Nguygn Kieu Linh T?p chi KHOA HOC & CONG NOH$ 99(10:85-89
VAI SUY NGHl VE M p T BAI TOAN TOI VV TRONG R^
Nguyen Kieu Linh Tru&ng Dgi hpc Khoa hpc - IDI I Thai Nguyen TOM TAT
Bai bdo nSy dl cdp tdi bdi todn t6i uu, thudng g$p trong iing dyng thyc tiln: Tim trin duong iron da cho mot diem sao cho ldng khodng cdch tir dd ldi hai diem cho tru&c a ngodi dudng iron Id nhd nhdt? Bdi t o ^ d^t ra tuy don gidn nhung vi$c tim Idi gidi cho nd bang gidi tich hay hlnh hpc thuc khong dl. No Id mot md rflng tryc tilp ciia bdi toin quen bilt sau ddy, nhung don gidn hem vd da c6 ldi giai dep: 77m trin du&ng thdng cho Iruoc mgt diem sao cho long khodng cdch lir nd tai hai diem dS cho a ngodi dudng thdng Id nho nhatl Trong bai vilt ndy chiing toi trinh bAy hai cdch tiep can bai toin dya tr€n kiln thiic toi uu vd cdc tinh chdt hinh hgc cua ellipse,
Tii'kh6a: Bdi lodn loi uu. cdch tiep cgn gidi lich. cdch tiep cgn hinh hgc...
NQI DUNG BAI TOAN VA Y NGHIA THUC TE
Xet bdi toan tdi uu sau day trong mat phang:
Bai toan. Tun tren ducmg Iron da cho mot diem sao cho long khodng cdch tir do ldi hai diem cho trudc a ngodi duung iron la nho nhdtl Bai toan nay la mdt md rdng ciia bai loan quen biet sau day: 77m tren dirdng thdng cho tric&c mot diem sao cho long khodng cdch tir no tai hai diem da cho & ngodi du&ng thdng id nho nhdtl
Cd the gidi thich y nghTa ciia bdi toan dat ra theo mdt sd each nhu sau:
a. Gia sir dien ludi dugc truyen dgc theo tuyin dudng H din mdt ngd ba trung tam, sau khi cap dien chieu sang vd sinh hogt cho vdng trdn trung tam, ngudn dien can dugc chuyen tilp tdi hai tuyen dudng tilp theo sau nga ba, bit dau d A va B (xem Hinh 1). Vdn de dat ra Id cdn tim mdt vj tri D tren vdng trdn trung tam dl tir dd dgt hai dudng cdp ngam chgy thdng tdi A vd B sao cho tong khoang each tir vi tri dugc chgn (tren vdng trdn trung tam) tdi A va B la nhd nhdt (tire Id tdn it cdng siic, vgt Ii6u vd di?n nang nhat)?
b. Gia sii cd mgt hd nude hinh trdn (tam I, bdn kinh r). Cd hai canh ddng ma A va B Id nguon cap nude cho mdi canh ddng. Can ddt mdt trgm bom d ven hd va cac dudng muong (hogc dng) dan nude thing tu do tdi A va B sao cho dd tdn dudng dan nhat?
Bdi todn ddt ra tuy don gidn nhung ham chiia npi dung todn hpc sau sde, bdi vi viec tim Idi gidi cho nd bang gidi tich hay hinh hpc thuc khdng de. Trong bai viet nay chiing tdi xin neu mdt vai suy nghT ve bdi loan dgt ra, mong dupe trao ddi vol ban dgc va ddng nghiep cd quan tdm tdi bai toan.
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Hinh 1. Minh hoa bdi lodn:
Tim D cdp dien cho AvdB TRU6NG HOP DE GIAI
Cdch gidi bai toan tiiy thugc vdo vj tri tuong ddi cua dudng trdn va hai diem da cho (ky hi?u A va B). Sau day la mpt sd trudng hpp digiai khac nhau,
a. Doan thdng AB tilp xiic vdi dudng trdn (tilp diem nam trong AB). Khi do tilp diem chinh la dilm cdn tim vd khoang each ngdn nhat bdng dp dai dogn AB (xem Hlnh 2a).
Luu y: neu dudng thdng qua A, B tiep xiic vdi dudng trdn d ngoai dogn AB thi tiep diem khdng la Idi giai'(xem each gidi cho trudng hgp tdng qudt).
85
Nguyin Kieu Linh Tgp chl KHOA HQC & C 6 N G N G H $ 99(11): 85-89
c) diem R
Hinh 2. Diem i b. Dogn thdng AB cat dudng trdn tai hai diem (ndm trong AB). Khi dd mdi giao diem deu la diem can lim va khoang cdch ngdn nhal bdng dp ddi doan AB (xem Hinh 2b). Luu y: neu dudng thang di qua A va B cdt dudng trdn tai hai diem (nam ngodi dogn AB) thi cdc giao diem cung khdng chdc chan Id Idi giai (xem cdch gidi cho trudng hgp tong quat).
c. Tam dudng trdn nam ngodi dogn AB, nhung d tren dudng thang di qua A va B. Khi do mgt trong hai giao diem ciia dudng thdng vdi dudng trdn (diem ndm gan A hodc B hon diem kia) se Id dilm cdn tim va khodng cdch ngan nhdt bang dg dai dogn AB cgng vdi khodng cdch tir giao diem tdi A hogc B (.\em Hinh 2c).
d. Mgt trudng hgp rieng dl gidi nO'a nhu sau:
Tam 1 ciia dudng trdn ndm tren dudng trung tryc ciia dogn thdng AB vdi O la diem giiia dogn AB. Gia sir 10 cat dudn^ trdn tgi diem K. Khi dd K chinh Id diem can tim, Cd the tinh tdng khodng cdch tii K tdi A \a B nhu sau: Nlu ddt h - 10, b = h - r, c = jAB|/2 va a
= v b + C thi long khoang each nhd nhat tir K tdi A va B bang 2a (xem Hinh 2d),
TRUONG HOP TONG QUAT
Neu khdng gdp mgt trong bdn trudng hgp ke tren thi ta can each tiep can khdc.
Bdi todn ddt ra dugc phdt bieu bdng Idi ma khdng diing din bat cii- mgt cdng thiic loan hgc nao. Day Ihyc chat Id mgt bai lodn tdi uu (tim eye lieu ham khodng each theo diem chgy tren dudng trdn). De cd the vgn dung dugc cdng cy toi uu, tnrdc het can ddt lai (hay mo hinh hoa) bdi todn bdng ngdn ngir toan hpc. Ciing mgt bai todn cd the cd nhieu cdch ddl md hinh loan hgc khac nhau vd each giai don gidn hay khdng phy thupc rdt nhieu vao mire dp thdnh cdng ciia \ i$c md hinh hda do.
Trong bai viet nay ch6ng toi md hinh hda bai todn nhu sau.
Ky hi?u dp dai dogn AB da cho Id 2c (c > 0), Lay dudng thdng di qua A vd B ldm true hoanh, dudnng thdng vudng gdc vdi tryc hoanh vd di qua trung diem dogn thang AB Idm true tung. Nhu vay goc tpa dp la dilm giifa doan thdng AB, ky higu dd Id dilm O vdi tpa d6 0
= (0', 0). Gia sii A ndm phia trai O cd tga do A = (- c. 0) vd B ndm phia phdi O cd tpa do B
= (c,0).
NguySn Kieu Linh Tgp chl KHOA HQC & CONG NGHE 99(11): 85- Gia sir tam I cua dudng tron da cho cd tpa dp
(p, q) vdi p, q e K, va ban kinh ciia dudng tron la r (r > 0). Ky hieu tga dp ciia diem D ndm tren dudng trdn da cho Id D = (x, y) (xem Hinh 3).
Khi dd tong khodng cdch tCr D tdi A vd B Id
f(x,y)= V(x + c)^ +y^ +
vd ta di den bai toan: 77m arc lieu hdm f(x, y) vdi dieu ki^n (x - p)" + (y - q)" = r".
Hinh 3. He true tpa do Day Id bdi toan tdi uu hai bien vdi mpt rang budc ddng thirc phi tuyen, Cd thl thay f(x, y) la mdt hdm ldi (trong gidi tich ldi ta da biet hdm khoang each lir mdt dilm trong W tdi mdt tap Idi Id mgt ham Idi). Hon niia, rang bupc ddng thiic da cho cd the thay thl bdng rdng budc bdt ddng thiic (x - p)^ + (y - q)^ < r^
ma khdng Idm thay ddi nghiem cua bdi todn.
Tir dd md hinh bai loan cd dang mdt bdi todn qui hogch loi (xem [ 1 ]):
min{fl:x,y)=V(x + c ) ^ + y ^ + V ( x ~ c ) ^ + y ^ : ( x - p ) ' + ( y - q ) ' < r = } . Do ham muc tieu f lien tyc vd tap rang huge compac, khac rdng (do r > 0) nen bai toan phai cd nghiem cue tiiu (Djnh ly Weierstrass). Vdn de Id xet xem diem eye tieu cd tinh chdt gi vd Idm thl nao tim dugc diem cue tiiu dd (bdng giai tich hogc hinh hgc)?
Tir luat phdn xg anh sang suy ra diem eye tiiu D cd tinh chdt: doan AD vd BD tgo vdi tiep tuyin dudng trdn tgi D hai gdc bdng nhau (gdc tdi = gdc phdn xg).
Cach tiep can giai tich
Bdng gidi tich ta cd the dung phuong phdp nhan tir Lagrange (xem [1], §8.2, tr, 229 - 240). Cdch Idm nhu sau: Dua vdo nhan tii X vd xet hdm Lagrange:
L(x, y, X) = ^{x + cf +y^ +
V ( x - c ) ^ + y ^ + ; i ( ( x - p ) ' + (y-q)--r-) Lay dgo hdm rieng cua L theo x, y, X vd cho bang 0 ta dugc h^ ba phucmg trinh ciia ba an sd (x, y vd X):
fit ^ 2 M \ - P ) = 0.
^2/.(\-<i) = i),
= (x-pJ- + (\-([)--r = 0
Rat tiec la ta khdng the giai tryc tiep he phuong trinh nay de tim ra cac diem dirng, tir do xac djnh diem eye tieu. Ta chi cd the giai gan diing bang so he nay.
Cach tiep can hinh hgc
Ta nhic Igi dinh nghTa va cdc tinh chat cua ellipse trong mat phdng.
Theo dinh nghTa, ellipse Id quT ti'ch tat cd nhiing diem cd tdng khoang each tii' no tdi hai diem cd dinh da cho (chdng hgn, A vd B) bang hang sd E > 0 cho trudc. Diem A va B ggi la cac liiu diem, do ddi doan thang AB = 2c gpi Id lieu cu ciia ellipse, Mdi ellipse cd mdt iruc Idn, dp dai ky hieu Id 2a vd indt iruc nho, dg dai ky hieu Id 2b (vdi a > b > 0), vudng gdc vdi nhau tai trung diem dogn thang AB. Ta cd mdi lien he: a^ = b' + c" va C = 2a (xem Hinh 4). Neu ve he tryc tpa dp vudng gdc vdi gdc tpa dg O tai diem giii'a hai tieu dilm ciia ellipse vd tryc hoanh song song vdi tryc Idn thi ellipse se dugc bieu dien bdi phuong trinh:
a b
Nhu vgy, nhirng diem nam phia trong dudng ellipse cd tong khoang each tdi A va B nhd hon t; nhij'ng dilm ndm phia ngoai dudng 87
Nguyen Kieu Linh Tgp chi KHOA HQC & CONG NGH$ 99(11): 85-89 ellipse cd long khoang cdch Idi A vd B Idn
hon f. Mpt tinh chat dang chii y nifa Id ellipse tuan theo lugt phdn xg anh sang: Tiep tuyen vdi ellipse tgi diem bat ky D tren ellipse tgo vdi AD vd BD hai gdc bdng nhau (gdc Idi bdng gdc phdn xg, nghTa hi liii sdng di tir A tdi D se phdn chieu tdi B vd ngirgc Igi). Khi A triing vdi B (c 0) thi ellipse Ird thdnh dudng trdn (tam O. bdn kinh a = b). I Iv so e = c/a ggi id h^ so do l('ch ldm hay tdm .sai cita ellipse (0
< c < l ) .
Hinh 4. Ellipse vdi a = 5. b =-I vd c ^ 3 (f = 10) Y>k md la hp ellipse nhan A vd B cd djnh ldm tieu dilm (|AB| = 2c) vd cd bdn tryc Idn a = a c ( a > 1) ta phdi cd: b^ = a^ - c^ = a^c^ - c^ = (a^ -I)c^ (b Id bdn tryc be). Khi dd phucmg trinh bieu diln hg ellipse ndy cd dgng (phy thudc thgm sd a ) :
x ' V^
- ^ ^ + ^ - r- = l ( a > l ) .
Khi a = 1 thi a = c va ellipse suy thodi thanh dogn thdng AB (b = 0). Khi a cdng Idn thi dien tich hinh ellipse cang Idn (di?n tich hinh ellipse bdng S = nab, nhung chieu dai dudng ellipse rat khd tinh!).
Cdch tiep can bang hinh hpc dya tren y tudng:
Diem eye tieu can tim D cd tinh chat: dogn AD va BD tgo vdi tiep tuyen dudng Iron tgi D hai goc bdng nhau (gdc tdi bdng gdc phan xg).
Nhu tren da thay, bat cu diem nao tren ellipse nhgn A va B Iam tieu dilm deu cd tinh chat ddi hdi. Nhu vay, tiep diem ciia duimg trdn
lam I bdn kinh r vdi mpt dudng ellipse trong hg trSn sfi Id diem can tim (xem Hinh 5).
Hlnh 5. Cdch tiep cgn hinh hoc Vi thl, bai todn ddt ra cd the diln dgt thanh:
Tim ellipse nhgn A va B Iam tieu diem sao cho nd tiep xiic vdi dudng trdn tam I bdn kinh r tgi mpt diem. Tiep diem ndy vd tiep tuyen chung ciia ellipse vdi dudng trdn se thda man yeu cau de ra, Ta cd the d^mg ellipse nay bang hinh hgc nhu sau.
Lay mpt dogn day khdn§ dan hdi cd dp dai Idn hon 2c mgt chiit. Gdn co dinh mpt ddu day d A, dau kia d B. Dyng ellipse gdm cdc diem cd tdng khoang cdch tir nd tdi A \a B bdng dp ddi dogn day da chpn (ddt dau but chl tua vdo mpt dilm bat ky tren day va di chuyen dau biit chi theo day sao cho dogn day luon cdng vd quay dii mgt gdc 360 ). Sau do keo dai dp ddi dogn day (neu ellipse khdrig cat dudng Iron) ho§c nit ngdn dg dai ddy (neu ellipse cdt dudng trdn tgi hai diem) sao cho ellipse nhgn dugc theo cdch nay tilp xiic voi dudng trdn dd cho (tam I, bdn kinh r). Tiep diem nhgn dugc se la diem cdn tim (diem tren dudng trdn cd tdng khoang each tdi A vd B nhd nhat). Cd the thdy ellipse cudi ciing thu dugc s€ cd di^n ti'ch Idn nhat trong sd cdc ellipse nhgn A, B Idm lieu diem, khdng chiia dudng trdn va khong cd qua hai diem chung vdi dudng trdn.
Nh^n xet. Cd thl md rgn^ bdi toan cho trudng hgp A va / hodc B nam trong (hogc tren) dudng trdn. Dudng tron ciing cd the thay bdng cac dudng cong bgc hai khac (ellipse, parabol, hyperbon ... ). Bai loan cdn cd thS md rgng trong khdng gian ba chieu, tiic la co thl thay dudng trdn bdi mat cdu, ellipse bdi ellipsoid.
Nguyin Kiiu Linh Tgp chl KHOA HQC & C 6 N G N G H 5 99( 11): 85 - 89 TAI LIEU THAM K H A O t^J- '^•^' ^^^ara et all. Nonlinear Programming, Theory and Algorithms. 3rd Edition. A John [1]. T. V. Thieu va N. T. T. Thily. Gido trinh ldi Willey & Sons, Inc., Publication, 2006.
uuphi luyen. Nxb Dai hgc Qudc gia Hd Ngl, 2011. [3]. Properties of Ellipses (Internet),
SUMMARY
SOME IDEAS ABOUT AN OPTIMIZATION PROBLEM IN PLANE Nguyen Kieu Linh College of Sciences - T^ U This paper deals with the following optimization problem usually encountered in many applications and containing interesting mathematical contents: Find on a given circle one point such that the sum of distance from u to two specified points out of the circle is smallest? The problem seems to be rather simple, but finding by analysis or geometry a solution for it is not easy.
The problem under consideration is a direct extension of the well-known following problem, but simpler and having a nice solution: Find on a given line in plane one point such that the sum of distance from it to two given points out of the line is smallest? In this paper we present some approach to the problem based on optimization knowledge and geometry properties of ellipses.
Key word: Optimization problem, analysis approach, geometry approach.
Ngdy nhdn bdi. 08/11/2012. ngdy phdn bien: 23/11/2012, ngdy duyH dang-10/12/2012
' Tel- 0985 059646. Email, [email protected]