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1 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I 6 LINEAR INEQUATINS AND THEIR APPLICATINS You have alread read about linear and quadratic equations in one or two variables. You have also learnt the methods of solving the linear and quadratic equations and have applied this knowledge in solving word problems from dail life. Some times, it is not possible to translate a word problem in the form of an equation. Let us consider the following situation: Alok goes to market with Rs. 0 to bu pencils. The cost of one pencil is Rs..60. If denotes the number of pencils which he bus, then he will spend an amount of Rs..60. This amount cannot be equal to Rs. 0 as is a natural number. Thus..60 < 0 (i) Let us consider one more situation where a person wants to bu chairs and tables with Rs. 50,000 in hand. A table costs Rs. 550 while a chair costs Rs. 50. Let be the number of chairs and be the number of tables he bus, then his total cost = Rs.( ) Thus, in this case we can write, < 50,000 or (ii) Statement (i) involves the sign of inequalit < and statement (ii) consists of two statements: +9 < 000, +9 = 000 in which the first one is not an equation: Such statements are called Inequations. In this lesson, we will discuss linear inequations and solve problems related to diet, business and to transportation using graphical method. MATHEMATICS 89 Get Discount Coupons for our Coaching institute and FREE Stud Material at

2 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I BJECTIVES After studing this lesson, ou will be able to: differentiate between a linear equation and a linear inequation; state that a planar region represents the solution of a linear inequation; represent graphicall a linear inequation in two variables; show the solution of an inequation b shading the appropriate region; solve graphicall a sstem of two or three linear inequations in two variables; define the terms such as objective function, constraints, feasible region, optimisation; formulate problems related to diet, business, transportation, etc; and solve above problems using graphical method. EXPECTED BACKGRUND KNWLEDGE Solution of linear equations in one or two variables. Graph of linear equations in one or two variables in a plane. Graphical solution of a sstem of linear equations in two variables. 6. INEQUATIN In this lesson we will discuss more about linear inequations and their applications from dail life. A statement involving a sign of equalit (=) is an equation. Similarl, a statement involving a sign of inequalit, <, >,, or is called an inequation. Some eamples of inequations are: (i) + 5 > 0 (ii) 7 < 0 (iii) a + b 0, a 0 (iv) a + b c, a 0 (v) + (vi) (vii) a + b + c > < 0 (v) and (vii) are inequations in two variables and all other inequations are in one variable. (i) to (v) and (vii) are linear inequations and (vi) is a quadratic inequation. In this lesson, we shall stud about linear inequations in one or two variables onl. 6. SLUTINS F LINEAR INEQUATINS IN NE/TW VARIABLES Solving an inequation means to find the value (or values) of the variable (s), which when substituted in the inequation, satisfies it. For eample, for the inequation.60<0 (statement) (i) all values of are the solutions. ( is a whole number) 90 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

3 Get Discount Coupons for our Coaching institute and FREE Stud Material at For the inequation + 6>0, where is a real number, all values of which are > 8 are the solutions. For the linear inequation in two variables, likea + b + c > 0, we shall have to find the pairs of values of and which make the given inequation true. Let us consider the following situation : Anil has Rs. 60 and wants to bu pens and pencils from a shop. The cost of a pen is Rs. 5 and that of a pencil is Rs. If denotes the number of pens and, the number of pencils which Anil bus, then we have the inequalit (i) Here, = 6, = 0 is one of the solutions of the inequation (i). Similarl = 5, =; =, = ; = 0, = are some more solutions of the inequation. In solving inequations, we follow the rules which are as follows :. Equal numbers ma be added (or subtracted) from both sides of an inequation. Thus (i) if a > b then a + c > b + c and a c > b c and (ii) if a b then a + d b + d and a d b d. Both sides of an inequation can be multiplied (or divided) b the same positive number. Thus (i) if a > b and c > 0 then ac > bc and a > b c c and (ii) if a < b and c > 0 then ac < bc and a b c c. When both sides of an inequation are multiplied b the same negative number, the sign of inequalit gets reversed. Thus (i) if a > b and d < 0 then ad < bd and a < b d d and (ii) if a < b and c < 0 then ac > bc and a b c c 6. GRAPHICAL REPRESENTATIN F LINEAR INEQUATINS IN NE R TW VARIABLES. In Section 6., while translating word problem of purchasing pens and pencils, we obtained the following linear inequation in two variables and : 5 + < (i) Let us now find all solutions of this inequation, keeping in mind that and here can be onl whole numbers. To start with, let = 0. MDULE - I MATHEMATICS 9 Get Discount Coupons for our Coaching institute and FREE Stud Material at

4 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I 9 Thus, we have < 60 or < 0, i,e the values of corresponding to = 0 can be 0,,,..., 0 onl Thus, the solutions with = 0 are (0,0), (0,), (0,)...(0, 0) Similarl the other solutions of the inequation, when =,,... are (,0) (,) (,)... (,8) (,0) (,) (,)... (,6) (0,0) (0,) (0,), (0,) (,0) (,) (,0) You ma note that out of the above ordered pairs, some pairs such as (0,0), (, 5), (6, 0), (9, 5), (,0) satisf the equation 5 + = 60 which is a part of the given inequation and all other possible solutions lie on one of the two half planes in which the line 5 + = 60, divides the - plane. If we now etend the domain of and from whole numbers to real numbers, the inequation will represent one of the two half planes in which the line 5 + = 60, divides the -plane. Thus we can generalize as follows : If a,b,c, are real numbers, then a + b + c = 0 is called a linear equation in two variables and, where as a + b + c < 0 or a + b + c 0, a + b + c>0 and a + b + c < 0 are called linear inequations in two variables and. The equation a + b + c = 0 is a straight line which divides the plane into two half planes which are represented b a+ b +c 0 and a + b + c 0. For eample + = 0 can be represented b line AB, in the - plane as shown in Fig. 6. The line AB divides the cordinate plane into two half - plane regions : (i) (ii) half plane region I above the line AB half plane region II below the line AB. ne of the above region represents the inequalit + < 0...(i) and the other region will be represented b + > (ii) ' A ' 5 + = Fig. 6. Fig. 6. II I + = B MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

5 Get Discount Coupons for our Coaching institute and FREE Stud Material at To identif the half plane represented b inequation (i), we take an arbitrar point, preferabl origin, if it does not lie on AB. If the point satisfies the inequation (i), then the half plane in which the arbitrar point lies, is the desired half plane. In this case, taking origin as the arbitrar point we have 0+0 <0 i.e <0. Thus origin satisfies the inequation + < 0. Now, origin lies in half plane region II. Hence the inequalt + < 0 represents half plane II and the inequalit + >0 will represent the half plane I Eample 6. Show on graph the region represented b the inequation + > 5. Solution : The given inequation is + > 5 Let us first take the corresponding linear equation + = 5 and draw its graph with the help of the following table : 5 0 Since (0,0) does not lie on the line AB, so we can select (0,0) as the arbitrar point. Since > 5 is not true The desired half plane is one, in which origin does not lie The desired half plane is the shaded one (See Fig. 6.) ' 5 ' A Fig. 6. B C + > Before taking more eamples, it is important to define the following : (i) (ii) Closed Half Plane: A half plane is said to be closed half plane if all points on the line separating the two half planes are also included in the solution of the inequation. The Half plane in Eample 6. is a closed half plane. An pen Half Plane : A half plane in the plane is said to be an open half plane if the points on the line separting the planes are not included in the half plane. Eample 6. Draw the graph of inequation 5 > 0 Solution : The given inequation is 5 > 0 MDULE - I MATHEMATICS 9 Get Discount Coupons for our Coaching institute and FREE Stud Material at

6 = Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I The corresponding linear equation is 5 = 0 we have the following table The line AB divides - plane into two half planes I and II. As the line AB passes through origin, we consider an other arbitrar point (sa) P (,) which is in half plane I. Let us see whether it satisfies the given inequation 5 > 0 Then 5 () > 0 or 0 > 0, or 7 > 0 which is not true The desired half plane is II Again the inequation is a strict inequation 5 > 0 Line AB is not a part of the graph and hence has been shown as a dotted line. Hence, the graph of the given inequation is the shaded region half plane II ecluding the line AB. Eample 6. Represent graphicall the inequation > 0 Solution : Given inequation is > 0 and the corresponding linear equation is = 0 or =0 or = which is represented b the line ABC on the - plane (See Fig. 6.5). Taking (0,0) as the arbitrar point, we can sa that 0 and so, half plane II represents the inequation 0 Eample 6. inequation + 0 Solve graphicall the Solution : Here the inequation is + 0 and the corresponding equation is + = 0 or = The line ABC represents the line = which divides the - plane into two half planes and the inequation + > 0 is represented b the half plane I. B 5 o 5 (0,0) Y Fig I II A B C Fig. 6.6 I I II A B C Fig. 6.5 A 5 5 > 0 II > 0 + > 0 = 9 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

7 Get Discount Coupons for our Coaching institute and FREE Stud Material at CHECK YUR PRGRESS 6. Represent the solution of each of the following inequations graphicall in two dimensional plane:. + > 8. < > 0. 8 > 5. > 6 6. > 0 7. < 8. > 8 9. < 5 0. < 8 6. GRAPHICAL SLUTIN F A SYSTEM F LINEAR INEQUATINS IN TW VARIABLES. You alread know how to solve a sstem of linear equations in two variables. Now, ou have also learnt how to solve linear inequations in two variables graphicall. We will now discuss the technique of finding the solutions of a sstem of simultaneous linear inequations. B the term solution of a sstem of simultaneons linear inequations we mean, finding all ordered pairs (,) for which each linear inequation of the sstem is satisfied. A sstem of simultaneous inequations ma have no solution or an infinite number of solutions represented b the region bounded or unbounded b straight lines corresponding to linear inequations. We take the following eample to eplain the technique. Eample 6.5 Solve the following sstem of inequations graphicall: + > 6 ; > 0. Solution : Given inequations are and + > 6... (i) > 0... (ii) We draw the graphs of the lines + = 6 and = 0 (Fig. 6.7) The inequation (i) represent the shaded region above the line + = 6 and inequations (ii) represents the region on the right of the line = 0 The common region represented b the double shade in Fig. 6.7 represents the solution of the given sstem of linear inequations. (0,6) 8 6 (0,0) B C + > 6 > 0 = 0 (,) 6 (6,0) + = 6 MDULE - I Fig. 6.7 MATHEMATICS 95 Get Discount Coupons for our Coaching institute and FREE Stud Material at

8 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I Eample 6.6 Find graphicall the solution of the following sstem of linear inequations : + 5, +, + 5 5,,. Solution : Given inequations are and... (i)... (ii)... (iii)... (iv)... (v) We draw the graphs of the lines + = 5, + =, + 5 = 5, = and = (Fig. 6.8) The inequation (i) represents the region below the line + = 5. The inequations (ii) represents the region on the right of equation + = and the region above the line + 5 = 5 represents the inequation (iii). Similarl after shading the regions for inequations (iv) and (v) we get the common region as the bounded regionabcde as shown in (Fig. 6.8) The co-ordinates of the points of the shaded region satisf the given sstem of inequations and therefore all these points represent solution of the given sstem. Eample 6.7 Solve graphicall the following sstem of inequations : +, +, 0, 0. Solution : We represent the inequations +, +, 0, 0 b shading the corresponding regions on the graph as shown in Fig. 6.9 Here we find that there is no common region represented b these inequations. We thus conclude that there is no solution of the given sstem of linear inequations. Eample 6.8 Solve the following sstem of linear inequations graphicall : <, + < 6; 0, 0. Solution : The given inequations are < + < 6 0 ; 0... (i)... (ii)... (iii) + 5 = 5 5 A B + < > 0 > = E Fig. 6.9 Fig. 6.8 D C = 5 + = + = 5 + > > 0 > 0 = + = 96 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

9 Get Discount Coupons for our Coaching institute and FREE Stud Material at After representing the inequations <, + < 6, > 0 and > 0 on the graph we find the common region which is the bounded region ABC as shown in Fig. 6.0 MDULE - I 6 C 5 CHECK YUR PRGRESS 6. Solve each of the follwing sstems of linear inequations in two variables graphicall :.,..,.. + 0, , +, 0, , + 8, 7 + 0, 6, 0, , +, 0, ; + 6, 0, 0. + = 6 Fig = ; +,, 0 ; 0 A B 6.5 INTRDUCTIN T LINEAR PRGRAMMING PRBLEM A to dealer goes to the wholesale market with Rs to purchase tos for selling. In the market there are various tpes of tos available. From qualit point of view, he finds that the to of tpe A and tpe B are suitable. The cost price of tpe A to is Rs. 00 each and that of tpe B is Rs. 50 each. He knows that the tpe A to can be sold for Rs. 5 each, while the tpe B to can be sold for Rs. 65 each. Within the amount available to him he would like MATHEMATICS 97 Get Discount Coupons for our Coaching institute and FREE Stud Material at

10 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I to make maimum profit. His problem is to find out how man tpe A and tpe B tos should be purchased so to get the maimum profit. He can prepare the following table taking into account all possible combinations of tpe A and tpe B tos subject to the limitation on the investment. A tpe B tpe Investment Amount after sale Profit on the (including investment the unutilised amount if an) Now, the decision leading to maimum profit is clear. Five tpe A tos should be purchased. The above problem was eas to handle because the choice was limited to two tpes, and the number of items to be purchased was small. Here, all possible combinations were thought of and the corresponding gain calculated. But one must make sure that he has taken all possibilities into account. A situation faced b a retailer of radio sets similar to the one given above is described below. A retailer of radio sets wishes to bu a number of transistor radio sets from the wholesaler. There are two tpes (tpe A and tpe B) of radio sets which he can bu. Tpe A costs Rs.60 each and tpe B costs Rs. 0 each. The retailer can invest up to Rs B selling the radio sets, he can make a profit of Rs. 50 on each set of tpe A and of Rs. 0 on each set of tpe B. How man of each tpe should he bu to maimize his total profit? Here we have to maimize the profit. Sometimes we come across a problem in which the costs are to be minimized. Consider the following problem : Two tailors A and B earn Rs.50 and Rs.00 per da respectivel. A can stitch 6 shirts and pants per da, while B can stitch shirts and 7 pants per da. How man das shall each work if the want to produce at least 60 shirts and 7 pants at a minimum labour cost? In this problem we have to minimise the labour cost. These tpes of problems of maimisation and minimisation are called optimisation problems. The technique followed b mathematicians to solve such problems is called Linear Programming. 98 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

11 Get Discount Coupons for our Coaching institute and FREE Stud Material at HISTRICAL BACKGRUND The technique of linear programming is of recent origin. It started during the second world war, when the war operations had to be planned to economise the ependiture, minimise losses and maimise the damage to the enem. The first problem in linear programming was formulated in 9 b the Russian mathematician L.Kantorovich and the American economist F.L. Hilcheock, both of them worked independentl. This is the well known transportation problem which forms a branch of linear programming. In 95, an English economist, G.Stigler, described et another linear programming problemthat of determining an optimal diet. Mainl this problem was to determine the quantities of 77 foods that are to be bought not onl at the minimum cost, but also to satisf minimum requirements of nine nutritive elements. In 97, an American economist G.B. Dantzig published a paper in the famous journal Econometrics wherein he formulated the general linear programming problem. Dantzig is also credited with using the term Linear Programming and for solution of the problem b analtical methods. In 97, L. Kantorovich was awarded the Noble Prize for Economis for his work on these problems together with another famous American mathematical-economist T.C. Koopmans. 6.6 DEFINITINS F VARIUS TERMS INVLVED IN LINEAR PRGRAMMING A close eamination of the eamples cited in the introduction points out one basic propert that all these problems have in common, i.e., in each eample, we were concerned with maimising or minimising some quantit. In Eamples and, we wanted to maimise the return on the investment. In Eample, we wanted to minimise the labour cost. In linear programming terminolog the maimization or minimization of a quantit is referred to as the objective of the problem BJECTIVE FUNCTIN In a linear programming problem. z, the linear function of the variables which is to be optimized is called objective function. Here, a linear form means a mathematical epression of the tpe a + a ann, where a, a,..., a n are constants and,,..., n are variables. MDULE - I In linear programming problems, the products, services, projects etc. that are competing with each other for sharing the given limited resources are called the variables or decision variables CNSTRAINTS The limitations on resources (like cash in hand, production capacit, man power, time, machines, etc.) which are to be allocated among various competing variables are in the form of linear MATHEMATICS 99 Get Discount Coupons for our Coaching institute and FREE Stud Material at

12 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I equations or inequations (inequalities) and are called constraints or restrictions NN-NEGATIVE RESTRICTINS All decision variables must assume non-negative values, as negative values of phsical quantities is an impossible situation. 6.7 FRMULATIN F A LINEAR PRGRAMMING PRBLEM The formulation of a linear programming problem as a mathematical model involves the following ke steps. Step : Identif the decision variables to be determined and epress them in terms of algebraic smbols such as,,,... Step : Identif all the limitations in the given problem and then epress them as linear equations or inequalities in terms of above defined decision variables. Step : Identif the objective which is to be optimised (maimised or minimised) and epress it as a linear function of the above defined decision variables. Eample 6.9 A retailer wishes to bu a number of transistor radio sets of tpes A and B. Tpe A cost Rs.60 each and tpe B cost Rs. 0 each. The retailer knows that he cannot sell more than 0 sets, so he does not want to bu more than 0 sets and he cannot afford to pa more than Rs His epectation is that he would get a profit of Rs.50 for each set of tpe A and Rs.0 for each set of tpeb. Form a mathematical model to find how man of each tpe should be purchased in order to make his total profit as large as possible? Solution : Suppose the retailer purchases sets of tpe A and sets of tpe B. Since the number of sets of each tpe is non-negative, so we have 0, 0, () () Also the cost of sets of tpe A and sets of tpe B is and it should be equal to or less than Rs.5760, that is, or + 8 () Further, the number of sets of both tpes should not eceed 0, so 0 + () Since the total profit consists of profit derived from selling the tpe A sets and tpe B 00 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

13 Get Discount Coupons for our Coaching institute and FREE Stud Material at sets, therefore, the retailer earns a profit of Rs.50 on tpe A sets and Rs.0 on tpe B sets. So the total profit is given b : MDULE - I z = (5) Hence, the mathematical formulation of the given linear programming problem is as follows : Find, which Maimise z = (bjective function) subject to the conditions , 0 Constraints Eample 6.0 A soft drink compan has two bottling plants, one located atp and the other at Q. Each plant produ ces three different soft drinks A, B, and C. The capacities of the two plants in terms of number of bottles per da, are as follows : Plants Products P Q A B C A market surve indicates that during the month of Ma, there will be a demand for 000 bottles of A, 6000 bottles of B and 8000 bottles of C. The operating cost per da of running plants P and Q are respectivel Rs.6000 and Rs.000. How man das should the firm run each plant in the month of Ma so that the production cost is minimised while still meeting the market demand. Solution : Suppose that the firm runs the plant P for das and plant Q for das in the month of Ma in order to meet the market demand. The per da operating cost of plantp is Rs Therefore, for das the operating cost will be Rs The per da operating cost of plant Q is Rs.000. Therefore, for das the operating cost will be Rs.000. Thus the total operating cost of two plants is given b : z = () MATHEMATICS 0 Get Discount Coupons for our Coaching institute and FREE Stud Material at

14 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I Plant P produces 000 bottles of soft drink A per da. Therefore, in das plant P will produce 000 bottles of soft drink A. Plant Q produces 000 bottles of soft drink A per da. Therefore, in das plant Q will produce 000 bottles of soft drink A. Total production of soft drink A in the supposed period is But there will be a demand for 000 bottles of this soft drink, so the total production of this soft drink must be greater than or equal to this demand or + () Similarl, for the other two soft drinks, we have the constraints or + 6 () and or + () and are non-negative being the number of das, so 0, 0 (5) Thus our problem is to find and which Minimize z = (objective function) subject to the conditions (constraints) and 0, 0 Eample 6. A firm manufactures two tpes of productsa and B and sells them at a profit of Rs. on tpe A and Rs. on tpe B. Each product is processed on two machines G and H. Tpe A requires one minute of processing time ong and minutes on H, tpe B requires one minute on G and one minute on H. The machine G is available for not more than 6 hours and 0 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

15 Get Discount Coupons for our Coaching institute and FREE Stud Material at 0 minutes while machine H is available for 0 hours during one working da. Formulate the problem as a linear programming problem so as to maimise profit. Solution : Let be the number of products of tpe A and be the number of products of tpe B. The given information in the problem can sstematicall be arranged in the form of following table : MDULE - I Machine Processing time of the products Available time (in minute) Tpe A ( units) Tpe B ( units) G 00 H 600 Profit per unit Rs. Rs. (in minute) Since the profit on tpe A is Rs. per product, so the profit on selling units of tpe A will be. Similarl, the profit on selling units of tpe B will be. Therefore, total profit on selling units of tpe A and units of tpe B is given b z = + (objective function) () Since machine G takes minute time on tpea and minute time on tpeb, therefore, the total number of minutes required on machineg is given b + But the machine G is not available for more than 6 hours and 0 minutes (i.e., 00 minutes). Therefore, + 00 () Similarl, the total number of minutes required on machineh is given b + Also, the machine H is available for 0 hours (i.e., 600 minutes). Therefore, () Since, it is not possible to produce negative quantities, so 0, 0 () MATHEMATICS 0 Get Discount Coupons for our Coaching institute and FREE Stud Material at

16 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I Thus, the problem is to find and which Maimize z = + (objective function) subject to the conditions , 0 Eample 6. A furniture manufacturer makes two tpes of sofas sofa of tpea and sofa of tpe B. For simplicit, divide the production process into three distinct operations, sa carpentar, finishing and upholster. The amount of labour required for each operation varies. Manufacture of a sofa of tpea requires 6 hours of carpentar, hour of finishing and hours of upholster. Manufacture of a sofa of tpe B requires hours of carpentar, hour of finishing and 6 hours of upholster. wing to limited availabilit of skilled labour as well as of tools and equipment, the factor has available each da 96 man hours of carpentar, 8 man hours for finishing and 7 man hours for upholster. The profit per sofa of tpea is Rs.80 and the profit per sofa of tpe B is Rs. 70. How man sofas of tpe A and tpe B should be produced each da in order to maimise the profit? Formulate the problems as linear programming problem. Solution : The different operations and the availabilit of man hours for each operation can be put in the following tabular form : perations Sofa of tpe A Sofa of tpe B Available labour Carpentar 6 hours hours 96 man hours Finishing hour hour 8 man hours Upholster hours 6 hours 7 man hours Profit Rs. 80 Rs. 70 Let be the number of sofas of tpe A and be the number of sofas of tpe B. In order to achieve a large profit, one need onl to manufacture a large number of sofas of tpe A and tpe B. But, owing to restricted availabilit of tools and labour, the factor cannot manufacture an unlimited quantit of furniture. Each row of the chart gives one restriction. The first row sas that the amount of carpentar required is 6 hours for each sofa of tpea and hours for each sofa of tpeb. Further, onl 96 man hours of carpentar are available per da. We can compute the total number of man hours of carpentar required per da to produce sofas of tpe A and sofas of tpe B as follows : 0 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

17 Get Discount Coupons for our Coaching institute and FREE Stud Material at Number of man - hours per da of carpentar = { (Number of hours carpentar per sofa of tpe A) (Number of sofas of tpe A) } + { (Number of hours carpentar per sofa of tpe B) (Number of sofas of tpe B) } MDULE - I = 6 + The requirement that at most 96 man hours of carpentar per da means or + < () Similarl, second and third row of the chart give the restrictions on finishing and upholster respectivel as + < 8 and () or + < 6 () Since, the number of the sofas cannot be negative, therefore > 0, > 0 () Now, the profit comes from two sources, that is, sofas of tpea and sofas of tpe B. Therefore, Profit = (Profit from sofas of tpe A) + (Profit from sofas of tpe B) = { (Profit per sofa of tpe A) (Number of sofas of tpe A) } + { (Profit per sofa of tpe B) (Number of sofas of tpe B) } z = (objective function) (5) Thus, the problem is to find and which Maimize z = (objective function) subject to the constraints (Constraints) 0, 0 MATHEMATICS 05 Get Discount Coupons for our Coaching institute and FREE Stud Material at

18 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I CHECK YUR PRGRESS 6.. A compan is producing two products A and B. Each product is processed on two machines G and H. Tpe A requires hours of processing time on G and hours on H; tpe B requires hours of processing time time on G and 5 hours onh. The available time is 8 hours and hours for operations on G and H respectivel. The products A and B can be sold at the profit of Rs. and Rs. 8 per unit respectivel. Formulate the problem as a linear programming problem.. A furniture dealer deals in onl two items, tables and chairs. He has Rs to invest and a space to store at most 60 pieces. A table costs him Rs. 50 and a chair Rs. 50. He can sell a table at a profit of Rs. 50 and a chair at a profit of Rs. 5. Assuming, he can sell all the items that he bus, how should he invest his mone in order that ma maimize his profit? Formulate a linear programming problem.. A dair has its two plants one located at P and the other at Q. Each plant produces two tpes of products A and B in kg packets. The capacit of two plants in number of packets per da are as follows: Plants Products P Q A B A market surve indicates that during the month of April, there will be a demand for 0000 packets of A and 6000 packets of B. The operating cost per da of running plants P and Q are respectivel Rs.000 and Rs How man das should the firm run each plant in the month of April so that the production cost is minimized while still meeting the market demand? Formulate a Linear programming problem.. A factor manufactures two articles A and B. To manufacture the article A, a certain machine has to be worked for hour and 0 minutes and in addition a craftsman has to work for hours. To manufacture the article B, the machine has to be worked for hours and 0 minutes and in addition the craftsman has to work for hour and 0 minutes. In a week the factor can avail of 80 hours of machine time and 70 hours of craftsman s time. The profit on each article A is Rs.5 and that on each article B is Rs.. If all the articles produced can be sold awa, find how man of each kind should be produced to earn the maimum profit per week. Formulate the problem as a linear programming problem. 6.8 GEMETRIC APPRACH F LINEAR PRGRAMMING PRBLEM Let us consider a simple problem in two variables and. Find and which satisf the following equations + = + = 06 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

19 Get Discount Coupons for our Coaching institute and FREE Stud Material at Solving these equations, we get = and =. What happens when the number of equations and variables are more? Can we find a unique solution for such sstem of equations? However, a unique solution for a set of simultaneous equations inn-variables can be obtained if there are eactl n-relations. What will happen when the number of relations is greater than or less then n? A unique solution will not eist, but a number of trial solutions can be found. Again, if the number of relations are greater than or less than the number of variables involved and the relation are in the form of inequalities. Can we find a solution for such a sstem? Whenever the analsis of a problem leads to minimising or maimising a linear epression in which the variable must obe a collection of linear inequalities, a solution ma be obtained using linear programming techniques. ne wa to solve linear programming problems that involve onl two variables is geometric approach calledgraphical solution of the linear programming problem. 6.9 SLUTIN F LINEAR PRGRAMMING PRBLEMS In the previous section we have seen the problems in which the number of relations are not equal to the number of variables and man of the relations are in the form of inequation (i.e., or ) to maimise (or minimise) a linear function of the variables subject to such conditions. Now the question is how one can find a solution for such problems? To answer this questions, let us consider the sstem of equations and inequations (or inequalities). ' ' Fig. 6. > 0 ' > 0 ' Fig. 6. MDULE - I We know that > 0 represents a region ling towards the right of - ais including the - ais. Similarl, the region represented b > 0, lies above the - ais including the -ais. The question arises: what region will be represented b > 0 and > 0 simultaneousl. MATHEMATICS 07 Get Discount Coupons for our Coaching institute and FREE Stud Material at

20 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I > 0, > 0 ' ' Fig. 6. bviousl, the region given b > 0, > 0 will consist of those points which are common to both > 0 and > 0. It is the first quadrant of the plane. Net, we consider the graph of the equation + < 8. For this, first we draw the line + = 8 and then find the region satisfing + < 8. Usuall we choose = 0 and calculate the corresponding value of and choose = 0 and calculate the corresponding value of to obtain two sets of values (This method fails, if the line is parallel to either of the aes or passes through the origin. In that case, we choose an arbitrar value for and choose so as to satisf the equation). Plotting the points (0,) and (8,0) and joining them b a straight line, we obtain the graph of the line as given in the Fig. 6. below. ' ' B(0,) + = 8 Fig. 6. A (8,0) 08 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

21 Get Discount Coupons for our Coaching institute and FREE Stud Material at We have alread seen that 0 and 0 represents the first quadrant. The graph given b + < 8 lies towards that side of the line + = 8 in which the origin is situated because an point in this region will satisf the inequalit. Hence the shaded region in the Fig. 6.5 represents 0, 0 and + 8 simultaneousl. Similarl, if we have to consider the regions bounded b 0, 0 and + 8, then it will lie in the first quadrant and on that side of the line + = 8 in which the origin is not located. The graph is shown b the shaded region, in Fig. 6.6 The shaded region in which all the given constraints are satisfied is called the feasible region Feasible Solution A set of values of the variables of a linear programming problem which satisfies the set of constraints and the non-negative restrictions is called a feasible solution of the problem ptimal Solution A feasible solution of a linear programming problem which optimises its objective functions is called the optimal solution of the problem. Note : If none of the feasible solutions maimise (or minimise) the objective function, or if there are no feasible solutions, then the linear programming problem has no solution. In order to find a graphical solution of the linear programming problem, following steps be emploed. Step : Formulate the linear programming problem. ' (0,) ' ' (0,) ' + = 8 Fig =8 Fig. 6.5 (8,0) (8,0) MDULE - I MATHEMATICS 09 Get Discount Coupons for our Coaching institute and FREE Stud Material at

22 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I Step : Graph the constraints (inequalities), b the method discussed above. Step : Identif the feasible region which satisfies all the constraints simultaneousl. For less than or equal to constraints the region is generall below the lines and for greater than or equal to constraints, the region is above the lines. Step : Locate the solution points on the feasible region. These points alwas occur at the verte of the feasible region. Step 5 : Evaluate the objective function at each of the verte (corner point) Step 6 : Identif the optimum value of the objective function. Eample 6. Minimise the quantit z = + subject to the constraints + 0, 0 Solution : The objective function to be minimised is z = + subject to the constraints + 0, 0 First of all we draw the graphs of these inequalities, which is as follows : B(0,) + = A(,0) Fig MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

23 Get Discount Coupons for our Coaching institute and FREE Stud Material at As we have discussed earlier that the region satisfied b 0 and 0 is the first quadrant and the region satisfied b the line + along with 0, 0 will be on that side of the line + = in which the origin is not located. Hence, the shaded region is our feasible solution because ever point in this region satisfies all the constraints. Now, we have to find optimal solution. The verte of the feasible region area (,0) and B (0,). The value of z at A = The value of z at B = Take an other point in the feasible region sa (,), (,0), (0,) etc. We see that the value of z is minimum at A (,0). MDULE - I Eample 6. Minimise the quantit z = + subject to the constraints + + 0, 0 Solution : The objective function to be minimised is z = + subject to the constraints + + 0, 0 First of all we draw the graphs of these inequalities (as discussed earlier) which is as follows : The shaded region is the feasible region. Ever point in the region satisfies all the mathematical inequalities and hence the feasible solution. E(0,) D( 0, 0.75) C( 0.5, 0.5) A(,0) + = B(.5,0) Fig = MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

24 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I Now, we have to find the optimal solution. The value of z at B (.5,0 ) is.5 The value of z at C ( 0.5, 0.5 ) is.5 The value of z at E (0,) is If we take an point on the line + = between B and C we will get and elsewhere in the feasible region greater than. f course, the reason an feasible point (betweenb and C) on + = minimizes the objective function (equation) z = + is that the two lines are parallel (both have slope ). Thus this linear programming problem has infinitel man solutions and two of them occur at the vertices. Eample 6.5 Maimise z = subject to the constraints , 0 Solution : The objective function is to maimise z = subject to the constraints , 0 First of all we draw the graphs of these inequalities, which is as follows : MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

25 Get Discount Coupons for our Coaching institute and FREE Stud Material at The shaded region ABC is the feasible region. Ever point in the region satisfies all the mathematical inequations and hence the feasible solutions. Now, we have to find the optimal solution. The value of z at A (60, 0) is 0.00 The value of z at B (90, 05) is D(0,0) C(0,50) The value of z at C (0, 50) is A(60,0) The value of z at (0, 0) is 0. + = 00 If we take an other value from the feasible region sa (60, 0), (80, 80) Fig = 80 etc. we see that still the maimum value is obtained at the verte B (90, 05) of the feasible region. Note : For an linear programming problem that has a solution, the following general rule is true. If a linear programming problem has a solution it is located at a verte of the feasible region. If a linear programming problem has multiple solutions, at least one of them is located at a verte of the feasible region. In either case, the value of the objective function is unique. Eample 6.6 In a small scale industr a manufacturer produces two tpes of book cases. The first tpe of book case requires hours on machine A and hours on machines B for completion, whereas the second tpe of book case requires hours on machinea and hours on machine B. The machine A can run at the most for 8 hours while the machineb for at the most hours per da. He earns a profit of Rs. 0 on each book case of the first tpe and Rs. 0 on each book case of the second tpe. How man book cases of each tpe should he make each da so as to have a maimum porfit? Solution : Let be the number of first tpe book cases and be the number of second tpe book cases that the manufacturer will produce each da. Since and are the number of book cases so B(90,05) E(00,0) MDULE - I 0, 0 () Since the first tpe of book case requires hours on machine A, therefore, book cases of first tpe will require hours on machine A. second tpe of book case also requires hours MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

26 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I on machine A, therefore, book cases of second tpe will require hours on machine A. But the working capacit of machine A is at most 8 hours per da, so we have + 8 or + < 6 () Similarl, on the machine B, first tpe of book case takes hours and second tpe of book case takes hours for completion and the machine has the working capacit of hours per da, so we have + < () Profit per da is given b z = () Now, we have to determine and such that Maimize z = (objective function) subject to the conditions , 0 constraints We use the graphical method to find the solution of the problem. First of all we draw the graphs of these inequalities, which is as follows : C(0, ) (0,6) B(,) (0,0) A(6,0) (7,0) + = Fig = 6 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

27 Get Discount Coupons for our Coaching institute and FREE Stud Material at The shaded region ABC is the feasible region. Ever point in the region satisfies all the mathematical inequations and hence known as feasible solution. We know that the optimal solution will be obtained at the vertices (0, 0), A (6, 0). B (,). Since the co-ordinates of C are not integers so we don t consider this point. Co-ordinates ofb are calculated as the intersection of the two lines. Now the profit at is zero. Profit at A = = 80 Profit at B = = = 00 Thus the small scale manufacturer gains the maimum profit of Rs.00 if he prepares first tpe book cases and second tpe book cases. Eample 6.7 Solve Eample 6.0 b graphical method. Solution : From Eample 6.0 we have (minimise) z = (objective function) subject to the constraints (constraints) 0, 0 First we observe that an point (, ) ling in the first quadrant clearl satisfies the constraints 0, and 0, Now, we plot the bounding lines + = B(0,) MDULE - I or 8 + = An point on or above the line + = satisfies the constraint + (Fig. 6.) A(8,0) Fig. 6. MATHEMATICS 5 Get Discount Coupons for our Coaching institute and FREE Stud Material at

28 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I Similarl an point on or above the line 6 + = satisfies the constraint 6 + (See Fig. 6. ) B(0,6) A(6,0) Again an point on or above the line + = satisfies the constraints + (See Fig. 6. ) Thus, combining all the above figures we get, the common shaded unbounded region. (See Fig. 6.) B(0,8) D(0,) (0,6) C(,) Fig. 6. Fig. 6. A(,0) (0,8) B(,) (8,0) (6,0) + = Fig. 6. A(,0) + = + = 6 6 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

29 Get Discount Coupons for our Coaching institute and FREE Stud Material at Now the minimum value of z = is at one of the points A (,0), B(,), C(,) and D(0,) At A : z = =,000 At B : z = = 88,000 At C : z = = 7,000 At D : z = = 96,000 MDULE - I Thus, we see that z is minimum at C (, ) where = and =. Hence, for minimum cost the firm should run plantp for das and plant Q for das. The minimum cost will be Rs.7, Eample 6.8 Maimize the quantit z = + subject to the constraints +, 0, 0 Solutions : First we graph the constraints +, 0, 0 The shaded portion is the set of feasible solution. Now, we have to maimize the objective function. The value of z at A(, 0) is. The value of z at B(0, ) is. If we take the value of z at an other point Fig. 6.5 from the feasible region, sa (, ) or (, ) or (5, ) etc, then we notice that ever time we can find another point which gives the larger value than the previous one. Hence, there is no feasible point that will makez largest. Since there is no feasible point that makesz largest, we conclude that this linear programming problem has no solution. Eample 6.9 Solve the following problem graphicall. Minimize z = 0 subject to the constraints MATHEMATICS 7 B(0,) A(,0) Get Discount Coupons for our Coaching institute and FREE Stud Material at

30 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I , 0 Solution : First we graph the constraints , 0 or 0, 5 5 = 0 The shaded region is the feasible region. 5 5 A, (0,) + 5 = 5 Fig. 6.6 = 0 Here, we see that the feasible region is unbounded from one side. feasible region But it is clear from Fig. 6.6 that the objective function attains its minimum value at the pointa which is the point of intersection of the two lines = 0 and + 5 = 5. 5 Solving these we get = = 5 5 Hence, z is minimum when =, =, and its minimum value is = 0. Note : If we want to find ma. z with these constraints then it is not possible in this case because the feasible region is unbounded from one side. 8 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

31 Get Discount Coupons for our Coaching institute and FREE Stud Material at CHECK YUR PRGRESS 6. Solve the following problems graphicall. Maimize z = +. Maimize = + subject to the conditions subject to the conditions MDULE - I , , 0. Minimize z = Maimize z = subject to the conditions , 0 subject to the conditions +, , 0, 0 5. Maimize z = Minimize z = subject to the conditions , 0 LET US SUM UP subject to the conditions , 0 A statement involving a sign of inequalit like, <, >,,, is called an inequation. The equation a + b + c = 0 is a straight line which divides the -plane into two half planes which are represented b a + b + c > 0 and a + b + c < 0 B the term, solution of a sstem of simultaneous linear inequations we mean, finding all values of the ordered pairs (,) for which each linear inequation of the sstem is satisfied. Linear programming is a technique followed b mathematicians to solve the optimisation problems. MATHEMATICS 9 Get Discount Coupons for our Coaching institute and FREE Stud Material at

32 Get Discount Coupons for our Coaching institute and FREE Stud Material at MDULE - I A set of values of the variables of a linear programming problem which satisfies the set of constraints and the non-negative restrictions is called a feasible solution. A feasible solution of a linear programming problem which optimises its objective function is called the ptimal solution of the problem. The optimal solution of a linear programming problem is located at a verte of the set of feasible region. If a linear programming problem has multiple solutions, at least one of them is located at a verte of the set of feasible region. But in all the cases the value of the objective function remains the same. SUPPRTIVE WEB SITES TERMINAL EXERCISE Solve each of the following inequations graphicall :. >. <.. <. > 5. 5 > 6. 5 <. 7. < > < 0. Solve each of the following sstems of linear inequations in two variables graphicall.. < <, < <.. + < 6, + < < 50,. + <, + < 6 + < 80 + < 0, > 0, > 0 < 5, > 0, > >, < < 5, < > 0, > 0 6. A dealer has Rs. 500 onl for a purchase of rice and wheat. A bag of rice costs Rs. 50 and a bag of wheat costs Rs. 0. He has a storage capacit of ten bags onl and the dealer gets a profit of Rs. and Rs. 8 per bag of rice and wheat respectivel. Formulate the problem as a linear programming problem to get the maimum profit. 0 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

33 Get Discount Coupons for our Coaching institute and FREE Stud Material at 7. A business man has Rs at his disposal and wants to purchase cows and buffaloes to take up a business. The cost price of a cow is Rs. 900 and that of a buffalo is Rs.00. The man can store fodder for the live stock to the etent of 0 quintals per week. A cow gives 0 litres of milk and buffalo gives 0 litres of milk per da. Profit per litre of milk of cow is 50 paise and per litre of the milk of a buffalo is 90 paise. If the consumption of fodder per cow is quintal and per buffalo is quintals a week, formulate the problem as a linear programming problem to find the number of live stock of each kind the man has to purchase so as to get maimum profit (assuming that he can sell all the quantit of milk, he gets from the livestock) 8. A factor manufactures two tpes of soaps each with the help of two machinesa and B. A is operated for two minutes andb for minutes to manufacture the first tpe, while the second tpe is manufactured b operating A for minutes and B for 5 minutes. Each machine can be used for at most 8 hours on an da. The two tpes of soaps are sold at a profit of 5 paise and 50 paise each respectivel. How man soaps of each tpe should the factor produce in a da so as to maimize the profit (assuming that the manufacturer can sell all the soaps he can manufacture). Formulate the problem as a linear programming problem. 9. Determine two non-negative rational numbers such that their sum is maimum provided that their difference eceeds four and three times the first number plus the second should be less than or equal to 9. Formulate the problem as a linear programming problem. 0. Vitamins A and B are found in two different foods E and F. ne unit of food E contains units of vitamina and units of vitaminb. ne unit of foodf contains units of vitamin A and units of vitamin B. ne unit of food E and F costs Rs.5 and Rs..50 respectivel. The minimum dail requirements for a person of vitamina and B is 0 units and 50 units respectivel. Assuming that anthing in ecess of dail minimum requirement of vitamina and B is not harmful, find out the optimal miture of foode and F at the minimum cost which meets the dail minimum requirement of vitamina and B. Formulate this as a linear programming problem.. A dealer has Rs.500 onl for the purchase of rice and wheat. A bag of rice costs Rs. 80 and a bag of wheat costs Rs. 0. He has a storage capacit of 0 bags onl and the dealer gets a profit of Rs. and Rs. 8 per bag of rice and wheat respectivel. How he should spend his mone in order to get maimum profit?. A machine producing either producta or B can produce A b using units of chemicals and unit of a compound and can produceb b using unit of chemicals and units of the compound. nl 800 units of chemicals and 000 units of the compound are available. The profits available per unit of A and B are respectivel Rs. 0 and Rs.0. Find the optimum allocation of units betweena andb to maimise the total profit. Find the maimum profit. MDULE - I MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at