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m251F04ex2

# m251F04ex2 - M MATH 251 Test II Name Section There are a...

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Unformatted text preview: M MATH 251 Test II October 26, 2004 Name Section There are a total of ﬁve problems. No calculators are allowed. 1. A function u = f (:13, y) is said to be homogeneous of degree n for sOme n > 0 --if f(t5’37ty) : tnf(x7y)‘ (a) Show that 2712016, y) + yfy(:c, y) = nf(:v, y). (5%) (b) Show that m2f\$\$<£7y) +y2fyy(\$7y) :n(n_ _ (c) Show that fy(tx,ty) = tn‘lfy(x,y) for all t > 0. l o I (5%) (d) Show that xfz(2a:, 23/) + yfy(2:17, 23/) = n - 2"_1f(x, ' (5%) (e) Let n: 5/2. Give a concrete example of a. function f (any) that is homogeneous of degree 5/2. (5%) [M1314 (a) F’vm 1L(+x,t.1):t"+cx,vp, mu it: 474E-{4-l-x, {11) =1 fl? [(-"fgxﬂd : Mfﬁym'j). ArplvwwmaﬁmWWA/mum: . ® fifwmm): _::_3{3}L¥ﬁ7)igx)+ iﬁix,m)%+v):ﬂ_aﬁbe)+jiﬁwv)= n+"jf‘x'tp‘ wt 97 9X I M tzl We OHM NM 967%Lx,‘1)+71§‘*'”}):"+“"9- Lb)Taug%4aMWo+®m: ' flaw/*7): fa[*%’”‘”’+v—£‘9 23' ﬂ=%<-c“+m> WW”. “"11 AW w wwtw W mmm: 4 'b‘Jf) 47x35) — .31.. ,i— 4 (ﬁ, 235%} 7+jﬁgj—JJ—thﬁ‘b‘ wwm'f" h’) ’ of ) = x [hxwxﬁ'ﬂ 12%) + t7t+xkj)§éwj+l1 09,1153) 0%") + “(10,197 mafia : X [ﬁghﬁ‘jﬂ + m1(-l:)<;l'\j) ‘jj+lj [qud’xﬂﬁﬂx +fl17(fx,f\j)ﬁ :xlz’ﬁx (ﬂit-7) +zx113§7 (f-x, +jlﬁ7 (76x, 1‘73): 0401-4) t’wths‘j). A-jmla) Ad‘ tit!5 Xufxx(x’9})+zxjﬁjcx,v) +jzj€17(£)7):n(n—I)f(x,j) .. (a) H _ (i1 _ 2 n ; n .3; erxﬁj) - ﬁtfxxj) 39;); ﬂaming WE {W19 t jam): #4” ln ff¢b<ky= 1-" ﬁx”) We) véro‘tL Mala, W 067,24,“ J3.(tth-n);f 3309:). >< ﬂamw) +747W‘I‘7): 74- ijicm) . > (8/) FW ML“) 1 +¢W= WW4" r/L Name 2- Given f(x,y) = (a: + 1)4 + 256(y — 2)4 + 16mg _ 2) + y] _ 90, (a) ﬁnd all the critical points of f; (10%) (b) determine the local extrema of f. (5%) [Hint: Calculations will be simpliﬁed by setting u 2 36+ 1, v = 4(y— 2), and by writing 16[x(y — 2) + y] in terms of u and 1).] MMSJ bit-XH ml tr=z,L(j.7_)_—nW [99014991]: l6[xu1~;)+(tr7«)+2j :lla[<x+l)u1-2)+Ll = 4-[@+‘)}[4014)l *3; 5'- 4’140‘4'37’ MA 41M): 144+ U4+4w+32401 “it V4+4W~§9§° Fwy), We we yeti-WA lament +<m> £11 (“m-«7th WM W of FOLﬁO. '5 .. 1E: 3 \$20 =‘> M44120 {n+4 ﬁuz‘v} a: 34M: 0 .. _ 0 :5 7f av- "4' 0: M3Hf=<~vﬁf+1f= “Va-t U30 0 = “177' U: (U‘tl'l)(v%+l) (UH) (711-00210 0’) l3“: 4'5'(X+02>2£97 +x = lé. 3c =z5-é-‘t‘3L7-‘Q , 7 4,1 9 LI“) dr(x17):(o/ a) At 0,7)44279) «FOAH"17)=(“ZIT), D: 12~(Is.4-3)-2%>o) ,_ D=IZ~ (6-4-3 «254le D-{, ]L -f—fzoqevzyuo. ( ) fxx=12>a ... xx ‘1“! 7 _, Malz>0 \ ~ maxi-04AM gﬂmrtma mm_ Name 3. (21) Find the maximum and minimum value of the function f(:c,y,z) =4x+2y+32 subject to the constraints 2x + 2y + z = 10, 4(y + 1)2 + 22 = 16. (14%) (b) What is the name of the surface 4(y + 1)2 + 22 = 16? (3%) (0) Give a sketch of the constraint set. . (3%) (4) 4d“ ﬁkxiv,}): zx+z7+3§ to, M7819): Q-Lg-anfité TL“ V-j—m m) ='- M73091, wakcmw) .4':)\'2.+)4'0: 11A.2 174- 3'01H)) 3=l“9‘+/b2% ‘ I An; {2=4+p-9m+~1) 5,) win-i? 5=4+jkzy : .1 L l 7- t __L_ 1 x6: 4-(t1-H)1+'5.2‘: 4:(~£—};) +(--;3_‘-*7) =(Z-4' 4. Pu 372 ...—! " [4 = :1’ —%~ "' 8 (‘7 =11 - =_-_L_.§2:-1— "' =-—'~——:.-=+2J; /‘ 5’ ‘j ‘ +4? I A1113 2w]: 0,-1.9):(5’4ﬁgzn) x = 41- (lo—z», 77,) =5—(-l-~I1)~-L-(+2J'i)= 5 +0.43% 4~ 6 Jae—4:) 43(142) = 22. + “‘2 (W) .. 2 - E _~ ,, Whig-é i=~l+i-;%=-t+n,g-4i<- g)~ 2m (X) 33’)=Léx‘l+ "2- 7, “2i‘(1°4H'3')=r%(10)“(-l+~]”{)--§;(~zﬁ)=6 7 4 “E; 4‘) that): «an zc-l+J:)+3(—2J2): 22:44: (W) Name 4. (a) Evaluate /00 e_(”2+bm+c)da: for a > 0,1) 6 R and c > 0. (8%) (b) Use part (a) above to determine /Oooe“(“\$2+c)dx, for a > 0. (4%) (c) Use part (b) above to evaluate / ooac2e"23‘2d:1:. (4%) I 0 (d) In part (a), can we evaluate the integral if c < 0? b g- M (4%) b 1, to. _(M+bx+o) 50¢ -(aqb+b7+6) ‘ me- ﬁx +3.71) “weak. 56401333— A’X'Me/ 4’7"...“ '3‘ -00 ,a. b -> 96 El" 3.121") d ) ' , M22 “" 2 UM] W M Ul m by. u 1’; ,0) 9° -atxz’w") v bt‘c’ .904 ,qji' W‘0 M 8 (II/X M -Ax ‘ _ '13 S a 51/17 = 6 : a x = S a ‘LX a Do 2- " ' -Do " 1 113a_(/) 1.1)” M r7, V \$ ,6 Sgedq rdvié L I M _' a o 2. 1 ,L' 1'2.“ ELC zlﬁeé) Tl’ 2(4—93-0) . 1 «our J -_—-. 6““ ). ﬂ— = “"5" a a 2m“ “30:6 no '94 L . . f b .. “111W 1» ~cax”+bx+o) 474\$} .1. _‘ 9:“ 30f 9/ ‘ olX: @ ox pa )E”+c) g f%:o)we.ltm :- 4“" 6 -7:'. ‘ p mwv a ti» m7” VW’J 8%“ (b) 547 A1 :Zof ' 4?— 7? w ems-7t,— v m 1 - 12. “"4"” «xi-Elli} (a) Sodxz’e‘a'xdx; 50 -——}i—§[email protected]’ax)d>(“ "Xenia AX «’04 236». .._,L_ 11 MMMAZZ) l “ 24 "oi? ( I __ L H 11; 50m%‘6”d%=‘§a)l—Ez “5‘4 2 ~ "tie. (cock «WM/MQCMAWA/Bi C WM gooe~CaX\$+bX+ylocx A ﬁt (+th A’ﬁnqt —04 Name 5. Consider 1 4 / / 003(3w2 + 12x — 5)dacdy. —2 (y—Z) NIH (a) Sketch the region where the integral is evaluated. ’ (b) Evaluate the integral by changing the order of integration. [4. S4 SWBXL+12X~§M><47 " -7- '51“) ill View“ 2X-99474X .L -1. l :5 WM‘HZX'S’) [‘6 ‘1 ~ (5%) (15%) ...
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m251F04ex2 - M MATH 251 Test II Name Section There are a...

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