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5/19/2018 Bai tap kinh te vi mo.pdf
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Kinh tVi m
Bi tp thc hnh 1
Cu 1 (F0203-PS1-1)
Hy xem xt thtrng bnh m. M tngn gn tc ng ca tng trng hp sau ln cu,cung, lng cn bng v gi cn bng ca bnh m. Minh ha cu trli bng thnu thyhu ch.1. Mt loi phn bn mi lm tng nng sut la m.2. Gi btng do mt cn bnh lm nh hng ti b.3. Ni lo ngi vcht phgia trong lng thc lm gim cu i vi ph.4. Mt vbi cng ca nhng ngi lm bnh m lm tng tin cng lao ng.5. htrgi bnh m, chnh phng mua tt csbnh m thng dv trcao hn
gi thtrng hin ti 10%.6. gip gim lm pht, chnh pht mt gi trn cho gi bnh m bng vi gi thp hn
tng tn ti cch y hai nm.
Cu 2 (F0203-PS1-2)
Cu v cung i vi vic i bc s(khng c bo him) tun tl:P = 100 0,1Qd
P = 10 + 0,1Qs1. Gi v lng cn bng l bao nhiu? Tng chi tiu cho vic i bc sl bao nhiu?2. Chnh phgii thiu bo him sc khe quc gia, mt chng trnh schi tr75% cho
mi ln i bc s. iu g sxy ra i vi gi v lng cn bng ca vic i bc s? Gido ngi tiu dng trl bao nhiu? Tng chi tiu cho vic i bc sl bao nhiu? Tngchi tiu ca ngi tiu dng l bao nhiu? (Gi : Anh chc thdch chuyn ng cuti P = 400 0,4Qd hoc ng cung ti P = 2.5 + 0.025Qs phn nh tc ngca bo him. Nu anh chdch chuyn ng cu, th ng cu mi biu din tng sgi cm cc bc snhn c, bao gm cphn do bo him tr; ng cu ban u tiptc biu din phn do ngi tiu dng trtrong ha n bc s. Nu anh chdch chuynng cung, th ng cung mi biu din phn do ngi tiu dng trtrong ha ntng; ng cung ban u tip tc biu din tng stin thanh ton m cc bc snhnuc.)
Cu 3 (F0607-PS1-4)
Hm scu v hm scung ca mt sn phm c cho di y:
Cu: P = (-1/2) QD+ 100
Cung: P = QS+ 10
(n vca P l ng, n vca Q l kg)
1. Hy tm im cn bng ca thtrng2. Hy tnh co gin ca cung v cu theo gi im cn bng3. Hy tnh thng dsn xut, thng dtiu dng, v thng dton x hi.4. Nu nh nc p t mc gi trn cho sn phm l 50 ng, hy tnh khon tn tht (mt
mt) v ch ca phc li x hi v hy gii thch ti sao li c khon tn tht ny.
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Cu 4 (F0607-PS1-5)
Tht ln (l) v tht g (g) l hai loi tht m gia nh chHoa thng n. Hm tha dng canh chHoa c dng Cobb Douglas U(l, g) = l.g, cn ngn sch chi tiu cho hai loi thc
phm ny ca gia nh chl 120 ng; gi thtrng ca tht ln v thg ln lt lpl= 3ng vpg= 4 ng.
1. Hy vng ngn sch cho gia nh chHoa2. Hy tm im tiu dng ti u (l*, g*) ca gia nh chHoa.3. By gigisdo dch cm g, gi ca tht g gim xung cn 2 ng. n gin ha
phn tch, gisgi ca tht ln khng i. Hy vng ngn sch v tm im tiudng ti u mi (l*1, g
*1) ca gia nh chHoa.
4. Hy phn tch cvmt nh tnh (bng th) v nh lng (bng con s) hiu ng thunhp, thay th, v tng hp l kt quca vic gi tht g gim t4 xung cn 2 ng.
Cu 4 (F0506-PS1-3)Dch cm gia cm bng pht Vit Nam vo nhng thng cui nm 2003 v u nm 2004,sau li ti pht vo nhng thng cui nm 2004. Skin ny tc ng n gi ccc loithc phm khc nhthno giai on y? Anh/chhy dng thcung, cu minh hocho cu trli ca mnh.
Cu 6 (F0506-PS1-4)
Hm scu v hm scung thtrng ca hng ho X c c lng nhsau :
(D) : PD = -(1/2)QD + 110.
(S) : PS
= QS
+ 20
(n vtnh ca QD, QS l ngn tn, n vtnh ca PD, PS l ngn ng/tn)
1. Hy xc nh mc gi v sn lng cn bng ca hng ho X.2. Hy xc nh thng dca ngi tiu dng, ca nh sn xut v tng thng d x hi.3. Hy xc nh co gin ca cu theo gi ti mc gi cn bng. Tmc gi ny, nu cc
nh sn xut thng nht vi nhau gim gi bn xung mt cht th tng chi tiu ca tt cnhng ngi mua dnh cho hng ho ny stng hay gim?
4. By gi, nu chnh phnh thugi trgia tng i vi ngnh X vi mc thusut l10% th sn lng cn bng, gi ngi mua phi tr, gi ngi bn nhn c sau khi npthul bao nhiu?
5. Ai l ngi gnh chu thuv chu bao nhiu tnh trn mi tn sn phm? Tng tin thuchnh phthu c tngnh X l bao nhiu?
6. Anh chhy tnh sthay i trong thng dca ngi tiu dng, thng dca nh snxut v tng thng dx hi.
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Kinh tVi mBi tp 2
Cu 1 (F0203-PS2-1)
Hy vng ng dng vsthch ca nhng c nhn sau y i vi nem v nc cam.
1. C Hoa thch nem nhng ght nc cam. D c bao nhiu nc cam i na, c lunthch c nhiu nem hn.
2. ng c bng quang gia gi ba nem hay gi hai nc cam. Sthch ca ng khng ikhi ng dng nhiu nem hoc nc cam hn.
3. C Hin n mt nem v n cng vi mt nc cam. C thm mt nc cam na, c skhng n thm nem.
4. ng Anh thch nem nhng dng vi nc cam. ng sau bng nu ung nc cam.
Cu 2 (F0203-PS2-2)C Thu c thu nhp hng thng $200, s tin m c phn bgia hai mn hng: tht vkhoai ty.
1.
Gis tht gi $4 mt kilogram v khoai ty gi $2 mt kilogram. Hy vng giihn ngn sch ca c.
2. Cng gisrng hm thodng ca c c cho bi phng trnh U(M,P) = 2M+P. Cnn mua hn hp tht v khoai ty nhthno tha dng ca c ln nht?
3. Siu thca c Thy c mt t khuyn mi c bit. Nu mua 20 kilogram khoai ty(vi gi $2 mt kilogram), c sc min ph 10 kilogram tip theo. Khuyn mi nychp dng vi 20 kilogram u tin c mua. Tt ckhoai ty ngoi s20 kilogram u
tin (khng kskhoai tng thm) vn l $2 mt kilogram. Hy vng gii hn ngnsch ca c.
4. Sbt pht mt thbnh ca khoai ty lm gi khoai tng ln $4 mt kilogram. Siu thkt thc t khuyn mi. ng gii hn ngn sch ca c by gi c hnh dng thno? C nn mua hn hp tht v khoai ty nh th no tha dng ca c lnnht?
Cu 3 (F0203-PS2-3)
Chng minh rng hai hm thodng cho di y to ra cc hm cu tng ti vi hng xv y.
(a) U(x,y) = log(x) + log (y)
(b) U(x,y) = (xy)0,5
Cu 4 (F0506-PS2-1)
Hm nay tri nng bc v Minh rt kht. Sau y l gi trm anh ta gn cho mi chai nc.Gi trca chai nc thnht: 7 ngn ngGi trca chai nc thhai: 5 ngn ngGi trca chai nc thba: 3 ngn ng
Gi trca chai nc tht: 1 ngn ng
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1. Hy lp biu cu ca Minh tnhng thng tin ny v vng cu ca anh ta vncng chai.
2. Nu gi mt chai nc l 4 ngn ng, Minh smua bao nhiu chai? Minh nhn cbao nhiu thng dtiu dng? Hy chra thng dtiu dng ca Minh trong thcaAnh/Ch.
3.
Nu gi gim xung cn 2 ngn ng, lng cu thay i nhthno? Thng dtiudng ca Minh thay i ra sao? Hy chra sthay i ny trong thca Anh/Ch.
Cu 5 (F0506-PS2-2)
Gisvi Minh, tht b l hng ho thng thng, tri li go l mt hng cp thp. Nu githt b gim, tiu dng go ca Minh thay i nhthno? Tiu dng tht b ca Minh thayi nhthno? Anh/Chhy phc tho mt ththch hp minh ho.
Cu 6 (F0506-PS2-3)Gi mt kg to l 20 ngn ng v mt kg cam l10 ngn ng. Mt ngi tiu dng lc umua 10 kg to v 5 kg cam. hu dng bin ca ngi tiu dng khi i vi mt kg tol 3 n vv i vi mt kg cam l 1 n v.
1. Ngi tiu dng ny c t ti u trong tiu dng khng? V sao?2. Nu cu trli ca Anh/Chl khng th ngi tiu dng phi iu chnh slng mua
mi loi tri cy trn y nhthno t ti u?
Cu 7 (F0506-PS2-4)
Mt ca hiu git i ln Thnh phHch Minh mi ngy cn n hng trm k bt git.B chca hiu ni: Ti cho rng hai loi bt git OMO v TIDE u tt nhnhau, v cahiu ti chthng dng hai loi ny. B cn ni: Tuy vy, c nhng lc ti chmua duynht mt trong hai loi bt git ni trn, v cng c nhng lc ti mua chai loi mt cch
ngu nhin, min sau slng ti cn
1. Theo Anh/Ch, hai cu ni ca b Chca hiu c mu thun nhau khng? Gii thchtht ngn gn
a. Da trn kin ca B chca hiu, Anh /Chtrli tip nhng cu di y.
2.
Quan hgia hai mt hng ny trong tiu dng l g?3. Theo Anh/Ch, tlthay thbin (MRS) gia hai mt hng ny c c im g?4. Theo Anh/Chchng ta c thvit c phng trnh ca ng ng ch trong trng
hp ny khng? Nu c, phng trnh l g?5. GisB chca hiu cn 120 kg bt git mi ngy. Khi no th B ta chmua bt git
OMO? Hy vng ng ch, ng ngn sch v chra phi hp ti u.6. By gi, gisB chca hiu cn 140 kg bt git mi ngy. Khi no th B ta mua hai
loi bt git OMO v TIDE mt cch ngu nhin? Hy vng ng ch, ng ngnsch v chra phi hp ti u.
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HB.
Kinh tvi m
Bi tp ng dng 3
L thuyt sn xut v chi ph sn xut
Cu 1 (F0405-PS4-1)
1) Mt doanh nghip k c hp ng tiu thsn phm l Q0. Vi phng n sdngvn v lao ng l k0v l0th doanh nghip thc hin c hp ng ny vi chi ph snxut l C0.Tuy nhin, vi phi hp (k0, l0) th nng sut bin mi ng chi cho vn li caohn nng sut bin mi ng chi cho lao ng (MPK/r > MPL/w). Theo anh/ ch, cphng n s dng vn v lao ng tt hn khng? Nu c, th phng n ny c slng vn v lao ng sdng khc vi phng n trn nhthno? Gii thch.
2) Mi hm sn xut di y thhin hiu sut tng dn, khng i hay gim dn theo quym?
a. Q = 1,2 KLb. Q = 3K + 4L
3) Nu u tthm mt n vvn th sn lng tng thm 8 n v, cn u tthm mtn vlao ng th sn lng chtng thm 3 n v(MPK = 8 > MPL = 3). Vy doanhnghip nn u tthm vn hay lao ng? Hy gii thch.
Cu 2 (F0405-PS4-2)
Hm ssn xut ca mt doanh nghip c dng: Q = F(K,L) = 2,5KL. Doanh nghip va k
c hp ng tiu th31.250 sn phm. n gi ca vn l r = 5 n vtin, n gi calao ng l w = 4 n vtin.1) Hy xc nh phi hp ti u i vi hai yu tsn xut trn y.2) Chi ph thp nht sn xut mc sn lng trn l bao nhiu?3) Nu n gi sn phm hp ng trn l 0,04 n vtin th li nhun doanh nghip t
c l bao nhiu?4) Nu doanh nghip sdng vn v lao ng u tng thm 24% th sn lng stng ln
ln hn 24%, t hn 24%, ng bng 24%, hay khng thxc nh c?5) By giginh rng n gi ca lao ng tng gp i, trong khi n gi vn vn khng
thay i th doanh nghip ssdng vn v lao ng nhthno thc hin hp ng
trn? Tng chi ph lc ny l bao nhiu?
Cu 3 (F0405-PS4-3)
Anh/Chhy in nhng sthch hp vo trng ca bng chi ph di yQ TFC TVC TC AFC AVC AC MC
0
5 20
10 28
15 240
20
10
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HB.
25 4 14
30 540
35
40 20,25
Cu 4 (F0405-PS4-4)
Sn lng sn xut ca cng ty NOVI ph thuc vo slng vn v lao ng a vo sdng v c th hin di dng mt hm sn xut Cobb-Douglas nh sau: Q = K1/3L1/3.Cng ty NOVI thu vn v lao ng trong thtrng cnh tranh vi n gi l r = 18 v w =2.
1) Hin ti cng ty ang sdng vn vi slng l k0= 125 n v.a. Anh/Chhy xc nh cc phng trnh chi ph ngn hn ca cng ty NOVI (bao
gm TFC, TVC, TC, AFC, AVC, AC, v MC)b. Anh/Chhy tnh tng chi ph ca cng ty NOVI khi sn xut cc mc sn lng
l 40, 52 v 60 n v.2) Trong di hn, cng ty c ththay i ng thi cvn v lao ng c tht chi ph
thp nht mi mc sn lng cho trc.a. Anh/Ch hy xc nh cc phng trnh chi ph di hn ca cng ty (bao gm
LRTC, LRAC, v LRMC)
b. Anh/Chhy tnh tng chi ph ca cng ty NOVI khi sn xut cc mc sn lngging nhphn trn l 40, 52 v 60 n v
c. Anh/Chhy so snh tng chi ph ngn hn v di hn v a ra nhn xt.
Cu 5 (F0405-PS4-5)
Mt doanh nghip cnh tranh hon ho c tng chi ph ngn hn phthuc vo sn lng snxut v c thhin bi phng trnh : TC = q3-10q2+100q +1000.1) Anh/chhy vit cc phng trnh chi ph trung bnh , chi ph bin i trung bnh v chi
ph bin ca doanh nghip ny.
2) Bng Excel, Anh/chhy vba chtiu trn ln cng mt thv nu nhn xt vmi quan hgia chng.
3) Doanh nghip ssn xut mc sn lng no v li nhun t c l bao nhiunu gi thtrng ca sn phm l 292 n vtin.
4) Nu gi sn phm hxung chcn 132 n vtin/sp th doanh nghip c sn xutkhng? Ti sao? Nu sn xut th doanh nghip ssn xut mc sn lng no? Linhun t c l bao nhiu? nhng mc gi no ca thtrng th doanh nghipsng ca?
5) Nu nh ph ca doanh nghip ny khng phi l 1000 m l 2500 n v tin thdoanh nghip c sn xut khng? Ti sao? Nu sn xut th doanh nghip ssn xut mc sn lng no? Li nhun t c l bao nhiu? mc nh ph no thdoanh nghip sng ca?
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HB.
6) Nu chnh phbuc doanh nghip phi ng thutheo sn lng l 49 n vtin/spth doanh nghip ssn xut mc sn lng no? Li nhun sau thul bao nhiu?
7) Anh/Chhy vng cung ngn hn ca doanh nghip.
Cu 6 (F0506-PS5-4)
Hm ssn xut ca doanh nghip X c dng: Q(K,L) = 2k1/2l1/2. n gi ca vn l r = 2 vn gi ca lao ng l w = 6 . Hin ti doanh nghip X u tvn vi slng k = k0= 100n v.
a) Anh/Chhy vit hm tng chi ph v chi ph bin ngn hn ca doanh nghip theobin sn lng (Q).
b) Nu gi thtrng ca sn phm l P= 9 v doanh nghip X hot ng trong ngnhcnh tranh hon ho th doanh nghip s sn xut bao nhiu n v sn phm? Linhun t c l bao nhiu?
c) Trong di hn, doanh nghip c thiu chnh c s lng vn v lao ng. Nudoanh nghip X vn sn xut vi mc sn lng nhcu b th ssdng bao nhiuvn v bao nhiu lao ng? Li nhun t c bao nhiu? Ln hay nhhn so vicu b trn y?
Cu 7 (F0607-PS3-1)
Cng ty Cao su Sao Vng chuyn sn xut lp xe p. Cng ty hin thi c mt nh my sn
xut, cng nghsn xut c c trng bi hm sn xut KLQ = . Trong Q l lng
lp xe c sn xut tnh bng nghn lp/ thng, K l slng my mc v L l slng laong c sdng cho qu trnh sn xut trong thng. Gisrng chi ph thu 1 my l 100
la/thng v lng cng nhn l 400 la/thng.1) Trong di hn cng ty c ththay i quy m sn xut. Hy lm mt vi tnh ton v tr
li cc cu hi sau y:a) Cng nghsn xut ny c trng bi nng sut tng, gim hay khng i theo quy
m?
b) Hy vng ng lng ng vi mc sn lng Q = 6 nghn lp/thng. Hy tnh vnu ngha ca tlthay thkthut bin ca lao ng i vi vn (MRTSLK) ti A(K=12, L=3)?
c) xy dng chin lc di hn, phng khoch ca cng ty c lng slng lpxe m cng ty cung cp bnh qun thng trong di hn sl 6 nghn lp. Giscngty c thla chn 4 quy m cho vic sn xut lp xe ng vi K = 6, K=12, K=24 vK =36 . Hy xc nh quy m p ng lng cu trn thtrng l 6 nghn chiclp/thng trong tng lai vi chi ph thp nht?
d) La chn trn c g thay i nu nh my c t ti mt vng khc vi githitlng cng nhn chcn l 100 la/thng v bqua chi ph vn chuyn?
e) Hy vng tng chi ph, chi ph trung bnh v chi ph bin di hn ng vi githitban u vi nhiu quy m khc nhau?
2) Trong ngn hn, vi lng my mc thit b u ttrong di hn (K=12) l khng ththay i c. Lc ny tng sn lng, cng ty chc thtng thm lao ng. Hy lmmt vi tnh ton v trli cu hi sau y:
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HB.
a) Hy vng tng sn phm, sn phm bin v sn phm trung bnh ca lao ng.Nhn xt vmi quan hgia sn phm bin v sn phm trung bnh ca lao ng.
b) Hy vng tng chi ph, chi ph trung bnh, chi ph bin trong ngn hn? Gisrng lc ny lng lp xe m cng ty cung cp l 24 nghn lp mi thng thay v l 6nghn lp nh dkin. Tng chi ph, chi ph trung bnh v chi ph bin trong ngn
hn ng vi mc sn lng ny l bao nhiu? ng chi ph bin ct ng chi phtrung bnh ti mc sn lng no?c) Nu gisrng gi ca mi lp xe trn thtrng l 1 la. Hy cho bit lng lp
xe m cng ty sn sng cung cp l bao nhiu ti a ha li nhun? Li nhuncng ty lc ny l bao nhiu?
3) Do lng lp xe cung cp trn thtrng tng ngoi dkin, cng ty mua thm mt nhmy sn xut lp xe khc vi cng cng nghv quy m (K) ln gp bn ln nh myhin thi.a) Nu cng ty mun ti thiu ha chi ph sn xut trong ngn hn, mc sn lng nn
phn bnhthno gia hai nh my?
b) Vi githit cng ty phn bsn lng gia hai nh my ti thiu ha chi ph snxut, hy xc nh tng chi ph, chi ph trung bnh v chi ph bin trong ngn hn cacng ty ng vi cc mc sn lng l 24 nghn lp/thng?
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Kinh tvi mBi tp ng dng 4
Phn tch thtrng khng c i th(cnh tranh hon ho v c quyn thun tu)
Cu 1 (F0203-PS4- 2)Gisnhng ng cu v cung ca thtrng cnh tranh hon ho vnh dnh cho gianh n nht c cho tun tl:P = 200 0,1Qd
P = 20 + 0,2Qs
Ngoi ra, ATC ca mt nh cung cp tiu biu trong ngnh ny c cho bi:
ATC = 360/q + 20 + 10qCho bit gi v xut lng hin ti ca ngnh, xut lng v li nhun ca mi cng ty, v slng cng ty.
Cu 2 (F0203-PS1P2- 1)
Mt nh c quyn i din trc ng cu P = 11 Q, trong P c tnh bng latrn mi n vv Q tnh bng ngn n v. Nh c quyn c chi ph trung bnh khng il $6 mt n v.(a)Hy vcc ng doanh thu trung bnh v doanh thu bin, v cc ng chi ph trungbnh v chi ph bin. Mc gi v sn lng ti-a-ha-li-nhun ca nh c quyn l baonhiu, v li nhun t c l bao nhiu? Hy tnh mc quyn lc c quyn ca cng tybng cch dng chsLerner.(b)Mt cquan qun l ca chnh pht gi trn l $7 mt n v. Slng sc snxut l bao nhiu, v li nhun ca cng ty l bao nhiu? Mc quyn lc c quyn sra
sao?
Cu 3 (F0506-PS5-4)
Hm ssn xut ca doanh nghip X c dng: Q(K,L) = 2k1/2l1/2. n gi ca vn l r = 2 vn gi ca lao ng l w = 6 . Hin ti doanh nghip X u tvn vi slng k = k0= 100
n v.a) Anh/Chhy vit hm tng chi ph v chi ph bin ngn hn ca doanh nghip theo bin
sn lng (Q).b) Nu gi thtrng ca sn phm l P= 9 v doanh nghip X hot ng trong ngnh cnh
tranh hon ho th doanh nghip ssn xut bao nhiu n vsn phm? Li nhun tc l bao nhiu?
c) Trong di hn, doanh nghip c thiu chnh cslng vn v lao ng. Nu doanhnghip X vn sn xut vi mc sn lng nhcu b th ssdng bao nhiu vn vbao nhiu lao ng? Li nhun t c bao nhiu? Ln hay nhhn so vi cu b trny?
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Cu 4 (F0506-PS6 - Li nhun c quyn)1. Mt cng ty c quyn i mt vi mt ng cu P = 300 4Q, bin ph trung bnh
khng i a =100, v chi ph cnh = 50. Tnh sn lng v gi ti a ho li nhun.Gii thch
2. Mt cng ty c quyn phi xem xt n ng cu m cng ty c gp phi ti aho li nhun cho cng ty. C ta nghe ni rng sc mnh c quyn cng ln thng cu cng khng co dn. C ta hi kin ca anh/chxem cng ty ca c ta cnn sn xut mt sn lng theo vi mt gi trn ng cu c co dn E = - 0. 5khng? Gii thch kin ca anh/ch.
Cu 5 (F0405-PS7&8- 2)
Mt doanh nghip c quyn hon ton i din trc hm scu thtrng l : P = -Q +200. Chi ph bin ca cng ty tng theo mc sn lng sn xut: MC = Q+20 . Tng chi phcnh l 1.400 n vtin.
a. Anh/chhy xc nh mc sn lng, mc gi bn, li nhun ca doanh nghip, thng dca ngi tiu dng, thng dca nh sn xut, tn tht x hi do c quyn v tng tinthuthu c ca chnh pht doanh nghip ny, trong ba trng hp di y.
i) Doanh nghip khng bng thu.ii) Doanh nghip bng thun vl t= 30 n vtin/ n vsn lngiii) Doanh nghip bng thugp T= 2.000 n vtin.
b. Nu chnh php dng thuthu nhp doanh nghip i vi doanh nghip ny v muntng tin thuthu c phi t bng vi tng tin thu trong trng hp doanh nghipbng thun vth mc thusut l bao nhiu?
c. Anh/Chhy lp bng tm tt theo mu di y.
Chtiu Khng c thu Thun v Thugp ThuTNDNP
Q
CS
PS
T
DWL
d. Tbng tm tt trn y, Anh/chhy c nhng nhn xt g?
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Kinh tVi m
Gi li gii bi tp thc hnh 1
Cu 1 (F0203-PS1-1)
1. a.Nu mt loi phn bn mi lm tngnng sut ca la m, iu slmgim chi ph sn xut bnh m. Vic nyslm cho ng cung tng ln (dchchuyn sang phi). Ti gi cn bng c,giy sc cung thng d. Nhvy,gi thtrng sgim. Khi gi thtrnggim, slng cu tng (dch chuyndc theo ng cu). Ti im cn bngmi, gi sthp hn v slng c
trao i scao hn.
Lng
CungCau
P
Q Q*
P*
Tang cung
Gia
b.Bv bnh m l hai thbsung chonhau trong tiu dng. Nhvy, gi btng, do bnh ca b, slm cho cubnh m gim (dch chuyn sang tri).Ti gi cn bng c, giy sc cungthng d. Nhvy, gi thtrng sgim. Khi gi thtrng gim, slngcung gim (dch chuyn dc theo ungcung). Ti im cn bng mi, cgi vlng c trao i sthp hn. Lng
Gia CungCau
P
Q
Giam cau
P*
Q*
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c.Bnh m v phl hai ththay thchonhau trong tiu dng. Nhvy, gim cui vi phslm tng cu i vi bnhm (dch chuyn sang phi). Ti gi cnbng c, giy sc cu thng d.
Nhvy, gi thtrng stng. Khi githtrng tng, slng cung tng (dchchuyn dc theo ng cung). Ti imcn bng mi, cgi v lng c traoi scao hn. Ln
Gia CunCau
P*
Q*
Tan cau
P
Q
d. Cng nhn bnh m l mt nhplng trong sn xut bnh m. Gi nhplng tng s lm gim cung (dch
chuyn sang tri). Ti gi cn bng c,giy s c cu thng d. Nh vy,gi th trng s tng. Khi gi thtrng tng, s lng cu gim (dchchuyn dc theo ng cu). Ti imcn bng mi, gi s cao hn v slng c trao i sthp hn. Ln
Gia
CunCau
P
Q
Giam cun
P*
Q*
e.Nu chnh phng mua ht bnhm thng d vi gi cao hn gi hinhu trn th trng 10%, iu nytng ng vi vic tng s lngngi tiu dng. Kt qul, cu i vibnh m kp tht stng (dch chuyn).Thc vy, ng cu tr thnh nmngang ti gi P* = (1,1)P phn nh vicchnh phsmua mt slng v hnti mc gi ny. Gi cn bng ln tiP*. Khi gi tng, s lng cung tng
(dch chuyn dc theo ng cung).Nhvy, clng cn bng v gi utng.
Ln
Gia CunCau
P
Q
Chnh phu mua
Q*
P*
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f.Gi trn sgigi bnh m di gitr cn bng. Khi gi gim, s lngcung gim v s lng cu tng (dchchuyn dc theo c hai ng). Slng c trao i s gii hn s
lng cung. iu ny gy ra cu thngd, nhng gi khng thtng loi bcu thng d. Kt qu l mt c chphn phi khc v d nh mt thtrng chen c thpht trin. Ln
Gia
CunCau
P
Q
Gia tran
Tran
Q Qds
P*
Cu 2 (F0203-PS1-2)
2. a.Cn bng xy ra ti mc gi m slng cu bng slng cung. Trong trnghp ny, Qd = Qs = Q. tm Q, cho ng cu v ng cung bng nhau.
100 - 0,1Q = 10 + 0,1Q =>
90 = 0,2Q =>
Q = 450
Tng cu hoc ung cung, khi Q = 450 th P = 55. Tng chi tiu l PxQ = 450 x 55 =24.750
b.Nu bo him chi tr75% tng chi ph y t, th c nhn chtr (P), vi P l gi tng. Biv cu ca ti c da trn stin ti tr, ng cu c thc vit li l (P) = 100 0,1Qd=> P = 400 - 0,4Qd. Ni cch khc, bo him lm xoay ng cu, khin cdc ln
tung gc tng theo tha s4. Trong trng hp ny, gi tiu biu cho gi tng phi tr, kcbo him. C nhn tr25% gi ny. (Cch khc, anh chc thcoi bo him ca chnh phl mt khon trgi 75% cho ngi sn xut. Trong trng hp ny, bo him lm gimng cung theo tha s4, ti P = 2,5 + 0,025Qs. Trong trng hp ny, gi cn bng l gido c nhn tr. Gi tng, kcbo him, sln hn gi ny 4 ln.) tm P v Q cn bngmi, ta t cung v cu mi bng nhau.
400 - 0,4Q = 10 + 0,1Q =>
390 = 0,5Q =>
Q = 780.
Tng cu mi hoc tng cung, P = 88, v ngi tiu dng tr25% hay 22. Tng chitiu vo y tl 88 x 780 = 68.640, trong khi c nhn tr22 x 780 = 17.160. Hy lu rnglng v tng chi ph tng, nhng chi ph i vi ngi tiu dng (cgi ln tng chi tiu) gim so vi trong phn (a) (mc d thuca hc thtng trang tri chi tiu ca chnhph). (Nu anh chdch chuyn ng cung, cn bng sxy ra ni
100 0,1Q = 2,5 + 0,025Q => Q = 780 v P = 22. Kt quny tng ng vi cu trlibng cch dch chuyn ng cu.)
Cu 3 (F0607-PS1-5)
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Hm scu v hm scung ca mt sn phm c cho di y:
Cu: P = (-1/2) QD+ 100
Cung: P = QS+ 10
(n vca P l ng, n vca Q l kg)
a)
Hy tm im cn bng ca thtrng
tm im cn bng ca thtrng, chng ta sdng mt tnh cht quan trng can, l lng cung v cu im cn bng phi bng nhau. Nhvy, tm gi cv sn lng ti im cn bng ca thtrng, ta chcn gii h2 phng trnh bcnht vi 2 n s. Kt qul, ti im cn bng QE= 60 v PE= 70.
b)
Hy tnh co gin ca cung v cu theo gi im cn bng
Vit li phng trnh hm cung v cu nhsau:
QD= 200 2P
Q
S
= P 10Xut pht tmc gi cn bng (PE= 70), gisgi tng 1% (tc tng 0,7), th Q
Dsgim 1,4 (tc 2,33%). Nhvy, ti im cn bng, co gin ca cu so vi gi l -2,33.
Tng tnhvy, xut pht tmc gi cn bng (PE= 70), gisgi tng 1% (tctng 0,7), th QSstng 0,7 (tc 1,17%). Nhvy, ti im cn bng, co gin cacu so vi gi l +1,17.
c) Hy tnh thng dsn xut, thng dtiu dng, v thng dton x hi.
Thng dsn xut = PS = 60 (70 10)/2 = 1.800
Thng dtiu dng = CS = 60 (100 70)/2 = 900Thng dton x hi = SS = PS + CS = 2.700
Lu l tlphn chia thng dx hi cho hai khu vc sn xut v tiu dng phthuc dng thc ca hm cung v cu (v do vy vo co gin ca cu v cung sovi gi.)
d) Nu nh nc p t mc gi trn cho sn phm l 50 ng, hy tnh khon tn tht(mt mt) v ch ca phc li x hi v hy gii thch ti sao li c khon tn tht ny.
mc gi trn ny, lng cung QS l 40 v lng cu QD l 60, nh vy c mtlng thiu ht l 100. Cha cn thc hin cc php tnh, chng ta cng c ththyrng thng dca ckhu vc sn xut (PS) v tiu dng (CS) v do ca ton xhi (SS) sbgim.
Khon tn tht phc li v ch = din tch hnh tam gic nh(c mt nh l im cnbng E) trn hnh v= (80 50) (60 40)/2 = 300.
Cu 4 (F0607-PS1-6)
Tht ln (l) v tht g (g) l hai loi tht m gia nh chHoa thng n. Hm tha dng canh chHoa c dng Cobb Douglas U(l, g) = l.g, cn ngn sch chi tiu cho hai loi thcphm ny ca gia nh chl 120 ng; gi thtrng ca tht ln v thg ln lt lpl= 3ng vpg= 4 ng.
a)
Hy vng ngn sch cho gia nh chHoa
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Dng thvng ngn sch
b) Hy tm im tiu dng ti u (l*, g*) ca gia nh chHoa.
p dng cng thc tm im tiu dng ti u trong trng hp hm tha dng c dngCobb Douglas trong = = 1, I = 120, pl= 3, pg= 4 ta c l
*= 20 v g*= 15. Mc
tha dng ca gia nh Hoa ti im tiu dng ti u ny l U
*
= 300.c) By gigisdo dch cm g, gi ca tht g gim xung cn 2 ng. n gin ha
phn tch, gisgi ca tht ln khng i. Hy vng ngn sch v tm im tiudng ti u mi (l*1, g
*1) ca gia nh chHoa.
Tng tnhcu (b), p dng cng thc tm im tiu dng ti u trong trng hp hmtha dng c dng Cobb Douglas trong = = 1, I = 120, pl= 3, pg= 2 ta c l
*=20 v g*= 30. Mc tha dng ca nh chHoa ti im tiu dng ti u ny l U1
*=600.
(Lu rng chng ta ginh (mt cch khng tht) rng hm tha dng ca nh chHoa khng i ngay ckhi c dch cm gia cm.)
d)
Hy phn tch cvmt nh tnh (bng th) v nh lng (bng con s) hiu ng thunhp, thay th, v tng hp l kt quca vic gi tht g gim t4 xung cn 2 ng.
xc nh hiu ng thay th, ta phi tm im tiu dng ti u mi (l*s, g*s) ti hai
iu kin sau c tha mn: (i) Mc tha dng bng U*= 300; v (ii) ng ngn schmi (vi mc gipg= 2) tip xc vi ng ng ch (hay ng dng) c. Tc l ta phigii hphng trnh sau:
300s s
l l
g g
l g
MU p
MU p
=
=
hay:
300s s
gs
s l
l g
pl
g p
= =
Gii hphng trnh ny ta tm c l*s= g*
s= 21,21 (sau khi lm trn). Nhvy hiu
ng thay thl: ( , ) ( , ) hay (20;15) (14,14; 21, 21)s s
l g l g
. Nhvy, khi gi
tht g gim t4 xung cn 2 ng, so vi trc tht g trnn rhn mt cch tngi so vi tht ln, v v vy t mc tha dng nhcmc gi minh chHoatng tiu dng thg 15 n vln 21,21; ng thi gim mc tiu dng thln t20 lnxung 14,14.
Kt hp vi kt quphn (c), hiu ng thu nhpl:
( , ) ( , ) hay (14,14; 21, 21) (20, 30)s s
l g l g
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Cu 4 (F0506-PS1-3)
Dch cm gia cm bng pht Vit Nam vo nhng thng cui nm 2003 v u nm 2004,sau li ti pht vo nhng thng cui nm 2004. Skin ny stc ng lm tng cu ccloi thc phm khc (ng cu dch chuyn sang phi) v kt qul gi tht heo, tht b, gic . tng ln trong thi gian ny.
Cu 6 (F0506-PS1-4)
Hm scu v hm scung thtrng ca hng ho X c c lng nhsau:
(D): PD = -(1/2)QD + 110.
(S) : PS = QS+ 20
(n vtnh ca QD, QS l ngn tn, n vtnh ca PD, PS l ngn ng/tn)
a) Thtrng cn bng khi QS= QD= Q0v PS= P
D= P0
=> Q0+ 20 = -(1/2)Q0 + 110
3/2 Q0 = 90 => Q0= 60 ngn tnv P0 = 80 ngn ng/tn
Q
P
S
D
D1
E0P0
E1
Q1Q0
P1
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b)
Thng dca ngi tiu dng l din tch tam gic AP0E0CS = *60*(110-80) = 900 triu ng
Thng dca nh sn xut l din tch tam gic BP0E0
PS = *60*(80 - 20) = 1.800 triu ng
Tng thng d x hi = CS + PS = 2.700 triu ng
c)
co gin ca cu theo gi ti mc gi cn bng.Ep= (dQ/dP) * (P/Q) = -2* (80/60) = -8/3
Ep < -1 : Cu co gin nhiu, tng chi tiu nghch bin vi gi nn tmc gi ny,nu cc nh sn xut thng nht vi nhau gim gi bn xung mt cht th tng chitiu ca tt cnhng ngi mua dnh cho hng ho ny stng.
d)
Nu c thuVAT, thtrng cn bng khi: QS= QD= Q1v PS+ thu= PD
PS+ 10% P
S= P
D hay 1,1 P
S = P
D
1,1(Q1+ 20) = -(1/2)Q1 + 110
1,6 Q1 = 88 => Q1= 55 ngn tn
Mc gi ngi mua phi trl PD1= -(1/2)55 + 110 = 82,5ngn ng/tn
Mc gi ngi bn nhn c sau khi np thu l PS1 = 55 + 20 = 75 ngnng/tn
e)
Ngi tiu dng chu 2,5ngn ng tin thu(82,5 80) v nh sn xut chu 5ngn ng tin thu(80 - 75) tnh trn mi tn sn phm.Tng tin thuchnh phthu c tngnh X l: 7,5* 55 = 412,5triu ng
f)
Thng dca ngi tiu dng gimCS = - * (60+55)*(82,5-80) = - 143,75triu ng
S
Q (ngn tn)
P (ngn ng/tn)
D
P0= 80
PD
1= 82,5
PS
1= 75
E0
110 A
20 B
F
C
Q0= 60Q1= 55
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( thhin trn thl din tch hnh thang PD1CE0P0)
Thng dca nh sn xut gim
PS = - * (60+55)*( 80 - 75) = - 287,5triu ng
( thhin trn thl din tch hnh thang PS
1FE0P0)
Khon thuthu c ca chnh phl
G = 7,5 * 55 = 412,5triu ng
( thhin trn thl din tch hnh chnht PD1CFPS
1)
Tng thng dx hi gim (phn gim ny thng gi l tn tht v ch hay mtmt v ch)
NW = CS + PS + G = - 18,75triu ng
( thhin trn thl din tch tam gic CFE0)
Ht
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Kinh tVi mLi gii gi bi tp 2
Cu 1 (F0203-PS2-1)
1. a)Nc cam l hng ha xu v nem l hng ha tt. C Hoa chchp nhn nhiu nccam hn i ly nhiu nem hn. Biu c hnh dng nhbn di (tutheo t nem haync cam trn trn trc Y.)
b)Hai mn ny l hng thay thhon ho trong tiu dng. ng ng dng l nhng ngsong song vi dc 2/3 (gisnc cam trn trc Y). Hy tham kho bi ging xemng ng dng i vi hng thay thhon ho.
c)Hai mn ny l hng bsung hon ho. thc trnh by di y.
d)Ta li c nem l hng ha tt v nc cam l hng ha xu. Biu cng tng tnhtrong phn a.
Nem
Nccam
U
U2
Nem
Nccam
U
U2
2
3 6
4
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Cu 2 (F0203-PS2-2)
2. a)Nu c Thy khng mua khoai ty c c thmua 50 kilogram tht; v nu khngmua tht, c c thmua 100 kilogram khoai ty. y, ng gii hn ngn sch c biudin bng mt ng thng gia hai chn la ny.
b)tha dng ca c Thy bng 100 nu c mua 50 kilogram tht m khng muakhoai ty hoc 100 kilogram khoai ty m khng mua tht. V hm tha dng l ng thng,ng ng dng s l nhng ng thng. V vy, ng ng dng ca c vi U=100trng vi ng gii hn ngn sch. Cc ng ng dng khc song song vi ng cU=100. Do , c mun tiu dng nhng gi thtrng sao cho tlthay thbin ca c bngvi tsgi. Do ng ng dng va ng gii hn ngn sch c cng dc nhnhau, cThy s ti a ha tha dng bng cch tiu dng bt k gi th trng no nm trnng ngn sch ca c. Ta khng thxc nh chnh xc hn c Thy sthc schn gith trng no, bi v tt c mi gi nm trn ng ngn sch u lm c tha mn nhnhau.
c)Vi t khuyn mi c bit, c mt ng gy khc trong ng gii hn ngnsch ca c Thy.
M
50
0 100 P
ng ngan sach
20 30 110
d)Vi gi khoai ty l $4, c Thu c thmua hoc 50 kilogram tht hoc 50kilogram khoai ty hoc hn hp gia hai loi. ng ng dng ca c khng thay i. Cti a ha tha dng vi U=100 khi mua 50 kilogram tht m khng mua khoai ty. y lgii php gc (ngha l, tlthay thbin khng thc sbng tsgi).
50 10 P
U=10
ngngn sch
I
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Cu 3 (F0203-PS2-3)
3.Sdng phng php hsnhn Lagrangian tm nhng hm cu thc t. Ta gii hmthhai v anh ch bit hm thnht c dng Cobb-Douglas v c gii trong biging. Cho I = thu nhp, Px l gi hng x v Pyl gi hng y. ng thc Lagrangian l
0.5 0.5, ( )X YL X Y I P X P Y=
Rt gn: Px/Py = MUx/MUy = Y/X (v MUx =0 .5 0 .50 .5
XM U X Y= v
0 .5 0 .50 .5YM U X Y
= )
V vy Px X = Py Y
Xt ng gii hn ngn sch I = 2PxX hay I = PYYng cu ca x l X = I/2Px v ng cu ca y l Y = I/2Py
y cng l kt quanh chtm c nu p dng trnh ttng tvo hm tha dng kia.
Cu 4 (F0506-PS2-1)
a. Biu cu ca Minh vnc ng chai.P (ngn ng/chai) Q (chai)
7 1
5 2
3 3
1 4
Q (chai)1 432
3
1
5
7
P (n n n /chai)
(D)
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ng cu ca Minh vnc ng chai l mt ng hnh bc thang nu slng lmt bin ri rc. Nu ginh slng l mt bin lin tc th c ththhin ng cul mt ng thng.
b. Nu gi mt chai nc l 4 ngn ng, Minh smua 2 chai nc. Minh nhn cthng dtiu dng l 4 ngn ng = (7-4)+ (5-4)
c. Nu gi gim xung cn 2 ngn ng, lng cu l 3 chai. Thng dtiu dng caMinh lc ny l 9 ngn ng = (7-2)+(5-2)+(3-2), tng mt lng l: CS = 9- 4 = 5ngn ng.
Cu 5 (F0506-PS2-2)
Gisvi Minh, tht b l hng ho thng thng, tri li go l mt hng cp thp. Nu githt b gim, tiu dng go ca Minh gim, tiu dng tht b ca Minh stng.
thminh hothch hp c thhin di y.
Q (chai)1 432
3
P = 4
1
5
7CS
P (ngn ng/chai)
P =2
CS
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Cu 6 (F0506-PS2-3)
Gi mt kg to l 20 ngn ng v mt kg cam l 10 ngn ng. Mt ngi tiu dng lc umua 10 kg to v 5 kg cam. hu dng bin ca ngi tiu dng khi i vi mt kg tol 3 n vv i vi mt kg cam l 1 n v.
a. Ngi tiu dng ny khng t ti u trong tiu dng v MUA/PA> MUO/PO(3/20>1/10)
b. t ti u trong tiu dng, ngi ny phi iu chnh slng mua mi loi tri
cy trn y theo hng tngmua to v gimmua cam tho mn iu kin tiu: MUA/PA= MUO/PO.Gii thch thm:cch mua ban u, hu dng bin mi ng mua to ln hn hu dngbin mi ng mua cam. Theo quy lut hu dng bin gim dn, phi chuyn bt tin tmua cam sang mua to cho n khi hu dng bin mi ng chi cho cc mt hng khcnhau phi bng nhau.
Cu 7 (F0506-PS2-4)
Mt ca hiu git i ln Thnh phHch Minh mi ngy cn n hng trm k bt git.B chca hiu ni: Ti cho rng hai loi bt git OMO v TIDE u tt nhnhau, v cahiu ti chthng dng hai loi ny. B cn ni: Tuy vy, c nhng lc ti chmua duy
nht mt trong hai loi bt git ni trn, v cng c nhng lc ti mua chai loi mt cchngu nhin, min sau slng ti cn
a. Hai cu ni ca b Chca hiu hon ton khng mu thun nhau. Li ch mi kgbt git bt khiu g l nhnhau nhng mua nh thno l cn tuthuc vogi tng i ca chng.MUO= MUThon ton cha thng tin a ra quyt nh mua hng m phiso snh MUO/POvi MUT/PTa ra quyt nh.
Nu MUO/PO> MUT/PTdo POPTth chmua TIDE.
Tht b
Go
I/PR
I/PB1 I/PB2
U1U2
B1B2
A1
A2A/
B2B1 B/
R/
R2
R1
Gi tht b gim:
Tc ng thay thlm tng mua tht b tB1ln B
/, lm gim mua go tR1xung
R
/
. Tc ng thu nhp lm tng mua tht b t
Tc ng thu nhp lm gimmua go tR/xung R2vgo l hng cp thp.
B
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Nu MUO/PO= MUT/PTdo PO=PTth mua hai loi mt cch ngu nhin
b. Quan hgia hai mt hng ny trong tiu dng l thay thhon ho.c. Tlthay thbin (MRS) gia hai mt hng ny l mt hng s. Nu mi bao bt
git cc loi c ng gi vi trng lng l 1kg, 5kg, hay 10kg nhnhau thMRS = 1. Nu OMO ng gi 5kg, TIDE ng gi 10kg th MRSOT=1/2 hocMRSTO= 2
d. Hai mt hng ny l thay thhon ho nn ng ng ch l mt ng thngdc xung nhng ngn sch v chng ta c thvit c phng trnh cang ng ch trong trng hp ny.U(O,T) = aO + bT. Trong O l sgi OMO v T l sgi TIDE.
(a = b nu trng lng mi gi ca hai loi l nhnhau, a = 2b nu trng lngmi gi OMO gp i mi gi TIDE).
e. Gi s B ch ca hiu cn 120 kg bt git mi ngy. B ta ch mua bt gitOMO khi POB1,do vy A l im tiu dng tiu (gii php gc)
A (im tiu dng ti u)
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OMO (kg)
TIDE (kg)
U120
140
140
B2
Khi n gi hai loi bt git bngnhau th ng ngn sch mua 140kg bt git l B2, trng vi ngng ch. Mi im trn ng B2
u l phi hp ti u nn B tamua hai loi bt git OMO v TIDEmt cch ngu nhin.
B1
B3
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Kinh tvi mGi li gii bi tp ng dng 3
L thuyt sn xut v chi ph sn xut
Cu 1 (F0405-PS4-1)
1) Vi phng n kt hp vn v lao ng (k0, l0) c MPK/r > MPL/w v chi ph l C0 th sc phng n khc tt hn l (k1, l1), vi MPK/r = MPL/w v chi ph l C1< C0. Phng nny c slng vn nhiu hn v lao ng t hn so vi phng n trn.phng n u, nng sut bin mi ng u tvo vn cao hn nng sut bin mi ngu tvo lao ng. Qu trnh chuyn dn tin t thu lao ng sang thu vn slm chotng sn phm tng ln. Nu u tsn xut vi tng chi ph nhcl C0th sn lng stng ln (Q1> Q0). Do vy, nu sn xut vi sn lng c, ng theo hp ng l Q0th chiph sgim (C1< C0).
2) a.Hm sn xut Q = 1,2 KL c hiu sut tng dn theo quy m. Nu sdng K v Ltng n ln th sn lng l Qn = 1,2. nK.nL = n2. 1,2KL = n2Q > n Q.
b. Hm sn xut Q = 3K + 4L c hiu sut khng i theo quy m. Nu sdng K vL tng n ln th sn lng l Qn = 3.nK + 4.nL = n(3K + 4L) = nQ3) Nu u tthm mt n vvn th sn lng tng thm 8 n v, cn u tthm mtn v lao ng th sn lng ch tng thm 3 n v (MPK = 8 > MPL = 3). Vy doanhnghip nn u t thm vn hay lao ng? Cha thtrli c v cn thiu thng tin vgicca hai yu tsn xut ny.
Cu 2 (F0405-PS4-2)
Q = F(K,L) = 2,5KL = 31.250 (1) r = 5 , w = 4.
1) Phi hp ti u ca hai yu tsn xut vi bt kmc sn lng cho trc no phithoiu kin: MPK/r = MPL/w
2,5L/5 = 2,5K/4 => L = (5/4)K (2)
Th(2) vo (1): 2,5K(5/4)K = 31.250 => K2 = 31.250* 4/12,5 = 10.000Suy ra K = 100 v L = 125
2) Chi ph thp nht sn xut mc sn lng trn l:TC = rK + wL = 5*100 +4* 125 = 1.000 n vtin
3) Tng doanh thu = 31.250* 0,04 = 1.250 n vtinTng li nhun = 1.250 - 1.000 = 250 n vtin
4) Nu doanh nghip sdng vn v lao ng u tng thm 24% th sn lng stngln ln hn 24% (+>1).
5) By giginh rng n gi ca lao ng tng gp i, trong khi n gi vn vnkhng thay i th doanh nghip ssdng vn v lao ng nhthno thc hinhp ng trn? Tng chi ph lc ny l bao nhiu?
MPK/r = MPL/w
2,5L/5 = 2,5K/8 => L = (5/8)K (3)
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Th(3) vo (1) : 2,5K(5/8)K = 31.250 => K2 = 31.250* 8/12,5 = 20.000Suy ra K = 141,42 v L = 88,40
Tng chi ph lc ny l:TC = rK + wL = 5*141,42 + 8* 88,40 = 1.414,30 n vtin
Cu 3 (F0405-PS4-3)
Q TFC TVC TC AFC AVC AC MC
0 100 0 100
5 100 100 200 20 20 40
10 100 180 280 10 18 28
15 100 240 340 6,67 16 22,67
20 100 290 390 5 14,5 19,5
25 100 350 450 4 14 18
30 100 440 540 3,33 14,67 18
35 100 560 660 2,86 16 18,86
40 100 710 810 2,5 17,75 20,25
Cu 4 (F0405-PS4-4)
Q = K1/3
L1/3
. r = 18 , w = 2.
1) Hin ti cng ty ang sdng vn vi slng l k0= 125 n v.Q = K
1/3L
1/3 => L
1/3 = Q/ K
1/3
L = Q3/ K = Q
3/125
a. Cc phng trnh chi ph ngn hn ca cng ty NOVI
TC = rk0+ wL = 18*125 + 2* Q
3
/125 = 2.250 + 2 Q
3
/125TFC = 2.250
TVC = 2 Q3/125
AFC = TFC/ Q = 2.250/Q
AVC = TVC/Q = 2 Q2/125
AC = TC/Q = 2 Q2/125 + 2.250/Q
MC = dTC/dQ = 6 Q2/125
b. Tng chi ph ca cng ty NOVI khi sn xut cc mc sn lng 40, 52 v 60 nv.
Q = 40, TC = 2.250 + 2.(40)3/125 = 3.274
Q = 52, TC = 2.250 + 2.(52)3
/125 = 4.499,73Q = 60, TC = 2.250 + 2.(60)
3/125 = 5.706
2) Trong di hn, cng ty c ththay i ng thi cvn v lao ng c tht chi phthp nht mi mc sn lng cho trc.
a. Xc nh cc phng trnh chi ph di hn ca cng tyQ = K
1/3L
1/3 => L
1/3 = Q/ K
1/3
L = Q3/ K (1)
Tng t, K = Q3/ L (2)Phi hp ti u ca hai yu tsn xut vi bt kmc sn lng cho trc no phi thoiu kin: MPK/r = MPL/w
L1/3
/ 54K2/3
= K1/3
/ 6L2/3
10
24
20
16
30
12
12
18
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L = 9K hay K = (1/9)L (3)
T(1) v (3) => L = Q3/ [(1/9)L] hay => L2 = 9Q3 => L = 3Q3/2 (4)T(2) v (3) => K = Q3/ 9K hay K2 = Q3/ 9 => K = (1/3) Q3/2 (5)Th(4) v (5) vo phng trnh tng chi ph: LRTC = rK + wL ta c c hm tng chi phdi hn: LRTC = 18. (1/3) Q3/2 + 2. 3Q3/2 = 12 Q3/2
LRAC = 12 Q1/2 v LRMC = 18 Q1/2
b. Tnh tng chi ph ca cng ty NOVI khi sn xut cc mc sn lng ging nhphn trn l 40, 52 v 60 n v.
LRTC = 12 Q3/2
Q = 40, LRTC = 12. (40)3/2
= 3.035,79
Q = 52, LRTC = 12. (52)3/2
= 4.499,73
Q = 60, LRTC = 12. (60)3/2 = 5.577,1
c. So snh tng chi ph ngn hn v di hn v a ra nhn xt.Chc mc sn lng duy nht Q = 52 th tng chi ph ngn hn v di hn bng nhau(SRTC = LRTC = 4.499,73). nhng mc sn lng khc th chi ph ngn hn luncao chi ph di hn. Trong ngn hn, p ng sthay i sn lng, doanh nghipchc thiu chnh s lng lao ng sdng cn lng vn sdng th cnhnn iu kin ti u: MPK/r = MPL/w khng c thomn v do vy chi ph scao hn.
Cu 5 (F0405-PS4-5)
1) Phng trnh chi ph trung bnh, chi ph bin i trung bnh v chi ph bin ca doanhnghip:
AC = TC/q = q2
10q + 100 + 1000/q.AVC = TVC/q = q
210q + 100.
MC = dTC/dq = 3q220q + 100.
2) Vthba chtiu AC, AVC v MC (xem bn di)Mi quan hgia MC v AC (AVC).- Khi MC < AC (AVC) th AC (AVC) gim dn.- Khi MC > AC (AVC) th AC (AVC) tng dn.- MC = AC (AVC) khi AC (AVC) min. (Trn th, MC ct AC v AVC ti
im cc tiu ca mi ng).3) t li nhun ti a, doanh nghip ssn xut ti mc sn lng thoiu kin:
MC = P. 3q2 20q + 100 = 292 hay 3q
2 20q -192 = 0.
Gii phng trnh bc hai ny ta c q = 12. (Xem thtrn)Tng doanh thu ca doanh nghip:
TR = P* q = 292 * 12 = 3.504 n vtin.Tng chi ph ca doanh nghip:
TC = 12310*122+100*12 + 1000 = 2.488 n vtin.Tng li nhun ca doanh nghip:
= TR - TC = 3.504 - 2.488 = 1016 n vtin.
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Nu gi thtrng ca sn phm hxung chcn 132 n vtin/sp th doanh nghip bl(P =132 < AC min =200), nhng doanh nghip vn tip tc sn xut v phn lst hn lng ca doanh nghip (P =132 > AVCmin = 75).ti thiu ho l, doanh nghip ssn xut ti mc sn lng c MC = P.
3q2 20q + 100 = 132 hay 3q2 20q -32 = 0.
Gii phng trnh bc hai ny ta c q= 8. (Xem thtrn)Tng doanh thu ca doanh nghip:
TR = P* q = 132 * 8 = 1.056 n vtin.Tng chi ph ca doanh nghip:
TC = 8310*8
2+100*8 + 1000 = 1.672 n vtin.
Mc lca doanh nghip:
= TR - TC = 1.056 - 1.672 = - 616 n vtin.
Vi mi mc gi th trng thp hn mc chi ph bin i trung bnh ti thiu (P 3q
220q + 149 = 292 hay 3q
220q 143 = 0.
Gii phng trnh bc hai ny ta c q = 11.Tng doanh thu ca doanh nghip:
TR = P* q = 292 * 11 = 3.212 n vtin.Tng chi ph gm cthuca doanh nghip:
TC = 11310*11
2+149*11 + 1000 = 2.760 n vtin.
Tng li nhun sau thuca doanh nghip:
0
50
100
150
200
250
300
350
400
0 2 4 6 8 10 12 14 16 18 20
$/q
q
ng bieu dien AC, AVC, MC.
MC
AC
AVC
292
132
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= TR - TC = 3.212 - 2.760 = 452 n vtin.6) ng cung ngn hn ca doanh nghip l mt phn ng MC nm trn AVC
hnh vtrn y.
Cu 6 (F0506-PS5-4)
Q(K,L) = 2k1/2l1/2 (1)
r = 2 ; w = 6 ; k = k0= 100.
a) T(1) => l = Q2/400TC = r*k0+ w*l = 2*100 + 6* (Q
2/400) = 200 + (3/200) Q
2
MC = dTC/dQ = (3/100)Q
b) t li nhun ti a, doanh nghip cnh tranh hon ho ssn xut vi mc snlng tho iu kin: MC = P(3/100)Q = 9 => Q = 300
TR = P*Q = 2.700; TC = 200 + (3/200) *3002= 1.550
= TR TC = 1.150
c) Trong di hn, doanh nghip c thiu chnh cslng vn v lao ng, phi hphai yu tt ti u khi: MPk/r = MPl/w
l1/2
/2k1/2
= k1/2
/6l1/2
=> k = 3l (2)
Th(2) vo (1): Q = 2*(3l)1/2*l1/2= 300 => l = 150/ 31/2 = 86,6k = 259,8
TC = 2* 259,8 + 6* 86,6 = 1.039,2
= TR TC = 2.700 - 1.039,2 = 1.660,8
Cu 7 (F0607-PS3-1)Cng ty Cao su Sao Vng chuyn sn xut lp xe p. Cng ty hin thi c mt nh my sn
xut, cng nghsn xut c c trng bi hm sn xut KLQ= . Trong Q l lng
lp xe c sn xut tnh bng nghn lp/ thng, K l slng my mc v L l slng laong c sdng cho qu trnh sn xut trong thng. Gisrng chi ph thu 1 my l 100 la/thng v lng cng nhn l 400 la/thng.
1) Trong di hn cng ty c ththay i quy m sn xut. Hy lm mt vi tnh tan v trli cc cu hi sau y:a) Cng nghsn xut ny c trng bi nng sut tng, gim hay khng i theo quy
m?Vi m >1, ( , ) ( )( )F mK mL mK mL m KL mQ= = =
Cng nghsn xut ny c trng bi nng sut khng i theo quy m.b) Hy vng ng lng ng vi mc sn lng Q = 6 nghn lp/thng. Hy tnh v
nu ngha ca tlthay thkthut bin ca lao ng i vi vn (MRTSLK) ti A(K=12, L=3)?
Phng trnh ng ng lng ng vi Q = 6 (nghn lp)
6),( == KLLKF
Nng sut bin ca lao ng1 1
2 21
2L
MP K L
=
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6
Nng sut bin ca vn1 1
2 21
2K
MP K L
=
Tlthay thbin ca vn cho lao ng:
LK
LK
LK
MPMPMRTS
K
LLK ===
2
1
2
1
2
1
2
1
2
12
1
Tlthay thbin ca vn cho lao ng ti A (K=12, L=3)
43
12===
L
KMRTSLK
duy tr mc sn lng nhc, hng c ththay th1 n vlao ng cho bn nvvn v nng sut bin ca lao ng gp 4 ln nng sut bin ca vn.
c) xy dng chin lc di hn, phng khoch ca cng ty c lng slng lpxe m cng ty cung cp bnh qun thng trong di hn sl 6 nghn lp. Giscngty c thla chn 4 quy m cho vic sn xut lp xe ng vi K = 6, K=12, K= 24, vK = 36 . Hy xc nh quy m p ng lng cu trn thtrng l 6 nghn chiclp/thng trong tng lai vi chi ph thp nht? Hy so snh chi ph sn xut 6nghn lp xe/thng vi 4 quy m trn? ti thiu ho chi ph sn xut mc sn lng 6 nghn chic lp/thng, trong dihn hng la chn kt hp u vo thomn iu kin:
(1)r
wMRTSLK=
(2) 6),( == KLLKF
Vi w =400 v r =100. T(1) ta c
(3)100
400=
L
Khoc K= 4L. Th(3) vo (2)
(4) 64)4(2 == LLL hoc L =3
Th(4) vo (3) ta nhn c K =12Vi quy m (K=12, L=3) hng sn xut p ng c quy m thtrng dkin l 6nghn lp/thng vi chi ph sn xut l thp nht.sn xut 6 nghn chic lp vi cng nghcho trc c c trng bi hm sn
xut KLQ= , la chn phi hp gia K v L thomn 6== KLQ
Tng chi ph sn xut 6 nghn chic lp ng vi cc quy m:K 6 12 24 36
L 6 3 1.5 1
SRTC 3.000 2.400 3.000 4.000
Chi tit tnh chi ph sn xut:Khi K = 6 th SRTC = 100*6 +400*6 = 3000.
Khi K = 12 th SRTC = 100*12 +400*3 = 2.400.
Khi K = 24 th SRTC = 100*24 +400*1,5 = 3.000.
Khi K = 36 th SRTC = 100*36 +400*1 = 4.000.
Chi ph thp nht ng vi quy m l K =12 v chi ph cao nht ng vi quy m
K = 36.
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d) La chn trn c g thay i nu nh my c t ti mt vng khc vi githitlng cng nhn chcn l 100 la/thng v bqua chi ph vn chuyn?Vi githit gi nhn cng w =100. Tng chi ph sn xut 6 nghn chic lp/thngng vi cc quy m:
K 6 12 24 36
L 6 3 1.5 1SRTC 1.200 1.500 2.550 3.700
Chi tit tnh chi ph sn xut:Khi K = 6 th LRTC = 100*6 +100*6 = 1200.
Khi K = 12 th LRTC = 100*12 +100*3 = 1.500.
Khi K = 24 th LRTC = 100*24 +100*1,5 = 2.550.
Khi K = 36 th LRTC = 100*36 +100*1 = 3.700.
Trong trng hp ny chi ph thp nht sn xut mc sn lng Q = 6 nghn chiclp ng vi quy m l K = 6 v chi ph cao nht ng vi quy m K = 36.
e) Hy vng tng chi ph , chi ph trung bnh v chi ph bin di hn ng vi githit ban u ?Tng chi ph di hn l tng chi ph thp nht ng vi cc mc sn lng khc nhaukhi quy m thay i. xc nh tng chi ph, trc ht ta phi xc nh cu cc yutsn xut ng vi cc mc sn lng. Cu cc yu tsn xut phi tha mn cciu kin sau y:
(1) 4==L
Khay
r
wMRTSLK
(2) QKLLKF ==),(
T(1) v (2) ta tnh c K = 2Q v L = Q/2.y chnh l cu cc yu tsn xutHm tng chi ph di hn ng vi githit ban uLRTC = wL+rK = 100*2Q + 400*Q/2 = 400Q
Hm chi ph trung bnh v chi ph bin
400400
===Q
Q
Q
LRTCLRAC
400)(
==dQ
LRTCdLMC
ng LRTC, LRAC v LRMC
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8
1 2 3 4
400
800
1200
1600
2000
x
y
LRTC = 400Q
LRAC = LRMC = 400
2) Trong ngn hn, vi lng my mc thit b u ttrong di hn (K=12) l khng th
thay i c. Lc ny tng sn lng, cng ty chc thtng thm lao ng. Hy lmmt vi tnh ton v trli cu hi sau y:a) Hy vng tng sn phm, sn phm bin v sn phm trung bnh ca lao ng.
Nhn xt vmi quan hgia sn phm bin v sn phm trung bnh ca lao ng.
ng vi K =12, lc ny hm sn xut trong ngn hn c dng LQ *12=
Sn phm bin ca lao ngLdL
dQMPL
3==
Sn phm trung bnh ca lao ngLL
QAPL
12==
th:ng TP, MPLv APL
ng Thnh phHCh Minh ng MPLv APL
2 4 6 8 10 12 14 16 18
2
4
6
8
10
12
14
x
y
Q = 121/2L1/2Q
L
2 4 6 8 10 12 14 16 18
0.5
1
1.5
2
2.5
3
x
y
APL
MPL
Trong bi tp ny, sn phm trung bnh ca lao ng lun gim dn v sn phm bin calao ng lun nhhn sn phm trung bnh ca lao ng.
b) Hy vng tng chi ph, chi ph trung bnh, chi ph bin trong ngn hn? Gi srng lc ny lng lp xe m cng ty cung cp l 24 nghn lp mi thng thay v l 6nghn lp nh dkin. Tng chi ph, chi ph trung bnh v chi ph bin trong ngnhn ng vi mc sn lng ny l bao nhiu? ng chi ph bin ct ng chi ph
trung bnh ti mc sn lng no?
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Chi ph bin i ca hng trong ngn hnTVC =w* L = 400*(Q
2/12)
Chi ph cnh trong ngn hnTFC = r*K = 100*12 = 1.200
Tng chi ph sn xut trong ngn hnTC = TVC +TFC = 1.200 + (100/3)Q2
Tng chi ph sn xut ng vi Q = 24 (nghn lp).TC = TVC +TFC = 1.200 + (100/3)24
2= 20.400
Chi ph trung bnh trong ngn hnATC = TC/Q = 1.200/Q + (100/3)*Q
Chi ph trung bnh trong ngn hn ng vi Q = 24 (nghn lp)ATC = TC/Q = 20.400/24 = 850 la/nghn chic
Chi ph bin trong ngn hn
3
200)( Q
dQ
TCdMC ==
Chi ph bin ngn hn ng vi Q = 24 (nghn lp).MC= 1.600 la/nghn chic
Giao im ca ATC v MC:Gii phng trnh ATC = MC 1.200/Q + 100Q/3 = 200Q/3, ta c nghim Q= 6. Vy ng chi ph bin ct ng chi ph trung bnh ti mc sn lng 6 (nghnchic), ti y ATC t gi trcc tiu.
thca ng STC, SATC, v SMC
ng STC ng SAC v ng SMC
2 4 6 8 10 12 14 16 18 20 22 24
x
y
SRTC = 1200 + (100/3)Q2
1200 FC = 1200
VC = (100/3)Q2
2 4 6 8 10 12
100
200
300
400
500
600
700
x
y
SAC = 1200/Q + (100/3)Q
SMC = (200/3)Q
Q
SAC
SMC
c) Nu gisrng gi ca mi lp xe trn thtrng l 1 la. Hy cho bit lng lp
xe m cng ty sn sng cung cp l bao nhiu ti a ha li nhun? Li nhuncng ty lc ny l bao nhiu?
Vi githit thtrng lp xe p l thtrng cnh tranh.ti a ho li nhun, hng sla chn mc sn lng sao cho P =MC. Vi gi lp xe trn th trng l 1 la/chichay 1.000 la/nghn chic, mc sn lng cung ng ti a li nhun l
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10
P= MC hay3
200000.1
Q=
Q = 12.000/800 = 15 nghn chic lp/thngTng doanh thu : R = P*Q = 1000*15=15.000 laTng chi ph : TC = TVC +TFC = 1.200 + 400*(152/12) = 8.700 laLi nhun := R TC = 15.000 la 7.500 la = 6.300 la
3) Do lng lp xe cung cp trn th trng tng ngoi dkin, cng ty mun mua thmmt nh my sn xut lp xe khc vi cng cng nghv quy m (K) ln gp bn ln nhmy hin thi.a) Nu cng ty mun ti thiu ha chi ph sn xut trong ngn hn, mc sn lng nn
phn bnhthno gia hai nh my?ti thiu ho chi ph sn xut, mc sn lng phn bgia hai nh my sao cho chi phbin gia hai nh my bng nhau. V rng gi nhn cng khng i l 100, nn chi phbin ca hai nh my bng nhau khi sn phm bin ca hai nh my bng nhauTrc ht, ta xc nh sn phm bin (nng sut bin) ca hai nh my. Sn phm bin calao ng i vi nh my c, gi l nh my s1, c xc nh l
1
1
3
LdL
dQMPL ==
V nh my mi mua c quy m gp 4 ln nh my trc nn hm sn xut trong ngn
hn c dng nhsau LQ 48=
Sn phm bin ca lao ng i vi nh my mi , gi l nh my s2
2
2
12
LdL
dQMPL ==
ti thiu ho chi ph sn xut, nh my phi phn bsn lng gia hai nh my sao chochi ph bin ca hai nh my bng nhau, v v chc L nh hng n chi ph bin nn iuny cng c ngha l sn phm bin ca lao ng ca hai nh my bng nhau. Khi sn phmbin ca hai nh my bng nhau th L2= 4 L1. V y l hm ng nht bc 1 nn khi ta tngu vo ln 4 ln th u ra cng tng 4 ln. Q2= 4 Q1V Q2+Q1= 24 nn Q1= 4,8 v Q2= 19,2.ti thiu ho nh my sn xut, nh my s1sn xut 4,8 nghn chic lp v nh my mi mua sn xut 19,2 nghn chic lp.
b) Vi githit cng ty phn bsn lng gia hai nh my ti thiu ha chi ph snxut, hy xc nh tng chi ph, chi ph trung bnh trong ngn ng vi cc mc snlng l 24 nghn lp/thng?
Tng chi ph sn xut 4,8 nghn chic lp ca nh my cTC1= TVC1+TFC1= 1.200 + 400*(4,8
2/12) = 1.968 la
Tng chi ph sn xut 19,2 nghn chic lp ca nh my miTVC2= wL = 400*(Q
2/48)
TFC2= 100*48 = 4.800
TC2= TVC2+TFC2= 100*48 + 400*(Q2/48) = 4.800 + 400*(Q
2/48)
= 4.800 + 400*(19,22/48) = 7.872
Tng chi ph trong ngn hn ng vi mc sn lng 24 nghn chic lp khi hai nh my thamgia vo sn xut. TC =TC1+TC2= 1.968 + 7.872 = 9.840 la ATC = TC/Q = 9.840/24 =410 la/ nghn chic.
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1
Kinh tvi mBi gii gi bi tp ng dng 4
Phn tch thtrng khng c i th(cnh tranh hon ho v c quyn thun tu)
Cu 1 (F0203-PS4- 2)Gi v xut lng ca ngnh xut hin ti ni giao nhau gia cc ng cung v cu cangnh.
200 0,1Q = 20 + 0,2Q => 180 = 0,3Q => Q = 600, P = 140.
m thnh vi ca cng ty, ta cn c ng MC ca cng ty. MC l tlthay i (tc o
hm) ca TC theo q. TC = ATC*q. Do :TC = 360 + 20q + 10q
2=> MC = 20 + 20q
Cc cng ty ring lshot ng ti P = MC trong cnh tranh hon ho. Do , vi P v MCca cng ty cho trc:140 = 20 + 20q => 120 = 20q => q = 6.
Li nhun l = TR - TC => = (P - ATC)q. , khi q = 6, ATC = 140.Vy
= (140 - 140)*6 => = 0.
Cui cng, nu Q = 600 v q = 6, phi c 100 cng ty trong ngnh.
Cu 2 (F0203-PS1P2- 1)(a)Bi v cu, hay doanh thu trung bnh (AR), c thc din tbng P = 11 Q nn tabit hm doanh thu bin l MR = 11 2Q. Ta cng bit rng nu chi ph trung bnh (AC)khng i th chi ph bin khng i v bng chi ph trung bnh. Nhvy, MC = 6. iu nyc thhin bng thdi y.
tm mc xut lng ti-a-ha-li-nhun, ta hy t doanh thu bin bng chi ph bin:MR = MC hay 11 2Q = 6. Vy Q = 2,5, tc xut lng l 2.500 n v, v P = 11 2,5 =$8,50. Li nhun bng tng doanh thu trtng chi ph.
= TR - TC = PQ - AC(Q) = (11 - Q)Q - 6Q = (11 2,5)2,5 - 6(2,5) = 6,25
Li nhun ca nh c quyn l 6,25 nhn $1.000. Mc quyn lc c quyn l
294,05,8
65,8=
=
P
MCP.
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2
11
6
06 11
AC=MC
D=ARMR
Li nhuan
$/n v
Q(1.000)
(b)Ti gi trn $7 th lng cu l 4.000 n v: Q = (11 P)1.000 bng (11 7)1.000. ycng l mc xut lng ti-a-ha-li-nhun ca nh c quyn, bi v mc ny MR =MC. Li nhun, bng tng doanh thu tr tng ph, l (11Q - Q2- 6Q)1000 = (11*4 - 16 6*4).1000 = $4.000. Giy, mc quyn lc c quyn l
143,07
67=
=
P
MCP
Cu 3 (F0506-PS5-4)Q(K,L) = 2k1/2l1/2 (1)
r = 2 ; w = 6 ; k = k0= 100.
a) T(1) => l = Q2/400TC = r*k0+ w*l = 2*100 + 6* (Q
2/400) = 200 + (3/200) Q2
MC = TC/Q = (3/100)Q
b) t li nhun ti a, doanh nghip cnh tranh hon ho ssn xut vi mc sn lngtho iu kin: MC = P(3/100)Q = 9 => Q = 300
TR = P*Q = 2.700; TC = 200 + (3/200) *3002= 1.550
= TR TC = 1.150
c) Trong di hn, doanh nghip c thiu chnh cslng vn v lao ng, phi hp haiyu tt ti u khi: MPk/r = MPl/w
l1/2
/2k1/2
= k1/2
/6l1/2
=> k = 3l (2)
Th(2) vo (1): Q = 2*(3l)1/2*l1/2= 300 => l = 150/ 31/2 = 86,6 v k = 259,8
TC = 2* 259,8 + 6* 86,6 = 1.039,2
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= TR TC = 2.700 - 1.039,2 = 1.660,8
Cu 4 (F0506-PS6 - Li nhun c quyn)1. iu kin u tin ti a ho li nhun: MR = MC
TR = PQ = 300Q 4Q2nn MR = 300 8Q
MC = 100 m MR = MC => 300- 8Q =100
Kt quQ = 25; P = 200
Li nhun= doanh thu- chi ph = 300(25) 4(25)2 - 100 (25) 50 =2450
2. Mun ti a ho li nhun cn phi c: MR = MC
TP = MC/ ( 1 + 1/Ed) (xem sch gio khoa) Ta c:
P = MR/ ( 1 + 1/Ed) => MR = P( 1 + 1/Ed)
Nu cu km co dn: Ed > -1 => ( 1 + 1/Ed) Q = 60.
Mc gi bn ca doanh nghip: P= -60 + 200 = 140.
Tng doanh thu ca doanh nghip:
TR = P* Q = 140 * 60 = 8.400 n vtin.
Tng chi ph ca doanh nghip:
TC = (1/2) 602+ 20 * 60 + 1.400 = 4.400 n vtin.
Tng li nhun ca doanh nghip:
= TR TC = 8.400 - 4.400 = 4.000 n vtin.
Thng dca ngi tiu dng l din tch tam gic AP*B.
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CS = (200 140)* 60/2 = 1.800 n vtin.
Thng dca nh sn xut l din tch tgic P*BCF.
PS = (120 + 60)* 60/2 = 5.400 n vtin.
Tn tht v ch do c quyn l din tch tam gic BCE th2.
DWL= (140 -80)*(90 -60)/2 = 900n vtin.
ii) Doanh nghip bng thun vl t = 30 n vtin.t li nhun ti a, doanh nghip sn xut ti mc sn lng tho:
MC + t = MR hay Q + 20 + 30 = -2Q +200 => 3Q = 150 => Q = 50.
Mc gi bn ca doanh nghip: P= -50 + 200 = 150.
Tng doanh thu ca doanh nghip: TR= P* Q = 150 * 50 = 7.500 n vtin.
Tng chi ph ca doanh nghip (gm cthu):
TC = (1/2) 502+ 50 * 50 + 1.400 = 5.150 n vtin.
Tng li nhun ca doanh nghip:
= TR - TC = 7.500 - 5.150 = 2.350 n vtin.
Thng dca ngi tiu dng l din tch tam gic AP/ H.
CS = (200 150)* 50/2 = 1.250 n vtin.
Thng dca nh sn xut l din tch tgic P/HJG.
PS = (100 + 50)* 50/2 = 3.750 n vtin.
o th
0
20
40
60
80
100
120
140
160
180
200
0 10 20 30 40 50 60 70 80 90 100 110 120Q
P
MR
D
MC
MC+t
A
B
F
C
E
P*
Q*
G
Q/
P/H
I
J
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Tn tht v ch do c quyn l din tch tam gic HIE th2.
DWL= (150 -70)*(90 -50)/2 = 1.600n vtin.
Tng tin thuthu c ca chnh ph(din tch tgic FGJI)
T = t Q/
= 30* 50 = 1.500n vtin.
iii) Doanh nghip bng thu gp T = 2000 n vtin.Cch nh thuny khng lm thay i MC hoc MR ca doanh nghip (thu c thcxem l mt khon nh ph tng thm), do vy hu ht cc ch tiu yu cu tnh ton uging phn i). Chc mt im khc bit nhl li nhun ca doanh nghip gim i 2000v trthnh khon thuthu c ca chnh ph.
b. Li nhun trc thuca doanh nghip l = 4.000n vtin (phn i). Tng tinthubng vi trng hp thu thun vl T = 1.500 (phn ii). Vy thusut thuthunhp doanh nghip l tc = (1.500/4.000)*100 = 37,5%.
c. Bng tm ttChtiu Khng c thu Thun v Thugp ThuTNDNP 140 150 140 140
Q 60 50 60 60
4.000 2.350 2.000 2.500
CS 1.800 1.250 1.800 1.800
PS 5.400 3.750 5.400 5.400
T 0 1.500 2.000 1.500
DWL 900 1.600 900 900
d. Tbng tm tt trn y, chng ta c tha ta nhng nhn xt sau: Thun v (hay mt loi thugin thu no khc nh thudoanh thu, thugi tr gia
tng) slm gim sn lng, tng gi bn, gim thng dtiu dng. iu ny c ngha lnh sn xut c thchuyn mt phn thu ln pha trc cho ngi tiu dng gnh chu.Ngoi ra, tn tht x hi cng stng thm.
Thugp hay thuthu nhp doanh nghip (thu trc thu ni chung) skhnglm thayi sn lng, gi bn v thng dtiu dng. iu ny c ngha l nh sn xut chu thumt mnh. Ngoi ra, thng dsn xut v tn tht x hi cng khng thay i. V thuc xem nh l khon chuyn giao mt phn li nhun ca doanh nghip sang ngnsch quc gia.