1:,y) restricted to the boundary of D, parameterized by a: = guest, :1; = —1+%sint, 0 51:5 211'. Find the absolute minimum value of f[:1:,y) or: D. (A) 0 (B) 1-5 ":3: . (C) 2-15 ‘° ::;;:;::::3::::::::::;::';::'"'-""'"“"-"“ 1 (D) 3 u 1 (E) 4 I II..—_:':::':::::::::.:_:: . (F) 4,25 11: "'-‘.'"-.."::".:27:31:: 5 .1 .__._,_-:.'::::.' (G) 7.2 a ' "j ""::"I"::":: ' (H) 73 {7,5 —— -—— - ——+———::-.——-'.---- n: --- ---... _:...“.... (I) 8 n 7.2.; -..1-“::::_. - 1”.--“ l — —7—.: .1. ....... ....... (.1) 10.25 '3: xiii. ‘2‘}; hyjnﬂzi-y]+yzv4rullﬂulldioxzt-[yvﬂazﬂ4 19. Refer to problem 18. Find the absolute maximum value of f (3,3,1) on D. (A) n (B) 1.5 (C) 2.15 (D) 3 (E) 4 (F) 4.25 (G) 7.2 (H) 7.3 (1) s {.11 10.25 2_' Find the minimal value of ﬂat. y) = 2m2 + 31:;2 subject to the constraint 9:2 + y2 = 4-- (A 6 :2); -. (H)11 (D)12 (1)13 E 14 ' (1)15 a. () atx‘m‘ﬂ’ﬂl 234:“: pm gem-m A? two -— Wu H m“ e42“ W": “M" °‘ 0 w:- would W 42... abswcﬂ-‘Jrﬂ 3. For the function f(a:, y) = a.“ + 3mg 1- 33:9;"2, fhas: 2 . O2 ‘73 “ 5:34- 3‘l = 36"”) (A) saddle pomts at (-1,1) and (I. — 1) '1- 31M") (B)Iocal max'ima men) and(— 1, ~ I) 0’ ’33 "- 3“ * 3" (C) local minima at (0,0) and (4,1) n.._ -3 ¢ X (D) a local maximum at (0,1) and a local minimum at (3, - 3) =3 x. "' a saddle point at (0,0) and a local minimum at (1,-1) a. l 1 _. (a a saddle point a: (0,1) and a local minimum at (0,0) 0" x (H) a saddle point at (4,3) and a local minimum at (0,0) (I) a local maximum at (-1,1) and a local minimum at (3,-1) , (I) no critical points 9“ 9.1.1 ”‘1”: '- @a saddle point at ( — 1,3) and a local maximum at (0,0) =5 2.1.9.“- )ﬁ ’0 2 ‘1 2. Suppose the temperature in degrees Celsius at a point (as, g) on a metal plate is given by the formula T(z, y) = lone-(31W). Which of the following unit vectors gives the direction in which T decreases most rapidly at the point (1,4)? (A)<—1,~1>/¢§ {(F)<3;4>25I (B) <1,4>!x/ﬁ (G) <2,3>/¢ﬁ (C)<—3,~I>/\/E (H)<—4,—3>/5 (D)<1,0> ' (I) <1,1>/\/§ (E) <0.1> (J) <5.2>N'§9 Dutch.“ 1'. “(VT)“.UX ,: + ‘00 (Mi) ___ +¢'+>{~§ LQTT‘U‘ \ lite (9.8)} 4) Find the Linimum value of the function f (x,y,z ) = 4x + 4y + 42 that lies on the intersection of the sphere :2 + 3:2 + 22 = 3 and the plane z+2y+z:4. F\ 3 A)-6 3):? .. 9?? 3% VP :(waw , [7942:932» Ifﬁg VA :- < 1/ 2/ ‘7 12 _. Fran (/7 \$19) Wire P?‘0 H} ,5: r) L/_ a) x +,u 11:14 1; «~— 3??) 459/" Fm 07+ a) M m (:3:- J I)? 3)L/:a>"2 ‘1'” Fro» itbﬂbrz) Q's-=9" [ii/ii From 8’ X"+(z~x)“+x" = 3* "’9’ ”:3 0” BXL-"qX'F' £0 Jo (EXPIYXH )io X;/ art/2v (Ix/ﬂ) am! +(4//;)=/L Mitt/Mt" aim-me w of the function le.y)= Ky Q the unit :circle x2+y2=1 (cg. not the inside). ( ) X13: / : 1 K : I {(593) aura anJ'f/‘GHT‘ 3 lb 4) Find the glinimum value of the ﬁmction f (x,y,z ) = 4x + 4y + 42 that lies on the intersection of the sphere :2 + y2 + 22 = 3 and the plane .z+2y+z:4. F\ B): 32% V‘F ~‘-‘(‘4/9/‘/> / [73:<2x,ab,a?> VA : <1,2,t? 12 r) L/: svx ‘7” From a; um) um»: 9750 H 25: . ﬂ I))§14 ”p.503 +50“ Pram aumeH/ag "’3'? ”enema +,or an Mme) Q‘s-=‘f—‘9X 3 9‘26 From 8' y‘: (Man X’" = axis/x +‘/= 3 0” 3X"—-‘/X+ I =0 30 (3% ~! TX" VD X::/ are}?! (bl/f) am/ ‘F(’///U:/L MIA/Mr" ...
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