Abhilasha Classes Class- XII Date: SOLUTION (Chap - 9,10,12) MM 50 Mob no

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1 hlsh Clsses Clss- XII Dte: 0- - SOLUTION Chp - 9,0, MM 50 Mo no-996 If nd re poston vets of nd B respetvel, fnd the poston vet of pont C n B produed suh tht C B vet r C B = where = hs length nd dreton rtos,, 6 Fnd the vet r The dreton osnes of r re s the poston vet of pont C 6, 6, 6 6 Sne r mkes n ute ngle wth -s, therefe os > 0 e, l > 0 So, dreton osnes of r re,, 6 r ˆ ˆ r 6ˆ 9 ˆ 8kˆ 6 kˆ [usng r r lˆ m ˆ nkˆ ] So, omponents of r long o, o nd oz re 6ˆ, 9 ˆ nd 8kˆ respetvel Fnd the ngle etween the vets wth dreton rtos,, 5 nd,, 5 Let = vet prllel to the vet hvng dreton rtos,, 5 = ˆ ˆ 5kˆ nd = vet prllel to the vet hvng dreton rtos,, 5 = ˆ + ˆ + 5kˆ Let e the ngle etween the gven vets Then os = Thus, the ngle etween the vets wth dreton rtos,, 5 nd,, 5 s 60 Fnd der nd degree of d 5 sn d d e d d d Dfferentl equton Order of DE Degree of DE Sol:

2 d d sn d d d d 5 e 5 Use vet methods to prove, n the usul notton, n trngle osc Let B, BC, C So tht + + = 0 = + 80 C C = + + os 80 C B gvng the result 6 F n two vets nd prove tht os where sn s the ngle etween nd ddng, os sn Show tht the vets k ; 8 k nd 0 k fm rght-ngled trngle k 8 k 0 k Hene,, fm the sdes of trngle lso k 8 k = = 0

3 Hene nd re perpendulr The trngle s therefe rght-ngled 8 Let ; k nd k k Determne vet suh tht, nd = 0 It s gven tht Vetll pre multpl oth sdes 0 5; ; Susttutng these vlues n we get, 5 k k, 6 k gvng k 8 9 If nd then prove tht 0 lso 0, nd 0,, re mutull perpendulr vets nd 0 mnufturer of lne of ptent mednes s preprng produton pln on mednes nd B There re suffent ngredents vlle to mke 0, 000 ottles of nd 0,000 ottles of B ut there re onl 5,000 ottles nto whh ether of the mednes n e put Further me, t tkes hours to prepre enough mterl to fll 000 ottles of, t tkes one hour to prepre enough mterl to fll 000 ottles of B nd there re 66 hours vlle f ths operton The proft s Rs 8 per ottle f nd Rs per ottle f B Fmulte ths prolem s lner progrmmng prolem Suppose the mnufturer produes ottles of mednes nd ottles of medne B Sne the proft s Rs 8 per ottle f nd Rs per ottle f B So, totl proft n produng ottles of medne nd ottles of medne B s Rs 8 + Let Z denote the totl proft Then, Z 8 Sne 000 ottles of medne re prepred n hours

4 So, tme requred to prepre ottles of medne 000 hours It s gven tht 000 ottles of medne B re prepred n hour Tme requred to prepre ottles of medne B = 000 hours Thus, totl tme requred to prepre ottles of medne nd ottles of medne B s hours But, the totl tme vlle f ths operton s 66 hours Sne there re onl 5,000 ottles nto whh the mednes n e put 5, 000 It s gven tht the ngredents re vlle f 0,000 ottles of nd 0,000 ottles of B 0,000 nd 0,000 Sne the numer of ottles n not e negtve Therefe, 0, 0 Hene, the mthemtl fmulton of the gven LPP s s follows: Mmze Z 8 Suet to 66,000; 5,000; 0,000; 0, 000 nd 0, 0 Solve the followng prolem grphll: Mnmze nd mmze Z 9 Suet to the onstrnts 60 0 v 0, 0 v Frst of ll, let us grph the fesle regon of the sstem of lner nequltes to v The fesle regon BCD s shown n the fgure Note tht the regon s ounded The odntes of the ner ponts, B, C nd D re 0, 0, 5, 5, 5, 5 nd 0, 0 respetvel Y 5 = D0, 0 5 C5, 5 0, 0 5 B5, 5 60, 0 X X , 0 + =60 + = 0 Y Cner Pont Crespondng vlues of Z = + 9 0, 0 90 B5, 5 60 C5, 5 80 D0, 0 80

5 We now fnd the mnmum nd mmum vlue of Z From the tle, we fnd tht the mnmum vlue of Z s 60 t the pont B5, 5 of the fesle regon The mmum vlue of Z on the fesle regon ours t the two ner ponts C 5, 5 nd D0, 0 nd t s 80 n eh se Remrk: Oserve tht n the ove emple, the prolem hs multple optml solutons t the ner ponts C nd D e, the oth ponts produe sme mmum vlue 80 In suh se, ou n see tht ever pont on the lne segment CD onng the two ner ponts C nd D lso gve the sme mmum vlue Sme s lso true n the se f the two ponts produe sme mnmum vlue F emple md-pont of lne onng ponts C nd D s P5/, 5/ whh les on CD Now Z t P5/, 5/ s 80, whh s sme s Z t C nd D Smlrl, ever pont on lne CD gves the sme vlue of Z n ol ompn requres,000, 0,000 nd 5,000 rrels of hgh-grde, medum grde nd low grde ol, respetvel Refner produes 00, 00 nd 00 rrels per d of hgh-grde, medum-grde nd low-grde ol, respetvel, whle refner B produes 00, 00 nd 00 rrels per d of hgh-grde, medum-grde nd low-grde ol, respetvel If refner osts Rs 00 per d nd refner B osts Rs 00 per d to operte, how mn ds should eh e run to mnmze osts whle stsfng requrements The gven dt m e put n the followng tulr fm: Refner Hgh-grde Medum-grde Low-grde Cost per d Rs 00 B Rs 00 Mnmum requrement,000 0,000 5,000 Suppose refneres nd B should run f nd ds respetvel to mnmze the totl ost The mthemtl fm of the ove LPP s Mnmze Z Suet to , , nd, 0 The fesle regon of the ove LPP s represented the shded regon n fgure = 0000 B 0, = 5000 B 0, 60 P60, 0 B 0, 50 0, 0 O 00/, 0 5, = 000 The ner ponts of the fesle regon re 0, 0, 60, 0 nd B 0, 50 P The vlue of the oetve funton t these ponts re gven n the followng tle: Pont, Vlue of the oetve funton Z , 0 Z

6 P 60, 0 B 0, 50 Z Z Clerl, Z s mnmum when = 60, = 0 Hene, the mhne should run f 60 ds nd the mhne B should run f 0 ds to mnmze the ost whle stsfng the onstrnts Determne grphll the mnmum vlue of the oetve funton Z 50 0 suet to the onstrnts: 5 v 0, 0 v Frst of ll, let us grph the fesle regon of the sstem of nequltes to v The fesle regon shded s shown n the fgure Oserve tht the fesle regon s unounded We now evlute Z t the ner ponts Y = 5 0, 5, 8, = 0 Cner Pont Z = , , 60, , 0 00 smllest X B0,, 0 6, 0 9, = X + = Y From ths tle, we fnd tht 00 s the smllest vlue of Z t the ner pont 6, 0 Cn we s tht mnmum vlue of Z s 00? Note tht f the regon would hve een ounded, ths smllest vlue of Z s the mnmum vlue of Z theem But here we see tht the fesle regon s unounded Therefe 00 m m not e the mnmum vlue of Z To dede ths ssue, we grph the neqult e, 5 0 nd hek whether the resultng open hlf plne hs ponts n ommon wth fesle regon not If t hs ommon ponts, then 00 wll not e the mnmum vlue of Z Otherwse, 00 wll e the mnmum vlue of Z s shown n the fgure, t hs ommon ponts Therefe, the gven onstrnts Z 50 0 hs no mnmum vlue suet to

7 d d Verf tht the funton e s soluton of the dfferentl equton 6 0 d d Gven funton s e Dfferenttng oth sdes of equton wth respet to, we get d e d Now, dfferenttng wth respet to, we hve d 9e d Susttutng the vlues of d, d d d LHS 9 e 6e 9e 9e 0 nd n the gven dfferentl equton, we get e = RHS Therefe the gven funton s soluton of the gven dfferentl equton 5 Solve the dfferentl equton d d e e Seprtng the vrles d d e e e d e d, ntegrtng, the soluton s e e e e C C s n rtrr onstnt 6 Solve the dfferentl equton d d Note:, + re homogeneous n nd of degree one Tkng = v, d d v dv d Susttutng n the gven equton v dv d dv d v v v v v v v v v v v Now, seprtng the vrles v dv v v d log v v log log [ v v ] = onstnt; v v = C = C s therefe the soluton where C s n rtrr onstnt

8 d Solve the dfferentl equton os d d Here, os d d os d ; ths s n the lner fm Integrtng ft e d log log e Multplng the ntegrtng ft, e d d os d os d d os d d os d os d sn, where s n rtrr onstnt sn d 8 Solve the dfferentl equton d d d On settng = v, the equton s v dv d v v Seprtng the vrles nd ntegrtng, d dv v log Ths smplfes to the fm = Ce / v 9 The surfe re of lloon eng nflted hnges t onstnt rte If ntll ts rdus s unts nd fter seonds, t s 5 unts, fnd the rdus fter t seonds Let r e the rdus nd S e the surfe re of the lloon t n tme t Then S r

9 ds dt dr 8 r dt It s gven tht ds dt onstnt = k s Puttng ds dt k n, we get dr k 8 r dt 8rdr kdt Integrtng oth sdes, we get r kt C We re gven tht t t 0, r nd t t, r 5 k0 C 6 nd 00 k C C 6 nd k Susttutng the vlues of C nd K n, we otn r t 6 r 8t 9 r 8t 9 0 The slope of the tngent to the urve t n pont s twe the dnte t tht pont The urve psses through the pont, Determne ts equton P, e n pont on the urve Let Then slope of the tngent t P s d d It s gven tht the slope of the tngent t P, s twe the dnte e, d d d d log log C Ce Sne the urve psses through, Therefe Puttng = nd = n, we get f 8 Ce C e 8 Puttng the vlue of C n, we get e 8 Ths s the requred equton of the urve Solve the dfferentl equtons d 0 d d os d d d

10 d d Integrtng, d d tn tn whh m e lso wrtten s tn tn where C s now rtrr onstnt n whh s lso sed The fm omnng the two terms on the LHS usng the fmul tn tn tn m e redued to C + + = os d d C d d os The soluton s C sn tn sn tn, where s n rtrr onstnt