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LINEAR PROGRAMMING

CONTENTS

~ Introduction 1

~ Linear Inequalities in two

Variables 1

Exercise 1 3~ Linear Programming 4

~ Convex Set of Feasible Solutions

4

~ Corner Point Method 5

Exercise 2 7

~ Answers to exercises 8

~ Solved Problems 9

i'~ Answers 16

SYLLABUS

eBSE: Introduction. definition of related

terminology such as constraints, ()~jective

function. ,optimization, difJerent types of linear

programming (L.P.) problems. mathematical. .formulation of L.P. problems. graphical method of

solution fOr problems in two variahles.feasihle

and infeasible regions. feasible and infeasible

solutions, optional ftasible solutions (up to three

non-trivial constraim).

Price Rs.: 300.00

.- '----

Introduction

A relation of the type f(x, y) > 0, f(x, y) < 0, f(x, y) ~ 0, or f(x, y) s 0 is called an inequality or an

inequation. It is also called a constraint or a condition. If f(x, y) is linear in x and y, then the

inequality is called a linear inequality or a linear inequation. For example

3x + 2y - 4 > 0, ax + by s c

are linear inequations in x and y. The solution set of a linear inequation is the set of all values of

(x, y) which satisfy the given inequation.

The above examples. involve two variables, namely x and y and the inequations are called linear

inequations in two variables x and y.

The relations ax + b s 0 or cx + d ~ 0 are linear inequations in one variable x.

Linear inequations in two variables

A person has Rs. 200/- and wants to buy some pens and pencils. The cost of a pen is Rs. 16/-

and that of a pencil is Rs. 6/-. If x denotes the number of pens and y, the number of pencils, then

the total amount spent is Rs. (16 x + 6 Y), and we must have

16x+6ys200. ...(i).

In this example x and yare whole numbers and can not be fractions or negative numbers. In this

case we find the pair of values of x and y which make the statement (i) true. The set of such pairs

is the solution set of (i).

To start with, let x = 0 so that

100 16y s 200 or y s - = 33 -.

3 3

As such the values of y corresponding to x = 0, can be 0, 1, 2, 3, .... , 33. Hence the

solutions of (i) are (0, 0), (0, 1), (0, 2) ,..... , (0, 33).

Similarly, the solutions corresponding to x = 1, 2, 3, ... , 12 are

(1,0), (1,1), , (1, 30),

(2,0), (2, 1), , (2,28),

(11,0), (11,1), ... , (11. 4),

(12,0), (12, 1).

We note that the values of xand y cannot be more than 12 and 33 respectively. We also note that

some of the pairs namely (5, 20), (8, 12), and (11, 4) satisfy the equation

16x + 6y = 200 which is a part of the given inequation.Let us now extend the domain of x and y from whole Y

number to the real numbers. Consider the equation

16 x + 6 y = 200and then draw the straight line represented by it. This

line divides the coordinate plane in two half planes.

For (0, 0), (i) yields, 0 s 200 which is true ,and hence

we conclude that (0, 0) belongs to the half plane X

represented by (i). Hence the solution of (i) consists of

all the points belonging to the shaded half plane which

consists of infinite number of points.

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Note:

In order to identify the half plane represented by an inequa~ion, it is sufficie t to take any known

point (not lying on the line) and check whether it satisfies the given inequatio I or not. If it satisfies,then the inequation represent that half plane, containing the known point, otherwise the

inequation represents the other half plane. . lIllustration 1: Draw the graph of the solution region S~tiSfied by the in-equa ion 3x + 5y +6 > O.

(0, 1.2)

ion may be

,(L2,0)

~

The boundary line is given by the

equality 3x + 5y + 6 = o. It is astraight line which meets the x-axis

at (-2, 0) and the y-axis at (0, 1.2).

Substituting (0, 0) in the given

expression, we have

0+0+6>0

which is true. Hence, the origin is a

point in the solution region. The

solution region is, thus, on the

same side of the line as the origin.

Since equality is not included in the

given inequation, the points on the

boundary of the line are not part of

the solution. The solution set is

shown in the figure.

For a set of linear inequations, in two variables, the solution re

(i) a closed region inside a polygon (bounded by straight lines)

(ii) an unbounded region (bounded partly by straight lines),

or (iii) empty.

Solution:

Illustration 2. Draw the graph of the solution region satisfied by the inequaliti s

2x + Y 5:4, Y 5:2, x ;? O.

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So.lution: The solution region is bounded by the straight lines

2x + y = 4 , (i)

y=2, , (ii)

'x = O. (iii)

. The straight line (i) meets the coordinate axes at (2, 0) and (0 4). Moreover, for

(0, 0), Jhe inequation 2x + y ~ 4 gives 0 ~4, which is true. Hehce the half plane

represented by 2x + y.~ 4, contains the origin (0, 0).

Also for (0, 0), y ~ 2 gives 0 ~ 2 which is

true, so that y ~ 2 also contains the origin.

With x ~ 0, the shaded region is the

required solution region. All the boul)dary

lines are part of the solution region.

Moreover, the solution region is an

unbounded region, bounded by the three

boundary lines.

Linear Programming P.1214-P7-CBSE.MA-LP-3

Illustration 3. Draw the graph of the solution region satisfied by the inequations

x + y ~ 1, 2x + y ~4, x ~O, y ~O.

Solution: The solution region is bounded by the straight

lines

x + y = 1, (1)2x + y = 4, (2)

x = 0, (3)

Y = O. (4)The straight line (1) meets the coordinates

axes at (1, 0) and (0, 1). For (0, 0),

x + y ~ 1 gives 0 ~ 1 which is not true.

Hence (0, 0) does not lie in the half plane represented by x + y ~ 1. The line (2)

meets the coordinate axes in (2, 0) and (0, 4) and 0 ~ 4.Hence (0, 0) lies in the

half plane 2x + y ~ 4. Hence the region bounded by x + Y ~ 1 and 2x + y ~ 4

belongs to either the fourth quadrant, or the first quadrant or the second

quadrant. But the point belonging to x ~ 0, y ~ 0, all lie in the third quadrant.

Hence no points satisfies all the four given inequation.

Hence the solution set is empty.

Illustration 4. Find the solution region satisfied by the inequalities 2x + y ~ 4, 3x + 3y ~1,

x - Y ~ 1, x ~ 0, y ~ 3.

y=3

Solution: The solution region is bounded

by the straight lines

2x+y=4, (1)

3x + 3y = 1, (2)

x - Y = h (3)x=O, (4)

Y = 3. (5)The points where the first three lines

meet the x-axis are (2, 0), (1/3, 0), (1, 0).

The points where the first three lines meet the y-axis are (0, 4),

(0, 1/3), (0, -1). Moreover, (0, 0) belongs to the half planes 2x + y ~ 4 and x - y ~

1, Y ~ 3 buf not to the half plane 3x + 3y ~ 1.

The solution region is the shaded region. All the boundary lines are part of the

solution region and the closed region inside a polygon.

Exercise 1.

Draw the graphs of the region satisfied by the solutions of

i) 2x+ 3y~3,

ii) 2x + Y > 0, x + 2y ~ 2,

iii) x-y~1, x+y~1,y>1,

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Linear Programming:

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Problems which seek to maximize or minimize a numerical function of a nu ber of variables (or

functions), where the variables (functions) are subject to certain co~straints are called

optimization or programming problems. The function which is to be optimized is called the

objective function; (no constant term appears in the objective function). Pr gramming problems

deal with determining optimal allocation of limited resources to meet the give objectives.

Linear programming deals with that class of optimization problems for WhiClall relations among

the variables are linea,r. The relations are linear both in constra,intsand in t e objective tunction, w~ich is to be optimized. .

Let X1, X2. ... ,Xr be r variables. Consider m inequalities (m may be greater tha ,equal to, or less

than r) of the form

a11x1 + a12X2 + + a1rX1

a21X1 + a22X2 + + a2rx2

{~, = ,:5:} b1{~, = , :5:} b2

{~, = ,:5:} br

... (1)

.. , (2)

... (m)

where for each constra,ints, only one of the signs ~,= ,:5: holds, but the sign may vary from one

constraint to another. Associated with these m constraints, are usually the con traints Xj ~ O.This

assumption is obvious in real life problems, since non,e of the variables rep senting resources

like money, material, men etc can be negative. The problem is to find the vall!les of the variables '\ ;

Xj satiSfying (1) - (m) and Xj ~ 0, which maximize' (or minimize) the linear objedtive functions.

z = C1X1 + C2X2 + ... + CrXr 'where Cj are known cqnstants.

Any set ,of val.iJes Xj which satisfy constraints (1) - (m) is called the solutioIof the LPP. Anysolution which further satisfies the non-negativity condition (Xj ~ 0) is called feasible solution.

The set of all Xj, for which feasible solution exist, is called the feasibles lution region. Any

feasible solution which optimizes the objective function is called an optima feasible solution.

Solving an LPP means finding an optimal feasible solution.

Convex set offeasible solutions:

Let the region of feasible solutions be a closed region. Then the set of points i this region is said

to be convex set; if the line joining any two points in this region lies entirely In this region. The

region shown in figure 1 is a convex region, whereas that is figure 2 is not a co vex region.

fig. 1

convex region'

fig.'2

not a convex region

This means that in the convex region there are no holes and there are no cuts ( r indentations) in

the boundary. In tne 2-dimensional case, the boundaries are straight lines. Ther are vertices and

edges (sides) of a polygon.

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Linear Programming

_Corner point Method:

P-1214-P7-CBSE-MA-LP.5

Let the feasible solution of a LPP be a convex polygon. Then the maximum or minimum values of

a linear function in two variables, occurs at some vertex of the polygon.

Illustration -5: Find the point at which the maximum value of z = 3x + 2y, subject to theconstraints

x + y ~2, x cO, Y cO, occurs.

Solution: Equation corresponding to x + Y s;2 is

x + y = 2 and the inequation x + y s; 2

represents the shaded region (since (0, 0)

satisfies x + y s;2).

Hence z = 3x+2y will have maximum value at

0, A orB.

/- Z]A = 3x + 2Y](2,0) = 6, z]s = 3x + 2y]{o,2)= 4, z]o = o.

Maximum value of 3x + 2y is, thus, 6 which

occurs at (2, 0).

Remarks:

x

(i) The feasible solutions of LPP (with two variables) is either a region inside a convex

polygon or a semi-infinite region bounded by straight lines.

(ii) If the optimal value of the objective function is finite, then the value occurs at atleast one

of the vertices of the convex polygon of the feasible region. If this value is finite and is

unique, then this value occurs only at one of the vertices of the polygon.

(iii) If the optimal value is not unique, then there are points other thana vertex at which the

value of the objective function is also optimal. If the optimal value is same at two

consecutive vertices of a convex polygon, then this optimal value occurs at every point on

the boundary line joining these two vertices:(iv) When the LPP has an unbounded solution, then this does not occur at any of the corners

of the feasible region; the unbounded solutions cannot lead to optimal values of the

objective function.(v) The LPP may have a bounded solution, even when the feasible solution region is

unbounded.

Illustration -6: Find the maximum value of z = 2x + y subject to the constraints2x + 3y ~ 30, x -2y ~ 8, xeD, yeO.

Solution: The corresponding equations are Y2x + 3y = 30 and x -2y =8, which

intersect the axes at A (15, 0),

B(O, 10), C(8, 0), 0(0, -4)

respectively.

Both the inequalities represent the

region containing the origin.

Shaded region is the region of

solution. This is convex, so that

maximum Z occurs at either of 0,

C, E, B; where E is the point of

intersection of the two lines.

E == (12, 2) and z]o = 0,

z]c = 2x + Y](S,O)= 2 x 8 + 0 = 16,

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Illustration -7:

Z)E = 2x + Y)(12, 2) = 26,

z)s = 2x + Y)(O, 10) = 10,Hence the maximum value of Z :: 26 which occurs at E (12, 2).

Find the minimum value of z = 2x + 7y subject to the conJraints '4x + 5y ~ 20,6x + 2y ~ 10, x, y ~ o.

ySolution: Corresponding equations are

4x + 5y = 20 and 6x + 2y = 10 which

intersect the axes at A (5, 0), S (0, 4);

C(~, 0). 0 (0, 5).Origin doesn't satisfy any of the two

inequations, so the region of solution

is shaded. This is YOPAX.

Minimum Z can occur at the vertices A

. (5 40)orPorO,whereP= -, -. 11 11

5 40 290and z)o = 35, z)p = 2 x - + 7 x - = --, Z)A :: 10.

11 11 11Clearly minimum value of Z is 10 which is at A (5, 0).

x

Illustration -8: Find the maximum value of z = 4x + 2y, subject to 4x + 2 s 46, x + 3y s 24, "x, y~O.

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x

X2

Xl -X2 =-1

X1 -X2 = -1 => -X1 + X2 = 1The shaded region is the

region of solution.

So, maximum Z can be at A

(0, 1) or S (0, 2) or P (2, 3).

Z)A :: 2, z)s = 4, z)p = 4and the

minimum value corresponds

to A(O, 1) and the maximum

value 4 occurs on the line SP.

Solution: Clearl~ the region of solution is as Y

shown by shaded area. The vertices

are A (~3,0), 0 (0, 8), P (9, 5)23

and Z]A = 4 x - = 46, .. 2

z)o=2x8=16,

z)p :: 4 x 9 + 2 x 5 = 46. .

We can see that any point on AS 0 Alegives value of Z = 46. .

Maximum value occurs at infinitely many points lying on line s. gment AP.

Illustration .;9: Find the optimal feasible solution of the linear programming roblem; maximize

z = 2X2-X1 subject to X1-x2~-1, -0.5X,' + X2S2; X1,X2~ O.

Solution:

Linear Programming P.1214.P7.CBSE.MA.LP.7 -

lI/ustration-10: For the L.P.P.,maximumz = x + y subject to x + y::; 1, 2x + 2y 2? 6, x, Y 2? 0, find

the numberof solution(s).

Solution: There is no common area of the two

inequations. So there is no region of

solution.Hence, there is no solution to this L.P.P.

y

x

Exercise 2:

(i). Find the solution of set of constraints 2x + 5y ::;30,x + 2y ::;24 and x 2? 0,Y 2? O.

(ii). Find the maximum value of 4x + 5y subject to the constraints x + y ::;20,x + 2y ::;'5,x -3y ::;12, x 2? 0, Y 2? O.

(iii). Find the maximum value of z = 4x + 2y subject to the constraints 2 x + 3 y ::;18,x + y 2?10;x, Y 2? O.

\

From/To A B

D 7 3

E 6 4

F 3 2

Miscellaneous Exercise:

(i) 2x + 3y ::; 6, x + Y 2? 1, 2x - Y - 4 ::; 0, y 2? O.

(ii) .Pindthe maximum value ofz = 3x+2y subject to x+ 2y 2? 2, x+2y ::;8; x, y 2?0.(iii) An oil company has two depots A and B with capacities of 70001 and 40001

\

respectively. The distances (in km) between the depots and the petrol pumps is given

in the following table:

The vitamin contents of one ka food is given below:

Food Vitamin A Vitamin B Vitamin C

X 1 2 3

Y 2 2 1

Assuming that the transportation cost of 10 /itres of oil is Rs. 1 per km, how should the

.delivery be scheduled in order that the transportation cost is minimum ? What is the

minimum cost?(iv) A dietician wishes to mix together two kinds of food X and Y in such a way that the

mixture contains atleast 10 units of vitamin A, 12 units of vitamin_Band 8 units of

vitamin C.

One kg of food X costs Rs. 16 and one kg of food Y costs Rs. 20. Find the least cost of

the mixture which will produce the required diet?

(v) An aeroplane can carry a maximum of 200passengers. A profit of Rs. 1000is made on

each executive class ticket and a profit of Rs. 600 is made on each economy class

ticket. The airline reserves at least 20 seats for executive class. However at least 4.times as many passengers prefer to travel by economy class than by the executive

class. Determine how many tickets of each type must be sold in order to maximize the

profit for the airline. What is the maximum profit?

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Exercise 1:

i)

ii)

(iii) Empty set or'null set.

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ANSWERS TO EXERCISES

Exercise 2:

(i) does not exist

(ii) 95

(iii) does not exist (unbounded)

Miscellaneous Exercise:

(i)

(ii) 24

l {"r.

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Linear Programmlng- P.1214.P7.CBSE-MA.LP.9

'.SOLVED PROBLEMS

Problem 1: Solve for x, the following system of inequations:

x + 2y.5'3, 3x + 4y;? 12, X;?0, y;? 1.

x=O

(0,3/2)

The solution region is bounded by the

straight lines

x+2y =3, (1)

3x + 4y = 12, (2)

x = 0, (3)

y=1.. .. (4)

The straight lines (1) and (2) meet the

x-axis in (3, 0) and (4, 0) and for (0, 0),

x + 2y ~ 3 => ~3, which is true.Hence (0, 0) lies in the half plane x + 2y ~ 3. Also the lines (1) and (2) meet the

Solution:

y-axis in

(0, 3/2) and (0, 3) and for (0, 0)

3x+4y~12

=> 0 ~ 12 which is not true. Hence (0, 0) doesn't belong to the half plane

3x + 4y ~ 12. Also x ~ 0, y ~~1 => that the solution set belongs to the first

quadrant. Moreover all the boundary lines are part of the solution. From the

shaded region, we find that there is no solution of the given system. Hence the

solution set is an empty set.

Problem 2. Draw the graph for the: solution region satisfied by the inequalities

5x + - Y ~ 5, x + 2y ~ 1, x - Y ~ 4, x ~ 0, y ~ 0.

2

Solution:. The solution regiorr lies in the first quadrant

and is bounded by the straight lines

5x+2"Y:= 5, (1)

x+2y:= 1, (2)

x- Y = 4\. (3)

The line (1) meets the coordinate axes at (5, 0)

and (0, 2). For (0, 0), x+~y ~ 5 => 0 ~ 5 which2

is true.

The line (2) meets the coordinate axes at (1, 0) and (0, 1/2). For (0, 0), 0 ~ 1

which is 110ttrue. Also the line (3) meets the coordinate axes at (4, 0) and (0, -4).

For (0, 0), 0 ~ 4 which is true.

The solUition set belongs to the shaded region, which is an enclosed region

bounded by the straight lines (1), (2), (3) and x = 0, y = o.

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Problem 3. Find the solution region satisfied by the inequalities

x +y :;;5, x:;; 4, y :;;4, X;? 0, y;? 0, 5x + Y ;? 5,x + 6y ;? 6.

Solution: We find that the solution set satisfies x ~ 0,

y ~ 0, x::;; 4, y::;; 4 so that the'solution region

lies within the square enclosed by the lines

x = 0, y =0, x = 4,

Y = 4. Moreover, the solution region is

bounded by the lines

x + y = 5, (1), (0,0)5x+y=5, (2)

x+6y=6. (3)

Line (1) meets the coordinate axes in (5, 0)

and (0, 5) and the lines x = 4 and y = 4 in

(4, 1) and (1,4), and 0 < 5 is true.

- Hence (0, 0) belongs to the half plane x + y ::;;5. But (0; 0) does not belong to the

half planes 5x + y ~ 5 and x + 6y ~ 6. The line 5x + y = 5, f.eets the coordinate

axes in (1,0) and (0, 5), and meets the line x = 4 in (4, 1), wrere as it meets the

line y = 4 in (1/5, 4). _ .

Si~ilarly x + 6y = 6 meets x = 4 in (4,1/3) and y = 4 in (-18, 4).

The solution is marked as the shaded region.

Problem 4. Draw the graph of the solution region of the inequations x ;? 0 y;? 0, X- Y ;?1,

x - y:;;-1.

Solution: Here the region belongs to the first

quadrant as x ~ 0, y ~ O.

Moreover, the region is bounded by the

straight lines

x-y=1, (1)

x - y = -1. (2)These are parallel lines.

Moreover, both 0 ~ 1 and 0 ::;;-1 are not

true. Hence the origin does not belong

to the half planes x - y ~ 1 and

x - y::;;-1

=> Both the half planes are away from

the origin and are disjointed.

Hence the solution set is empty,,--

y

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Linear Programming P-1214-P7-CBSE-MA-LP-ll

Problem 5:

Solution:

\

. .Find the maximum and minimum values of the objective function z = 3x + 4y

subject to the constraints 2x + 3y ~ 6, x + Y2 1, x 2 0, Y 2O.

The line 2x + 3y = 6 meets the

coordinate axes at (3, 0) and

(0, 2). Moreover, for the ofigin

(0, 0), 2x + 3y:s; 6 gives.

o + 0 :s;6 which is true.Hence the region of feasible

solution is on the same side of

the line as the origin (0, 0).

The line x + y ~ 1, meets thecoordinate axes at (1, 0) and

(0, 1). Also, for (0,0),0 + 0 ~ 1 (0,0)

is false. Hence the region

of feasible solution lies

on the opposite side of

the origin. With x ~ 0, y ~ 0,the feasible region is the shaded one. The vertices are at (1, 0), (3, 0), (0, 2) and

(0,1). At these vertices, the value of the objective function is:

At (1, 0); z = 3 + 0 = 3,at (3, 0); z = 9 + 0 = 9,at (0, 2), Z = 0 + 8 = 8,at (0, 1), Z =' 0 + 4 = 4.Hence the maximum value is 9 for x = 3, Y = 0, and the minimum value is 3 forx = 1, Y = O.

Problem 6: 'Maximize z = 2x + 5y subject to the constraints 2x + 5y ~ 10, x + 2y 2 1,

x-y~4;X2,Y20. \.

(0,0)

Solution: We find that the feasible

region is on the same side of

the line 2x + 5y = 10 as theorigin, on the same side of

the line x - y :s;4 as the origin

and on the opposite side of

the line x + 2y = 1 from theorigin. Moreover, the lines

meet the coordinate axes at

~,~, ~,~; 0,~,~, 1/~~ and (4, 0). The lines x - y = 4

and 2x + 5y = 10 intersect at

(370, i).Th.e values of the objective function at the vertices (5) of the pentagon are:

.. 5 5 60 10(A) 0 + 2' = 2' (8) 2 + 0 = 2 (C) 8 + 0 = 8 (D) 7 +7" = 10

(E)0+10=10

The maximum value 10 occurs at the points De70: i) and E(O, 2). Since 0 and-E are adjacent vertices, the objective function has the same maximum value 10

at all the points on the line DE.

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Problem 7:. Minimize z = 3x + 4y subject to the constraints x + y ::; 5, 5x + y;c 5, x + 6y ;c 6,x ::;4, y ::;4, x ;c 0, y ;c O.

Solution: The points where the boundary

lines_ meet the coordinate axes

are:

x-axis; (5, 0), (1, 0), (6, 0), (4, 0)

y-axis;(O, 5), (0, 5), (0, 1) ... ,(0,4).

Moreover, the position of the

feasible region and the origin with

respect to each boundary line is

(a) x + y::; 5 same side

(b) 5x + Y~ 5 opposite side

(c) x + 6y ~ 6 opposite side

(d) x::; 4 same side

(e) y::; 4 same side

The feasible region is the shaded one. The vertices of the pentagon are

(24 25) ( 1)) (1 _)A 29' 29 ,B 4'6 ,C(4,1 ,0(1,4), E 5,4 .

At - the vertices the value of the objective function is respectively,

172 38 16 19 and 83 l29,' 3" 5

Hence the minimum val~e of the objective function = ~92 for x = 2:' y = ~: ;

/

Find wnfJther the LPP: minimize z = 2x + 3y subject to x - y ::;1, + Y ;c 1, x ;c 0,y ;c 0, x ::;4 has a bounded solution.

Problem 8:

Solution: The 'points where the boundary

lines meet _the coordinate axes

are (1, 0), (-1, 0), (1, 0), (0, 1),

(4,0).

The corresponding positions of

the feasible region and the origin

(0, 0) with respect to the

boundary lines are:

x - y = 1; same sidex + y = 1; opposite sidex = 4 same side.

The feaSible region is as shown.

We find that the feasible region is

unbounded in the y-direction.

The vertices A, B, C of a portion of the feasible region are. A(1, 0), B(4, 3) and

C(O, 1). The lowest value of the objective function at these vertibes occurs at

A(1, 0) which is z = 2 + 0 = 2. ~

As y takes larger and larger values, the value of the objective fun ion increases

indefinitely. Hence the minimum value of the objective func ion is 2 for

x=1,Y=O I

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Linear Programming P-1214-P7-CBSE-MA-LP-13 -

Problem 9: Maximize z = 2x - y subject to the constraints x - y ~ 1, x ;c 0, y ;c 0, 2x - Y ~ 6.

Solution: The points where the boundary lines

meet the coordinates axes are

x-axis; (1, 0), (3, 0)

y-axis; (0, -1), (0, -6).

Moreover, the feasible region is on

the same side of the origin, for both

the lines x - y = 1 and 2x - y = O.These two lines intersect at the point

(5, 4). The feasible region is shown

as the shaded one.

The region is unbounded in both x

and y directions in such a way that

2x - y s; 6.

Hence for all points in the feasible

region,

2x - y s; 6.

Or the maximum value of 2x - Y is 6. .

Hence the maximum value of the objective function z = 2x - Y is 6.

Problem 10: A shop-keeper deals in the sale of TVs and VCPs. He has Rs 5.2 lacs to invest.

He has only space for 50 pieces. A TV costs Rs 20, 000/ and a VCP costs Rs

B,OOO/-. From a TV and a VCP he earns a profit of Rs 1500/- and Rs BOO/-respectively. Assuming that he sells all the items that he purchases, find the

number of TVs and VCPs he should buy in order to maximize his profit.

\..

Solution: Let x and y denote the

number of TV and VCP

respectively. From the given

data, we have x ~ 0, y ~ a, x+ y s; 50. He has Rs 5.2 lacs

to invest. Hence

20000 x + 8000 y s; 520000.

The profit he earn is t500x

+ 800y . Hence the LPP is:

Maximize: z = 1500x + 800ysubject to

x ~ 0, y ~ 0, x + y s; 50 and

5x + 2y s; 130.

The boundary lines meet the

coordinate axes at :

x-axis; (50, 0), (26, 0)

y-axes; (a, 50); (a, 65)The boundary lines x + y = 50 and 5x + 2y = 120 intersect at (10, 40). The

feasible region is shown as shaded.

The quadrilateral has four vertices namely 0(0, a), A(26, a), 8(10, 40), (C(O, 65).

The maximum occurs at 8(10, 40). Hence he should store 10TVs and 40 VCPs.

He earns a profit of Rs 1500 x 10 + 800 x 40 = 47,000

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website: www.fllt.jee.com.

p.1214-P7-CBSE-MA-LP.14 P;lnnac/e Study paCkaJ.

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4.

6.

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ASSIGNMENT PROBLEMS

1. Represent the following inequations graphically in the two dimensional pane:

(i) x - 2y + 4 sO,

(ii) Y + 8 ~ 2x,

(iii) 2x s 6 - 3y.

2. Represent the following systems of inequations graphically:

- (i) x + y s 9, y > x, x ~ 1.(ii) 3x +4y s 60; x + 3y's 30, x ~ 0, y ~ O.

(iii) 2x + Y ~ 4, x + y s 3, 2x - 3y s 6.

(iv) 3x + 2y ~ 24, 3x + y s 15, x ~ 4.

(v) 3x +2y s 24, x + 2y s 16, x + y s 10, x ~ 0, y ~ O.

An aeroplane can carry a maximum of 360 passengers. The airlines re erves at least 20

seats for Dusiness class. But at least 8 times as many passengers ,tefer to travel by

economy class than by business class. Find the possible graph c distribution of

passengers in the two classes.

A company makes chairs and tables. The company has two different types of machines

on which the chairs and tables are manufactured. The time reqUt'red in hours. for

manufacturing each item and total available time in a week is given belo :

Chairs Tables Total vailable time (in

hours)

Machine 1 5 2 24

Machine 2 4 8 64

Show graphically the sets of chairs and tables can possilbly be manufac~ured in a week.

How many chairs and tables should the company manufacture to maximize profit. if the

profit on the sale of a chair and table is respectively Rs 60/- and Rs 100V-.

Find the maximum and minimum values of 3x + 2ysLibject to the conttrants x + y ~ 1,_

x - 3y s 0, Y s 4, x ~ 0, y ~ O.

Find the maximum and minimum values of 10x + 6y subject to x + y 1, Y s 6, x s 4,

2x + 3y s 12, x ~ 0, y ~ O.

7. Maximize 2x + 5y subject to 3x - 4y + 12 ~O, x - y + 1 ~ 0, x s 4, Y s 4, ~ 0, Y ~ O.

8. Minimize8x + 5y subject to the constraints2x + 3y s 12, x + 5y ~ 5, 3x + 4y ~ 12,

x~ 0, Y ~ O.

9. Maximize 5x + 3y subject to x ~ 0, y ~ 0, x s 4, x +y s 6, x - Y + 1 ~ 0, 5y - x ~ 0,. .

x + y ~1.

10. A company manufactories electric cooking ranges and mixers. The company employee a

total of 400.. workers. The cooking range requires 2 man-week of labtr and the mixers

requires 1 man-week labour. The product are produced and sent 0 the market on

weekly basis. The weekly production of cooking ranges cannot excjed 150 and the

production of mixers cannot exceed 200. How many cooking ranges and mixers be

produced in a week so as to maximize profit, if the profit for each of the cooking ranges is

Rs 250/- and that from a mixer Rs 150/-.

FIITJULtd., FIITJEE Houae, 29.A, Xalu Saral, Sarvaprlya Vlhczr, New Delhl.llOO16, Ph:46106000, 26 69493, Fa. 2661:1942

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Linear Programming p.1214.P7.CBSE.MA.LP.15

11. Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively.

They supply to 3 ration shops. 0, E and F, whose requirements are 60, 50 and 40

quintals respectively. The cost of tra!1sportation per quintal from the godowns to the

shops are given in the following table:

Transportation cost per quintal (in Rs.)

\\From ITo A B

0 6 4

E 3 . 2

F 2.50 3

How woulg the supplies be transported in order that the transportation cost is minimum.

What is the minimum cost?

12. One kind of cake requires 200 g of flour and 25 g of fat, and another kind of cake requires

100 g of flour and 50 g of fat. Find the maximum number of cakes which can be madefrom 5 kg of flour and 1 kg of fat assuming that there is no shortage of. the other

ingredients, used in making the cakes.

13. An oil company requires 13,000, 20,000 and 15,000 barrels of grade, median grade and

low grade oil respectively. Refinery A produces 100, 300 and 200 barrels per day of high,

medium and low grade oil respectively whereas the refinery B produces 200, 400 and

10Q barrels per day respectively. If A costs Rs. 400 per day and B costs Rs 300 per dayto operate, how many days should each refinery be run to minimize the cost of meeting

requirements.

14. A company manufactures two types of toys A and B. Type A-requires 5 minutes each for

cutting and 10 minutes each for assembling. Type B requires 8 minutes each for cutting

and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours

available for assembling in a day. The profit is Rs. 50 each on type A and R. 60 each on

type B. How many toys of each should the company manufacture in a day to maximize

the profit?

15. A pharmaceutical company requires everyday 10, 12 and 12 units of chemicals C1, C2

and C3respectively. A liquid product contains 5,2 and 1 units of chemicals C1, C2 and C3

respectively per box and it costs Rs. 200 per box. How much quantity of either of the two

products should be purchased daily in order to minimize the cost.

FIIIJII Ltd., FnTJEE House, 29-A, Ka/u Saral, SarrJapriya Vihar, Nf!W Delhl.ll0016, Ph:46106000, 26&69493,Fax 26S~3942

website: www.fll~ee.com.

P-1214-P7-CBSE-MA-LP-16 . Plnnac.le Study Packa e

.,..... ~ANSWERS TO ASSIGNMENT f;1RbBLEMS

1. (i)

(iii) y

2. , (i)

(iii) Y

(iv)

(v)

y

FlIIIEE Ltd., FDTJEE House, 29-A, Kalu Sarai, SanJapriya Vlhar, New Delhi-110016, Ph:46106000, 26569493, Fax 26513942

website: www.flltjee.com.

Linear Programming

3. x = number in business class

y = number in economy class

::;> x ~ 0, y ~ 0, x + y:::; 360, x ~ 20, y ~ x.

4. x = number of chairs, y = number of tables::;> x ~ 0, y ~ 0, 5x + 2y :::;24, 4x + 8y :::;64;

2,7, Rs 8201-

5. 44,2

6. 48,6

7. 28

8. 15

9. 26.

10. 100, 200, Rs 550001-

P-1214-P7-CBSE-MA-LP-17

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webstte: www.fltt.jee.com.

IL.~~~,

NOTES

I

Chandigarh

Jaipur

Gwalior

Bhopal

Vadodara

Indore

Mumbai (Kandivali)

Mumbai (Chembur)

Pune

FliT JEE Junior Colleges

Hyderabad (Saifabad)

Hyderabad (Dilsukhnagar)

Hyderabad (Kukatpally)

Coimbatore

Delhi (South)

Delhi (Punjabi Bagh)

Delhi (East)

Delhi (Dwarka)

Noida (Delhi-NCR)

Gurgaon (Delhi-NCR)

Faridabad (Delhi NCR)

Ghaziabad (Delhi-NCR)

Ranchi

Bokaro

Bhilai

Raipur

FliT JEE Junior College

Visakhapatnam

FliT JEE Junior College

Vijayawada

Chennai (Adyar)

Chennai (Kilpauk)

Chennai (Sriperumbudur)

Chennai (KK Nagar)

Varanasi

Patna

Bhubaneswar

IIII\

III

FIITJEE Associate Schools

o FIITJEE Junior Colleges

-

..

Pune

Varanasl.

Patna

DhanbadBokaro

Kolkata (Rabindra Sadan)

Kolkata (Ultadanga-Salt Lake)

Durgapur

Bhubaneswar

Deihl (R hini)

Deihl (J nakpuri)

Deihl (0 arka)

Nolda ( elhi-NCR)

Gurgao (Delhi-NCR)

Farldabad (Delhi-NCR)

Ghazla ad (Delhi-NCR)

Visakhapatnam

VIJayawada

Ran hi (Lalpur)

Ra]hl (SOP Doranda)

Shaktl agar NTPC

Chennal (Nungambakkam

Chennai (Kilpauk)

Chennai (Adyar)

Chennal (Sriperumbudur)

Chennal (KK Nagar)

Lucknow (Gomti Nagar)

Lucknow (Aashlana)

Kanpur

Meerut

Deihl (South)

Delhi (Punjabi Bagh)

Deihl (East)

Coimbatore

Bhopal

Gwalior

Nagpur

Indore

Vadodara

Jaipur (J.L.N. Marg)

Jaipur (Vaishali)

Chandigarh

Amritsar

Mumbai (Chembur)

Mumbai (Kandivali)

Mumbai (Andheri)

Mumbai (Navi Mumbai)

Mumbai (Thane)

Hyderabad (Saifabad)

Hyderabad (Dilsukhnagar)

Hyderabad (Kukatpally)

Bangalore (Seshadripuram)

Bangalore (HSR Layout)

IFIITJEEIL I M I T E D

National AdmissionsOffice: FIITJEEHouse, 29-A, Kalu Sarai, SarvapriyaVihar, New Delhi - 110016

Ph. : 0 I I - 46106000/1 0/13/15 All India Toll Free No. : 1800 1I 4242 Fax: 0 I I - 6513942 Web: www.fiitjee.com

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