CHAPTER 3 INEQUALITIES. Copyright -The Institute of Chartered Accountants of India

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1 CHAPTER 3 INEQUALITIES Cpyright -The Institute f Chartered Accuntants f India

2 INEQUALITIES LEARNING OBJECTIVES One f the widely used decisin making prblems, nwadays, is t decide n the ptimal mix f scarce resurces in meeting the desired gal. In simplest frm, it uses several linear inequatins in tw variables derived frm the descriptin f the prblem. The bjective in this sectin is t make a fundatin f the wrking methdlgy fr the abve by way f intrductin f the idea f : u u u develpment f inequatins frm the descriptive prblem; graphing f linear inequatins; and determinatin f cmmn regin satisfying the inequatins. 3.1 INEQUALITIES Inequalities are statements where tw quantities are unequal but a relatinship exists between them. These type f inequalities ccur in business whenever there is a limit n supply, demand, sales etc. Fr example, if a prducer requires a certain type f raw material fr his factry and there is an upper limit in the availability f that raw material, then any decisin which he takes abut prductin shuld invlve this cnstraint als. We will see in this chapter mre abut such situatins. 3.2 LINEAR INEQUALITIES IN ONE VARIABLE AND THE SOLUTION SPACE Any linear functin that invlves an inequality sign is a linear inequality. It may be f ne variable, r, f mre than ne variable. Simple example f linear inequalities are thse f ne variable nly; viz., x > 0, x < 0 etc. x > 0 x The values f the variables that satisfy an inequality are called the slutin space, and is abbreviated as S.S. The slutin spaces fr (i) x > 0, (ii) x 0 are shaded in the abve diagrams, by using deep lines. Linear inequalities in tw variables: Nw we turn t linear inequalities in tw variables x and y and shade a few S.S. y y y y x >O x >O x >O y >O x >O y >O x x x x Let us nw cnsider a linear inequality in tw variables given by 3x + y < COMMON PROFICIENCY TEST Cpyright -The Institute f Chartered Accuntants f India

3 The inequality mentined abve is true fr certain pairs A Y f numbers (x, y) that satisfy 3x + y < 6. By trial, we may arbitrarily find such a pair t be (1,1) because = 4, (0, 6) and 4 < 6. (x, y) Linear inequalities in tw variables may be slved easily by extending ur knwledge f straight lines. { (2, 0) X Fr this purpse, we replace the inequality by an equality O B and seek the pairs f number that satisfy 3x + y = 6. We may write 3x + y = 6 as y = 6 3x, and draw the graph f this linear functin. Let x = 0 s that y = 6. Let y = 0, s that x = 2. Any pair f numbers (x, y) that satisfies the equatin y = 6 3x falls n the line AB. Nte: The pair f inequalities x 0, y 0 play an imprtant rle in linear prgramming prblems. Therefre, if y is t be less than 6 3x fr the same value f x, it must assume a value that is less than the rdinate f length 6 3x. All such pints (x, y) fr which the rdinate is less than 6 3x lie belw the line AB. The regin where these pints fall is indicated by an arrw and is shaded t in the adjining diagram. Nw we cnsider tw inequalities 3x + y 6 and x y 2 being satisfied simultaneusly by x and y. The pairs f numbers (x, y) that satisfy bth the inequalities may be fund by drawing the graphs f the tw lines y = 6 3x and y = 2 + x, and determining the regin where bth the X inequalities hld. It is cnvenient t express each equality O with y n the left-side and the remaining terms in the right side. The first inequality 3x + y 6 is equivalent t y 6 3x and it requires the value f y fr each x t be B less than r equal t that f and n 6 3x. The inequality is therefre satisfied by all pints lying belw the line y = 6 3x. The regin where these pints fall has been shaded in the adjining diagram. We cnsider the secnd inequality x y 2, and nte that this is equivalent t y 2 + x. It requires the value f y fr each x t be larger than r equal t that f 2 + x. The inequality is, therefre, satisfied by all pints lying n and abve the line y = 2 + x. The regin f interest is indicated by an arrw n the line y = 2 + x in the diagram belw. Fr x = 0, y = = 2; Fr y = 0, 0 = 2 + x i.e, x = 2. A Y 6 3x y = 6 3x MATHS 3.3 Cpyright -The Institute f Chartered Accuntants f India

4 INEQUALITIES y y = 2 + x (0, 2) (-2, 0) 0 x By superimpsing the abve tw graphs we determine the cmmn regin ACD in which the pairs (x, y) satisfy bth inequalities. A Y y = 2 + x C D O y = 6 3x X We nw cnsider the prblem f drawing graphs f the fllwing inequalities Nte: [1] x 0, y 0, x 6, y 7, x + y 12 and shading the cmmn regin. The inequalities 3x + y 6 and x y 2 differ frm the preceding nes in that these als include equality signs. It means that the pints lying n the crrespnding lines are als included in the regin. [2] The prcedure may be extended t any number f inequalities. 3.4 COMMON PROFICIENCY TEST Cpyright -The Institute f Chartered Accuntants f India

5 We nte that the given inequalities may be gruped as fllws : x 0 y 0 x 6 y 7 x + y 12 Y Y 9 = x y =7 O x > 0, x < 6 X O y > 0, y < 7 X Y x + y = 12 X x + y < 12 By superimpsing the abve three graphs, we determine the cmmn regin in the xy plane where all the five inequalities are simultaneusly satisfied. Y (0, 7) (5,7) 0,0 O (6, 6) (6,0) X Example: A cmpany prduces tw prducts A and B, each f which requires prcessing in tw machines. The first machine can be used at mst fr 60 hurs, the secnd machine can be used at mst fr 40 hurs. The prduct A requires 2 hurs n machine ne and ne hur n machine tw. The prduct B requires ne hur n machine ne and tw hurs n machine tw. Express abve situatin using linear inequalities. MATHS 3.5 Cpyright -The Institute f Chartered Accuntants f India

6 INEQUALITIES Slutin: Let the cmpany prduce, x number f prduct A and y number f prduct B. As each f prduct A requires 2 hurs in machine ne and ne hur in machine tw, x number f prduct A requires 2x hurs in machine ne and x hurs in machine tw. Similarly, y number f prduct B requires y hurs in machine ne and 2y hurs in machine tw. But machine ne can be used fr 60 hurs and machine tw fr 40 hurs. Hence 2x + y cannt exceed 60 and x +2y cannt exceed 40. In ther wrds, 2x + y 60 and x + 2y 40. Thus, the cnditins can be expressed using linear inequalities. Example: A fertilizer cmpany prduces tw types f fertilizers called grade I and grade II. Each f these types is prcessed thrugh tw critical chemical plant units. Plant A has maximum f 120 hurs available in a week and plant B has maximum f 180 hurs available in a week. Manufacturing ne bag f grade I fertilizer requires 6 hurs in plant A and 4 hurs in plant B. Manufacturing ne bag f grade II fertilizer requires 3 hurs in plant A and 10 hurs in plant B. Express this using linear inequalities. Slutin: Let us dente by x 1, the number f bags f fertilizers f grade I and by x 2, the number f bags f fertilizers f grade II prduced in a week. We are given that grade I fertilizer requires 6 hurs in plant A and grade II fertilizer requires 3 hurs in plant A and plant A has maximum f 120 hurs available in a week. Thus 6x 1 + 3x Similarly grade I fertilizer requires 4 hurs in plant B and grade II fertilizer requires 10 hurs in Plant B and Plant B has maximum f 180 hurs available in a week. Hence, we get the inequality 4x x Example: Graph the inequalities 5x 1 + 4x 2 9, cmmn regin. Slutin: We draw the straight lines 5x 1 + 4x 2 = 9 and x 1 = 3. Table fr 5x 1 + 4x 2 = 9 Table fr x 1 = 3 x 1 0 9/5 x x 2 9/4 0 x Nw, if we take the pint (4, 4), we find 5x 1 + 4x 2 9 i.e., r, 36 9 (True) x 1 3 i.e., (True) x 1 3, x 1 0 and x 2 0 and mark the 3.6 COMMON PROFICIENCY TEST Cpyright -The Institute f Chartered Accuntants f India

7 Hence (4, 4) is in the regin which satisfies the inequalities. We mark the regin being satisfied by the inequalities and nte that the crss-hatched regin is satisfied by all the inequalities. Example: Draw the graph f the slutin set f the fllwing inequality and equality: x + 2y = 4. x y 3. Mark the cmmn regin. Slutin: We draw the graph f bth x + 2y = 4 and x y 3 in the same plane. The slutin set f system is that prtin f the graph f x + 2y = 4 that lies within the half-plane representing the inequality x y 3. x 2 = x 2 x 1 = 0 5x +4x +4 = =9 =9 =9 =9 x1 +x 2 =3 =3 =3 =3 = x 1 Fr x + 2y = 4, x 4 0 y 0 2 y Fr x y = 3, x 3 0 y 0 3 x y=3 x+2y=4 x Example: Draw the graphs f the fllwing inequalities: x + y 4, x y 4, x 2. and mark the cmmn regin. MATHS 3.7 Cpyright -The Institute f Chartered Accuntants f India

8 INEQUALITIES Fr x y = 4, x 4 0 y 0 4 y x y = 4 Fr x + y = 4, x 0 4 y 4 0 x The cmmn regin is the ne represented by verlapping f the shadings. Example: Draw the graphs f the fllwing linear inequalities: 5x + 4y 100, 5x + y 40, 3x + 5y 75, x 0, y 0. and mark the cmmn regin. Slutin: x = 2 x + y = 4 5x + 4y = 100 3x + 5y = 75 r, r, y x + = y x + = x + y = 40 r, y x + = Pltting the straight lines n the graph paper we have the abve diagram: The cmmn regin f the given inequalities is shwn by the shaded prtin ABCD. Example: Draw the graphs f the fllwing linear inequalities: 5x + 8y 2000, x 175, x 0. 7x + 4y 1400, y 225, y 0. and mark the cmmn regin: Slutin: Let us plt the line AB (5x +8y = 2,000) by jining 3.8 COMMON PROFICIENCY TEST Cpyright -The Institute f Chartered Accuntants f India

9 the pints A(400, 0) and B(0, 250). Similarly, we plt the line CD (7x + 4y = 1400) by jining the pints C(200, 0) and D(0, 350). x y x y Als, we draw the lines EF(x = 175) and GH (y = 225). The required graph is shwn alngside in which the cmmn regin is shaded. Example: Draw the graphs f the fllwing linear inequalities: x + y 1, 7x + 9y 63, y 5, x 6, x 0, y 0. and mark the cmmn regin. Slutin: x + y = 1 ; x 1 0 y0 1 ; 7x + 9y = 63, x 9 0 y0 7. We plt the line AB (x + y = 1), CD (y = 5), EF (x = 6), DE (7x + 9y = 63). Given inequalities are shwn by arrws. Cmmn regin ABCDEF is the shaded regin. Example: Tw machines (I and II) prduce tw grades f plywd, grade A and grade B. In ne hur f peratin machine I prduces tw units f grade A and ne unit f grade B, while machine II, in ne hur f peratin prduces three units f grade A and fur units f grade B. The machines are required t meet a prductin schedule f at least furteen units f grade A and twelve units f grade B. Express this using linear inequalities and draw the graph. MATHS 3.9 Cpyright -The Institute f Chartered Accuntants f India

10 INEQUALITIES Slutin: Let the number f hurs required n machine I be x and that n machine II be y. Since in ne hur, machine I can prduce 2 units f grade A and ne unit f grade B, in x hurs it will prduce 2x and x units f grade A and B respectively. Similarly, machine II, in ne hur, can prduce 3 units f grade A and 4 units f grade B. Hence, in y hurs, it will prduce 3y and 4y units Grade A & B respectively. The given data can be expressed in the frm f linear inequalities as fllws: 2x + 3y 14 (Requirement f grade A) x + 4y 12 (Requirement f grade B) Mrever x and y cannt be negative, thus x 0 and y 0 Let us nw draw the graphs f abve inequalities. Since bth x and y are psitive, it is enugh t draw the graph nly n the psitive side. The inequalities are drawn in the fllwing graph: Fr 2x + 3y = 14, x y Fr x + 4y = 12, x 0 12 y x+4y 12 2x+3y In the abve graph we find that the shaded prtin is mving twards infinity n the psitive side. Thus the result f these inequalities is unbunded. Exercise: 3 (A) Chse the crrect answer/answers 1 (i) An emplyer recruits experienced (x) and fresh wrkmen (y) fr his firm under the cnditin that he cannt emply mre than 9 peple. x and y can be related by the inequality (a) x + y 9 (b) x + y 9 (c) x + y 9 (d) nne f these (ii) On the average experienced persn des 5 units f wrk while a fresh ne 3 units f wrk daily but the emplyer has t maintain an utput f at least 30 units f wrk per day. This situatin can be expressed as (a) 5x + 3y 30 (b) 5x + 3y >30 (c) 5x + 3y 30 (d) nne f these (iii) The rules and regulatins demand that the emplyer shuld emply nt mre than 5 experienced hands t 1 fresh ne and this fact can be expressed as (a) y x/5 (b) 5y x (c) 5 y x (d) nne f these 3.10 COMMON PROFICIENCY TEST Cpyright -The Institute f Chartered Accuntants f India

11 (iv) The unin hwever frbids him t emply less than 2 experienced persn t each fresh persn. This situatin can be expressed as (a) x y/2 (b) y x/2 (c) y x /2 (d) x 2y (v) The graph t express the inequality x + y 9 is (a) (b) (c) (d) nne f these (vi) The graph t express the inequality 5x + 3y 30 is (a) (b) (c) (d) nne f these MATHS 3.11 Cpyright -The Institute f Chartered Accuntants f India

12 INEQUALITIES (vii) The graph t express the inequality y 1/2x is indicated by (a) (b) (c) (d) (viii) L1 : 5x + 3y = 30 L2 : x+y = 9 L3 : y = x/3 L4 : y = x/2 The cmmn regin (shaded part) shwn in the diagram refers t (a) 5x + 3y 30 (b) 5x + 3y 30 (c) 5x + 3y 30 (d) 5x + 3y > 30 (e) Nne f these x + y 9 x + y 9 x + y 9 x + y < 9 y 1/5 x y x/3 y x/3 y 9 y x/2 y x/2 y x/2 y x/2 x 0, y 0 x 0, y 0 x 0, y COMMON PROFICIENCY TEST Cpyright -The Institute f Chartered Accuntants f India

13 2. A dietitian wishes t mix tgether tw kinds f fd s that the vitamin cntent f the mixture is at least 9 units f vitamin A, 7 units f vitamin B, 10 units f vitamin C and 12 units f vitamin D. The vitamin cntent per Kg. f each fd is shwn belw: A B C D Fd I : Fd II: Assuming x units f fd I is t be mixed with y units f fd II the situatin can be expressed as (a) 2x + y 9 (b) 2x + y 30 (c) 2x + y 9 (d) 2x + y 9 x + y 7 x + y 7 x + y 7 x + y 7 x + 2y 10 x + 2y 10 x + y 10 x +2 y 10 2x +3 y 12 x + 3y 12 x + 3y 12 2x +3 y 12 x > 0, y > 0 x 0, y 0, 3. Graphs f the inequatins are drawn belw : L1 : 2x +y = 9 L2 : x + y = 7 L3 : x+2y= 10 L4 : x + 3y = 12 The cmmn regin (shaded part) indicated n the diagram is expressed by the set f inequalities (a) 2x + y 9 (b) 2x + y 9 (c) 2x + y 9 (d) nne f these x + y 7 x + y 7 x + y 7 x + 2y 10 x +2 y 10 x +2y 10 x +3 y 12 x + 3y 12 x +3 y 12 x 0, y 0 MATHS 3.13 Cpyright -The Institute f Chartered Accuntants f India

14 INEQUALITIES 4. The cmmn regin satisfied by the inequalities L1: 3x + y 6, L2: x + y 4, L3: x +3y 6, and L4: x + y 6 is indicated by (a) (b) (c) (d) nne f these 5. The regin indicated by the shading in the graph is expressed by inequalities 3.14 COMMON PROFICIENCY TEST Cpyright -The Institute f Chartered Accuntants f India

15 (a) x 1 2 (b) x 1 2 (c) x 1 2 (d) x 1 2 2x 1 + 2x 2 8 x x + x x + 2x x + 2x > x 1 0, x 2 0, 6. (i) The inequalities x 1 0, x 2 0, are represented by ne f the graphs shwn belw: (a) (b) (c) (d) (ii) The regin is expressed as (a) x 1 x 2 1 (b) x 1 1 (c) x 1 1 (d) nne f these MATHS 3.15 Cpyright -The Institute f Chartered Accuntants f India

16 INEQUALITIES (iii) (a) The inequality x 1 + 2x 2 0 is indicated n the graph as (b) (c) (d) nne f these COMMON PROFICIENCY TEST Cpyright -The Institute f Chartered Accuntants f India

17 The cmmn regin indicated n the graph is expressed by the set f five inequalities (a) L1 : x 1 0 (b) L1 : x 1 0 (c) L1 : x 1 0 (d) Nne f these L2 : x 2 0 L2 : x 2 0 L2 : x 2 0 L3 : x 1 1 L3 : x 1 +x 2 1 L3 : x 1 1 L4 : x 1 x 2 1 L4 : x 1 x 2 1 L4 : x 1 x 2 1 L5 : x 1 + 2x 2 0 L5 : x 1 +2x 2 0 L5 : x 1 +2x A firm makes tw types f prducts : Type A and Type B. The prfit n prduct A is Rs. 20 each and that n prduct B is Rs. 30 each. Bth types are prcessed n three machines M1, M2 and M3. The time required in hurs by each prduct and ttal time available in hurs per week n each machine are as fllws: Machine Prduct A Prduct B Available Time M M M The cnstraints can be frmulated taking x 1 = number f units A and x 2 = number f unit f B as (a) x 1 12 (b) 3x 1 + 3x 2 36 (c) 3x 1 + 3x 2 36 (d) nne f these 5x 1 + 2x x 1 + 2x x 1 + 2x x 1 + 6x x 1 + 6x x 1 + 6x 2 60 x 1 0, x 2 0 x 1 0, x 2 0 x 1 0, x The set f inequalities L1: x 1 12, L2: 5x 1 + 2x 2 50, L3: x 1 + 3x 2 30, x 1 0, and x 2 0 is represented by (a) (b) MATHS 3.17 Cpyright -The Institute f Chartered Accuntants f India

18 INEQUALITIES (c) (d) nne f these 10. The cmmn regin satisfying the set f inequalities x 0, y 0, L1: x+y 5, L2: x +2y 8 and L3: 4x +3y 12 is indicated by (a) (b) (c) (d) nne f these ANSWERS 1. (i) b (ii) c (iii) a,c (iv) b,d (v) a (vi) c (vii) d (viii) e 2. d 3. c 4. a 5. a 6. (i) b (ii) c (iii) 7. b 8. c 9. b 10. a a 3.18 COMMON PROFICIENCY TEST Cpyright -The Institute f Chartered Accuntants f India

19 ADDITIONAL QUESTION BANK 1. On slving the inequalities 2x + 5y 20, 3x + 2y 12, x 0, y 0, we get the fllwing situatin (A) (0, 0), (0, 4), (4, 0) and ( 20, 36 ) (B) (0, 0), (10, 0), (0, 6) and ( 20, 36 ) (C) (0, 0), (0, 4), (4, 0) and (2, 3) (D) (0, 0), (10, 0), (0, 6) and (2, 3) 2. On slving the inequalities 6x + y 18, x + 4y 12, 2x + y 10,, we get the fllwing situatin (A) (0, 18), (12, 0),, (4, 2) and (7, 6) (B) (3, 0), (0, 3),, (4, 2) and (7, 6) (C) (5, 0), (0, 10),, (4, 2) and (7, 6) (D) (0, 18), (12, 0), (4, 2), (0, 0) and (7, 6) ANSWERS 1) A 2) A MATHS 3.19 Cpyright -The Institute f Chartered Accuntants f India