161Linear Programming
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09Linear Programming
Syllabus
Linear Programming Problem General Form of
Linear Programming Problem Different Types of
Linear Programming Problems Solution of LPP
by Graphical Method
The term 'programming' means 'planning' and it
refers to a particular plan of action from amongst
several alternatives for maximising or minimising
a function under given restrictions as maximising
profit or minimising cost, etc. The term 'linear'
means that 'all inequations' or 'equations used'
and the function to be maximised or minimised
are linear. Thus, linear programming is a technique
for resource utilisation.
Linear Programming Problem
A linear programming problem may be defined as
the problem of maximising or minimising a linear
function subject to linear constraints. The
constraints may be equalities or inequalities.
Important Terms Related to LPP
There are many terms related to a linear
programming problem. Definition of these terms
are given below:
Constraints
The linear inequations or inequalities or restrictions
on the variables of a linear programming problem
are called constraints. The conditions x > 0, y > 0
are called non-negative restrictions.
Optimisation Problem
Aproblem which seeks to maximise or mini mise
a linear function subject to certain constraints as
determined by a set of linear inequalities is called
an optimisation problem. Linear programming
problems are special type of optimisation
problems.
Objective Function
A linear function of two or more variables which
has to be maximised or minimised under the given
restrictions is called an objective function.The
variables used in the objective function are called
decision variables.
Optimal Value
The maximum or minimum value of an objective
function is known as the optimal value of LPP.
Feasible Region
The common region determined by all the
constraints including non-negative constraints
x,y > 0 of a linear programming problem is called
the feasible region or solution region. Each point
in this region represents a feasible choice. The
region other than feasible region is called an
infeasible region.
Bounded and Unbounded Region
A feasible region of a system of linear inequalities
is said to be bounded, if it can be enclosed within
a circle. Otherwise, it is said to be unbounded
region i.e. the feasible region does extend
indefinitely in any direction.
Feasible Solution
Points within and on the boundary of the feasible
region represent feasible solution of the
constraints. Any point outside the feasible region
is called an infeasible solution.
Optimal Feasible Solution
A feasible solution at which the objective function
has optimal value (maximum or minimum) is
called the optimal solution or optimal feasible
solution of the linear programming problem.
Optimisation Technique
The process of obtaining the optimal solution of
the linear programming problem is called
optimisation technique.
Mathematical Form of LPP
The general mathematical form of a linear
programming problem may be written as follows:
Objective function, z = c1 x + c
2y
Subject to constraints are alx + b
1y
2 < d
1,
a2x + b
2y < d etc.
and non-negative restrictions are x > 0, y > 0.
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General Form of Linear Programming
Problem
The mathematical statement of the general form
of a linear programming problem (abbreviated
LPP) may be written as follows:
Optimise (maximise or minimise)
Z = c1x
1 + c
zx
z + ... + c
nx
n
subject to constraints are
a11
x1 + a
12x
2 +.... + a
1nx
n (<, = >} b
1
a21
x1 + a
22x
2 + ... + a
2nx
n (<, = >} b
2 ...(i)
and am1x
1 + a
m2x
2 + ...+ a
mnx
n ((<, = >} b
m
where, x1,x
2 ..... ,x
n > 0, ...(ii)
i. x1, x
2, ..., Xn are the variable who e values we
wish to determine and are called the decision or
structural variables.
ii. The linear function Z which is to be maximised
or minimised is called the objective function of
the general LPP.
iii. The inequalities (i) are called the constraints of
the general LPP.
iv. The set of inequalities (ii) is known as the et of
non-negative restrictions to the general LPP.
v. The constant cj (j=1,2, ..., n) repre en the
contribution (profit or cost) to the objective
function of the jth variable.
vi. The coefficient aij (i = 1,2, ...,m; j = 1,2 ..., n) are
referred to as the technological or subtitution
coefficients.
vii. bi (i = 1, 2 ...., m) is the constant repre enting the
requirement or availability of the jth constraint.
viii. The expression (< =, >) means that onI one of
the relationship is the set (<, =, > ) should hold
for a particular constraint
Example 1
A firm has to transport 1200 packages using large
vans which can carry 200 packages each and small
van which can take 80 packages each. The cost
for engaging each large van is Rs. 400 and each
small van is Rs. 200. Not more than Rs. 3000 is
to be spent on the job and the number of large
vans cannot exceed the number of small vans.
Formulation of this problem as a LPP gi en that
the objective is to minimise cost is
a. minirni e = Z = 200y + 400x,
subject to constraints
5x + 2y > 30,
2x + y < 15,
x < y,
x, y, > 0
b. minimise z = 400x + 200y, subject to constraints
5x + 2y < 5 : 30
2x + y < 5 : 15,
x < y,
x, y > 0
c. minirniseZ = 200Y + 400x, subject to constraints
5x + 2y < 5 : 30
2x + y : 5 < 15,
x < y,
x,y > O
d. rninimise Z = 400x + 200y, subject to constraints
5x+ 2y < 5:30
2x+ y : 5 < 15,
x < y,
x,y < 0
Sol (b) Let the firm has x number of large vans and y
number of small vans. From the given information,
we have following corresponding con traint table.
Large vans (x) Small vans(y) Maximum / Minimum
Packages 200 80 1200
Cost 400 200 3000
Thus, we see that objective function for minimum
cost is
200 = 400x + 80y > 1200y.
Subject to con traints
200x + 80y > 1200 [package constraint)
5x + 2y > 30 ...(i)
and 400x + 200y < 3000 [cost constraint)
2x + y < 15 ... (ii)
and x < y [van constraint) ... (iii)
and x > 0, y > 0 [non-negativeconstraints).. (iv)
Thus, required LPP to minimise cost is
mini mise Z = 400x + 200y, subject to constraints
5x + 2y > 30,
2x + y < 15,
x < y,
x > 0, y > 0
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Different Types of Linear Programming
Problems
Some important types of linear programming
problems are given below:
To solve such problems, we first formulate the
mathematical form of LPP and then solve by using
corner point method, which will be discussed later.
Diet Problems
In diet problems, we have to determine the amount
of different kinds of constituents/nutrients which
should be included in a diet so as to minimise the
cost of the desired diet such that it contains a
certain minimum amount of each constituent/
nutrient.
• Example 2
(Diet problem) Adietician wishes to mix two types
of foods f1 and f
2 in such a way that the vitamin
contents of the mixture contain atleast 6 units of
vitamin A and 8 units of vitamin f1. Food II
contains 2 units/kg of vitamin A and 3 units/kg of
vitamin B while food 12 contains 3 units/kg of
vitamin A and 2 units/kg of vitamin B. Food f2
cost Rs. 50 per kg and food f2 cost Rs. 75 per kg.
Formulate the problem as an LPP to minimise
the cost of mixture.
a. min. Z = 50x + 75y
b. min. Z = 50x + 75y
subject to constraints subject to constraints,
2x + 3y > 6 2x+3y:5:6
3x + 2y > 8 3x + 2y < 8
and x, y > 0 and x, y > 0
c.min.Z=50x+75y d.Noneofthese
subject to constraints
2x + 3y > 8
3x + 2y < 6
and x, y > 0
Sol (a) Let the mixture contains food
f1 = x kg and food f
2 = y kg.
Here, cost of food 11 = Rs. 50 per kg and cost of
food 12 = Rs. 75 per kg and we have to minimise
the cost of mixture. So, the objective function is
to minimise, Z = 50x + 75y
Here, food f1
contains 2 units/kg and food f2
contains 3 units/kg of vitamin A. Also, mixture
contain atleast 6 units of vitamin A. So, first
constraint is 2x + 3y > 6.
Similarly, food 11 contains 3 units/kg and food f2
contains 2 units/kg of vitamin B. Also, mixture
contain atleast 8 units of vitamin B. So, second
constraint is 3x + 2y > 8.
Also, foods f1 and f
2 cannot be negative, so
x > 0, y – 0.
Thus,
Formation of this LPP is, objective function,
Minimum, Z = SOx + 7sy
subject to constraints, 2x + 3y > 6
3x + 2y >8 and x,y > 0
Manufacturing Problems
In manufacturing problems, we have to determine
the number of units of different products which
should be produced and sold by a firm, when
each product requires a fixed manpower, machine
hours, labour hour per unit of product, warehouse
space per unit of the output, etc. in order to make
maximum profit.
Example 3
A manufacturer considers that men and women
workers are equally efficient and so he pays them
at the same rate. He has 30 and 17 units of workers
(male and female) and capital respectively, which
he uses to produce two types of goods A and B.
To produce 1unit of A, 2 workers and 3 units of
capital are required while 3 workers and 1 unit of
capital is required to produce 1 unit of B. If A and
B are priced at Rs.100 and Rs. 120 per unit
respectively. Form the above as an LPP to
maximise the revenue.
a. maximise Z=100x+ 120y, subject to constraints
2x +3y < 30,
3x + y < 17,
x,y > o
b. maximise Z = 80x + 120y, subject to constraints
5x+ 2y < 16
x + y < 15,
x < y,
x, y > 0
c. maximise z = 100x+ 80 y, subject to constraints
2x + 5y < 30
2x + y <15,
x < y
x,y > 0
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d. maximise Z = 40x + 120y, subject to constraints
3x + 2y < 30
2x + y < 10,
x < y,
x,y > o
Sol (a) Let the manufacturer produces x units of goods
A and y units of goods B.
Now, formulate a table for given data
A B Required capacity
Workers 2 3 30
Capital 3 1 17
Profit function (z) 100 120
The required LPP becomes,
Maximise z =100x + 120y
Subject to constraints are 2x + 3y < 30 ....(i)
3x + y < 17 ...(ii)
and x, y > 0 .... (iii)
which is required formulation of the given
manufacturing problem.
Transportation Problems
In transportation problems, we have to determine
a transportation schedule in order to find the
minimum cost of transporting a product from
plants/factories situated at different locations to
different markets.
Example 4
There is a factory located at each of the two places
P and Q.
From these locations, certain commodity is
delivered to each of the three depots situated atA,
Band C. The weekly requirements of the depots
are respectively 5, 5 and 4 units of the commodity
while the production capacity of the factory at P
and Q are respectively, 8 and 6 units. The cost of
transportation per unit is given below:
To/From Cost (in Rs')
A B C
P 16 10 15
Q 10 12 10
Formulate the above LPP mathematically such
that the transportation cost is minimum.
a. minimise Z = x –7y+190, subject to constraints·
5x + 2y > 30,
2x + y < 15,
x < y,
x, y < O
b. minimise Z= x–7y + 190, subject to constraints
x + y < 8,
x + y > 4,
x < 5, y < 5
x, y > 0
c. minimise Z =x–7Y + 190, subject to constraints
x + y < 4
x + y > 4,
x, < 5, y > 5
x,y > 0
d. minimise Z=x – 7y + 190, subject to constraints
x + y < 4,
2x + 3y > 6,
x < 5, y < 5,
x,y > 0
Sol (b) Let the unit supplied from P to A be x and the
unit supplied form P to B be y.
Then, the unit supplied from P to C be 8 - (x + y)
and the unit supplied from Q to A be 5 – x and the
unit supplied from Q to B be 5 – y and the unit
supplied from Q to C be x + y – 4.
The objective function is to
Minimise Z = 16x + 10y + 15(8 – x – y) + 10
(5 – x) + 12(5 – y) + 10 (x + y – 4)
= 16 x + 10 + 120 –15x –15y + 50 – 10x
+ 60 –12y + 10x + 10y–40
= x – 7y + 190
Subject to the constraints are
x + y < 8 ... (i)
x + y > 4 ... (ii)
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x < 5, y < 5 ...(iii)
and x,y > 0 ... (iv)
Which is required formulation of the given
transportation problem.
Solution of LPP by Graphical Method
There are two graphical methods:
i. Iso - profit or Iso-cost method
ii. Corner point method
Iso-profit or Iso-cost Method
If the objective function is of maximisation type,
then this initial objective line z1 = ax + by is called
iso-profit line. If the objective function is of
minimisation type, then this initial objective line
z1 = ax + by is called iso-cost line.
Working Rule to Find Solution of LPP by Iso-
profit or Iso-cost Method
i. Graph the constraints and identify the feasible
region.
ii. Draw the initial objective zl = ax + by which
passes through a point belonging to the feasible
region.
iii. If objective function is of maximisation type,
then move the iso-profit line z1=ax+ by parallel
to it self away from the origin, till the feasible
extreme points are located. This iso-profit line
gives maximum value of z.
iv. If the objective function is of minimisation type,
then move the iso-cost line parallel to itself
towards the origin till the feasible extreme points
are located. At this point(s) the objective
function will have minimum value.
Example 5
Solve the following linear programming problem
graphically: Maximize z = 50x + 15y
Subject to 5x + y < 100
x + y < 60
x, y > 0
a. 1200 b. 1240
c. 1260 d. 1250
Sol (d) We nrst convert the inequations into equations
to obtain the lines 5x + y = 100,
x + y = 60, x = 0 and y = 0.
The line 5x + y = 100 meets the coordinate axes
at A1 (20, 0) and B
1 (0,100). Join these points to
obtain the line 5x + y = 100.
The line x + y = 60 meets the coordinate axes at
A2 (60, 0) and B
2(0, 60). Join these points to
obtain the line x + y = 60.
Also, x = 0 is the Y-axis and y = 0 is X-axis.
The feasible region of the LPP is shaded in figure.
The coordinates of the corner points of the feasible
region OA1, PB
2 are 0 (0, 0), A
1(20, 0),
P(10, 50) and B2 (0, 60).
Now, we take a constant value, say 300 (i.e, 2
times the 1 cm of 50 and 15)for Z. Then,
300 = 50x + 15y
This line meets the coordinate axes at P1(6,0) and
Q1 (0,20). Join these points by a dotted line. Now,
move this line parallel to itself in the increasing
direction i.e. away from the origin. P2Q
2 and P
3Q
3
are such lines. Out ofthese lines locate a line which
is farthest from the origin and has at least one
point common to the feasible region.
Clearly, P3Q
3 is such line and it passes through
the vertex P(10, 15) the convex polygen OAl: PB
2.
Hence, x = 10 and y = 50 will give the maximum
value of Z. The maximum value of Z is given by
Z = 50 × 10 + 15 × 50 = 1250
Corner Point Method
Working Rule
i. Convert the inequality into equality .
ii. Find the solution set of the given constraints
(inequality) i.e. feasible region.
iii. Find the point of intersection of the lines.
iv. Find the corner points of the feasible region.
v. Find the value of objective function at each
corner point.
vi. Find the optimum solution i.e. maximum or
minimum value.
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Example 6
The feasible region for a LPP is shown in the
following figure. Evaluate Z = 4x + yat each of
the corner points of this region. Then the minimum
value of Z, is (if it exists).
a. 9 b. 16
c. 3 d. one of these
Sol (c) From the shaded region, it is clear that
feasible region is unbounded with the corner points
A (4, 0), B (2,1) and C (0,3).
Also, we have Z = 4x + y.
[since, x+2y=4 and x + y = 3 Y = 1 and x = 2]
Comer points Corresponding value of Z
(4.0) 16
(2,1) 9
(0,3) 3 Minimum
Now, we see that 3 is the smallest value of Z at
the corner point (0,3). Note that here we see that,
the region is unbounded, therefore 3 may or may
not be the minimum value of Z.
To decide this issue, we graph the inequality
4x + y < 3 and check whether the resulting open
half plane has no point in common with feasible
region etherwlse. Z has no minimum value.
From the shown graph above, it is clear that there
is no point in common with feasible region and
hence Z has minimum value 3 at (0,3).
Special Cases of LPP
These special cases of LPP are given below:
i. Infinite Number of Optimal Solution
The LPP may have more than one solution
because of the nature of the objective function.
ii. Unbounded Solution
In some linear programming problem the solution
which are not bounded.
iii. Infeasible Solution
In some linear programming problem the feasible
solution does not exist.
Example 7
The linear programming problem
Maximise Z = x1 + x
2
Subject to constraints
x1 + 2x
2 < 2000, x
1 + x
2 < 1500, x
2 < 600
x1 > 0 has
a. no feasible solution
b. unique optimal solution
c. a finite number of optimal solutions
d. infinite number of optimal solutions
Sol (d) Given constraints are
x1 + 2x
2 < 2000, x
1 + x
2 < 1500, x
2 < 600 and
x1 > 0
Now, draw these inequalities, we have
The feasible region is OACDEO.
Given, Z = xl + x
2
AtO(0,0), Z = 0 + 0 = 0
At A (1500,0), Z = 1500 + 0 = 1500
At C (1000, 500), Z = 1000 + 500 = 1500
At D (800, 600), Z = 800 + 600 = 1400
At E (0,600), Z = 0 + 600 = 600
Here, Z is maximum on the segment AC.
Hence, there are infinite optimal solutions.
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Formulation of LPP
1. Shaded region is represented by
a. 4x – 2y < 3 b. 4x – 2y < – 3
c. 4x – 2y > 3 d. 4x – 2y > 2 – 3
2. In case of a linear programming problem, feasible
region is always
a. a convex set
b. a concave set
c. a bounded convex set
d. a bounded concave set
3. The feasible region for the following constraints
L1 < 0, L
2 > 0, L
3
= 0 x > 0 , y > 0 in the diagram
shown is
a. area DHF b. area AHC
c. line segment EG d. line segment GI
e. line segment IC
4. A vertex of a feasible region by the linear
constraints 3x + 4y < 18, 2x + 3y > 3 and x, y > 0
is
a. (0, 2) b. (4.8, 0)
c. (0, 3) d. None of these
5. In linear programming problem, the linear function
Z subject to certain conditions determined by a
set of linear inequalities with variables as non-
negative, is
a. maximised only b. minimised only
c. Both (a) and (b) d. None of these
6. A problem which seeks to maximise or minimise
a linear function (say, two variables x and y)
subject to certain constraints as determined by a
set of linear inequalities is called alan
a. optimisation problem
b. functional problem
c. numerical problem
d. computer problem
7. The linear inequalities or equations or restrictions
on the variables of a linear programming problem
are called
a. linear relations b. constraints
c. functions d. objective functions
9. Consider the following statements
I. The term linear implies that all mathematical
relations used in the problem are linear relation
II. The term programming refers to the method
of determining a particular programme.
Choose the correct option.
a. Only I is true b. Only" is true
c. Both I and " are true d. Neither I nor" is true
10. A furniture dealer deals in only two items-tables
and chair. He has Rs. 50000 to invest and has
storage space of atmost 60 pieces. A table costs
Rs. 2500 and a chair Rs. 500. Then, the
constraints of the above problem are (where, x is
number of tables and y is number of chairs)
a. x > 0, y > 0 b. 5x + y < 100
c. x + y < 60 d. All of these
11. If a furniture dealer estimates that from the sale
of one table he can make a profit of Rs. 250 and
from the sale of one chair of a profit of Rs. 75
and if x is the number of chairs and y is the number
of tables, then its linear objective function is
a. Z = 75x + 250y b. Z = 75x + 25y
c. Z = 250x + 75y d. Z = 25x + 75y
12. A linear programming problem is one that is
concerned with finding the ... A. .. of a linear
function called ... B.... function of several
variables (say x and y), subject to the conditions
that the variables are ... C. .. and satisfy set of
linear inequalities called linear constraints.
Here, A, Band C are respectively
a. objective, optimal value, negative
b. optimal value, objective, negative
c. optimal value, objective, non-negative
d. objective, optimal value, non-negative
Exercise
(Topical Problems)
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13. The conditions x > 0, y > 0 are called
a. restrictions only
b. negative restrictions
c. non-negative restrictions
d. None of these
13. Which of the following is a linear objective
function?
a. Z = ax + by b. Z < ax + by
c. Z > ax + by d. Z ax + by
14. The variables x and y in a linear programming
problem are called
a. decision variables b. linear variables
c. optimal variables d. None of these
15. The shaded region for the inequality x + 5 Y s 6 is
a. to the non-origin side of x + 5y = 6
b. to the either side of x + 5y = 6
c. to the origin side of x + 5y = 6
d. to the neither side of x + 5y = 6
16. Which of the following statements is false?
a. The feasible region is always a concave region
b. The maximum (or minimum) solution of the
objective function occurs at the vertex of the
feasible region
c. If two corner points produce the same
maximum (or minimum) value of the objective
function, then every point on the line segment
joining these points will also give the same
maximum (or minimum) two values
d. All of the above
17. The corner point method for bounded feasible
region comprises of the following steps
I. When the feasible region is bounded, M and
m are the maximum and minimum values of
Z.
II Find the feasible region of the linear
programming problem and determine its corner
points.
III.Evaluate the objective function Z = ax + by at
each corner point. Let Mand m respectively
be the largest and smallest values of these
points.
The correct order of these above steps is
a. III.I, II b. II, III, I
c. II, I, III d. I, III, II
Terminologies Related to the Solution of
LPP and Solution of LPP by Graphical
Method
18. Maximum value of 12x +3y subject to constraints
x > 0, y > 0, x + y < 5 and 3x + y < 9 is
a. 15 b. 36
c. 60 d. 40
19. The coordinate of the point at which minimum
value of z = 7x – 8y, subject to the conditions
x + y – 20 < 0, y > 5, x > 0, y < 9 is attained, is
a. (20, 0) b. (15, 5)
c. (0, 5) d. (0, 20)
20. The minimum value of the objective function
Z = 2x + 10y for linear constraints x > 0,
y > 0, x – y > 0, x – 5y < – 5 is
a. 10 b. 15
c. 12 d. 8
21. The minimum value of Z = 5x – 4y subject to
constraints x + y < 10, y < 4 ; x, y > 0 will be at
the point
a. (10,4) b. (–10,4)
c. (6,4) d. (0,4)
22. The optimal value of the objective function is
attained at the point is
a. given by intersection of inequations with axes
only
b. given by intersection of inequations with X-
axis only
c. given by corner points of the feasible region
d. None of the above
23. Variables of the objective function of the linear
programming problem are
a. zero b. zero or positive
c. negative d. zero or negative
24. A furniture dealer deals in only two items namely
tables and chairs he has Rs. 5000 to invest and
space to store atmost 60 pieces. A table cost is
Rs. 250 and a chair Rs. 60. He can sell a table at
a profit of Rs. 15. Assume that, he can sell all the
items that he produced. The number of constraints
in the problem are
a. 2 b. 3
c. 4 d. 1
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25. The maximum value of Z = 10x + 6y subject to
constraints x > 0, y > 0, x + y < 12, 2x + y < 20
is
a. 72 b. 80
c. 104 d. 110
Direction (Q. Nos. 26-31) Afurniture dealer deals
in only two items- tables and chairs. He has Rs.
50000 to invest and has storage space of atmost
60 pieces. A table costs Rs. 2500 and a chair Rs.
500. He estimates that from the sale of one table,
he can make a profit of Rs. 250 and that from the
sale of one chair a profit of Rs. 75. He wants to
know how many tables and chairs he should buy
from the available money so as to maximise his
total profit, assuming that he can sell all the items
which he buys.
On the basis of the above information, answer
the following questions.
26. The above stated optimisation problem is an
example of
a. linear programming problem
b. functional programming problem
c. computer programming problem
d. None of the above
27. Which of the following are constraints for the given
problem?
a. Maximum investment is ~50000 and storage
space is for maximum of 60 pieces
b. Minimum investment is ~50000 and storage
space is for minimum of 60 pieces
c. Maximum investment is f 50000 and storage
space is for minimum of 60 pieces
d. None of the above
28. If he buys tables only, then his profit will be
a. Rs. 500 b. Rs. 5000
c. Rs. 7500 d. Rs. 2500
29. Suppose he chooses to buy chairs only. Then, his
maximum profit is
a. Rs. 4500 b. Rs. 7500
c. Rs. 4000 d. Rs. 5700
30. If he choose 10 tables and 50 chairs, then his
maximum profit is
a. Rs. 6200 b. Rs. 6500
c. Rs. 6250 d. Rs. 6520
31. If x is the number of tables and y be the number
of chairs, then the given problem, mathematically
reduces to
a. maximise Z = 250x + 75y
subject to the constraints
5x + y < 100
x + y < 60
x.y > 0
b. minimise Z = 250x + 75y
subject to the constraints
5x + y < 100
x + y < 60
x.y > 0
c. minimise Z = 250x + 75y
subject to the constraints
5x + y > 100
x + y > 60
x.y > 0
d. maximise Z = 250x + 75y
subject to the constraints
5x + y > 100
x + y > 60
x.y > 0
Direction (Q. Nos. 32-34) The feasible solution
for a LPP is shown as below:
Let Z = 3x – 4y be the objective function. Then,
32. Maximum of Z occurs at
a. (5,0) b. (6, 5)
c. (6,8) d. (4, 10)
33. Minimum of Z occurs at
a. (0,0) b. (0, 8)
c. (5,0) d. (4, 10)
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34. Maximum value of Z + + Minimum value of Z)
is equal to
a. 13 b. 1
c. –13 d. –17
35. The maximum value of Z = 4x + 3y, if the feasible
region for an LPP is shown in following figure, is
a. 112 b. 100
c. 72 d. 110
36. The objective function of an LPP is
a. a constraint
b. a function to be optimised
c. a relation between the variables
d. None of the above
37. The maximum value of P = x + 3y such that
2x + y < 20, x + 2y < 20, x > 0, y > 0 is
a. 10 b. 60
c. 30 d. None of these
Special Cases of LPP
38. The optimal value of the objective function is
attained "at the point
a. given by intersection of inequations with axes
only
b. given by intersection of inequations with X-
axis only
c. given by corner points of the feasible region
d. None of the above
39. Maximum value of Z = 3x + 4y subject to
x - y > – t –x + y < 0; x, y > 0, is given by
a. 1 b. 4
c. 6 d. no feasible region
40. The area of the feasible region for the following
constraints 3y + x > 3,x M < 0, y > 0 will be
a. bounded b. unbounded
c. convex d. concave
41. The linear programming problem
Maximize Z = x1 + x
2
Subject to constraints
xl + 2x
2 < 2000
x1 + x
2 : < 1500
x2 < 600
x1 > 0 has
a. no feasible solution
b. unique optimal solution
c. a finite number of optimal solutions
d. infinite number of optimal solutions
42. Consider the linear programming problem
Maximise Z = 4x + y.
Subject to constraints x + y < 50, x + y > 100 and
x, y > 0. Then, maximum value of Z is
a. 0 b. 50
c. 100 d. Does not exist
43. Which of the following sets are not convex?
a. {(x,y) : 8x2 + 6y2 < 24}
b. {(x, y) : 6 < x2 + y2 < 36}
c. {(x,y) : y > 3,y > 30}
d. {(x,y) : x2 < y]
44. One of the important class of optimisation problem
is
a. functional programming problem
b. linear programming problem
c. numerical programming problem
d. None of the above
45. The problems which seek to maximise (or
minimise) profit (or cost) from a general class of
problems called
a. optimisation problems
b. customisation problems
c. Both (a) and (b)
d. None of these
46. The wide applicability of linear programming
problem is in
a. industry b. commerce
c. management science d. All of these
47. An optimisation problem may involve finding
a. maximum profit
b. minimum cost
c. minimum use of resources
d. All of the above
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48. If a young man rides his motorcycle at 25 km/h,
he has to spend Rs. 2 per km in petrol and if he
rides it at 40 km/h, the petrol cost rises to Rs. 5
per km. He has Rs. 100 to spend on petrol and
wishes to find the maximum distance, he can travel
within one hour. If x and y denote the distance
travelled by him (in km) at 25 km/h and 40 km/h,
respectively.
The inequations represent the data are
a. 2x + 5y < 100, x y
25 40 > 1, > 0, y > 0
b. 2x + 5y > 100, x y
25 40 > 1 x > 0, y > 0"
c. 2x + 5y < 100, x y
25 40 < 1,x > 0, y > 0
d. 2x + 5y < 100, 25x + 40y < 1 x > 0, y > 0
49. The maximum and minimum values of the
objective function Z = 3x–4y.
subject to the constraints,
x – 2y < 0
–3x + y < 4
x – y – 6
x.y > 0
are respectively
a. 12,10 b. 10,12
c. 12, does not exist d. 5,12
50. The maximum value of Z = 4x + 2y subject to
the constraints 2x + 3y < 18, x + y > 10
x, y > 0 is
a. 20
b. 36
c. 40
d. None of these
1. The point which provides the solution of the linear
programming problem, maximise Z = 45x + 55y.
Subject to constraints x, y > 0, 6x + 4y < 120 and
3x + 10y < 180, is
a. (15,10) b. (10,15)
c. (0, 18) d. (20, 0)
2. The maximum value of z = 10x + 6y, subject to
constraints x > 0, y > 0, x + y < 12, 2x + y < 20
is
a. 72 b. 80
c. 104 d. 110
3. If x + y < 2, x > 0, y > 0 is the point at which
maximum value of 3x + 2y attained will be
a. 72 b. (0, 0)
c. 104 d.1 1
,2 2
4. Maximum value of Z = 12x + 3y, subject to
constraints x > 0, y > 0, x + y < 5 and 3x + y < 9
is
a. 15 b. 36
c. 60 d. 40
5. Solve the linear programming problem.
Maximise Z = x + 2y
Subject to constraints x –y < 10, 2x+3y < 20 and
x > 0, y > 0
a. Z = 10 b. Z = 30
c. Z = 40 d. None of these
6. For an LPP, minimise Z = 2x + y subject to
constraints 5x + 10y < 50, x + y > 1, y < 4 and
x, y > 0, then Z is equal to
a. O b. 1
c. 2 d. 12
7. Consider the inequalities x1+x
2 < 3, 2x
1+5x
2> 10;
x1, x
2 2 > 0.Which of the point lies in the feasible
region?
a. (2,2) b. (1,2)
c. (2,1) d. (4,2)
8. Z = 4x + 2y, 4x + 2y 2: 46, x + 3y s 24 and x and
y are greater than or equal to zero, then the
maximum value of Z is
a. 46 b. 96
c. 52 d. None of these
9. The minimum value of z = 2x1 + 3x2 subjected
to the constraints
2x1+7x
2 22, x
1 + x
2 > 5x
1+x
2 > 10 and x
1,x
2 2y > 0
is
Exercise 2
(Miscellaneous Problems)
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a. 14 b. 20
c. 10 d. 16
10. The maximum value of = 3x + 4y, subjected to
the conditions x + y < 40, x+2y < 60; x, y > 60, is
a. 130 b. 140
c. 40 d. 120
11. Consider the inequalities 5x1 + 4x
2 > 9, x
1+x
2 < 3,
x1 > 0, x
2 > 0. Which of the following point lies
inside the solution set?
a. (1, 3) b. (1, 2)
c. (1, 4) d. (2, 2)
12. The minimum and maximum values of Z for the
problem, minimise and maximise Z = 3x + 9 Y
subject to the constraints
x + 3y < 60
x + y > 10
x < y
x > 0, y > 0
are respectively
a. 60 and 180 b. 180 and 60
c. 50 and 190 d. 190 and 50
13. The linear programming problem minimise
Z = 3x+ 2y
subject to the constraints
x + y < 8
3x + 5y < 15
x > 0, y > 0 has
a. one solution
b. no feasible solution
c. two solutions
d. infinitely many solutions
14. The maximum and minimum values of the
objective function Z = x + 2y
subject to the constraints
x + 2y > 100
2x – y < 0
2x + y < 200
x,y 2 > 0
occurs respectively at
a. one point and three points
b. two points and one point
c. one point and infinitely points
d. one point and one point
15. The maximum value of the objective function
Z = 3x+4y
subject to the constraints
x + y < 4
x > 0, y > 0 is
a. 16 b. 18
c. 20 d. 25
16. Let x and yare the number of tables and chairs
respectively, on which a furniture dealer wants to
make profit for the constraints
Maximise Z = 250x + 75Y
5x + y < 100
x + y < 60
x > 0
y > 0
Consider the following graph
Then, the maximum profit to the dealer results
from buying
a. 10 tables and 50 chairs
b. 50 tables and 10 chairs
c. 0 table and 60 chairs
d. 20 tables and 40 chairs
17. The graphical solution of linear inequalities
x + y > 5 and x – y < 3, where L1 x + y = 5 and
L2 x– y = 3,is
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18. By graphical method, the solutions of linear
programming problem maximise Z = 3xl + 5x
2
subject of contraints 3xl + 2x
2 < 18
x1 < 4, y
2 < 6 x
1 > 0, x
2 > 0 are
a. xl = 2, x
2 = 0, z = 6
b. x, = 2,x2 = 6, Z = 36
c. x1 = 4, x
2 = 3, z = 27
d. x1 = 4, x
2 = 6, z = 42
19. A toy manufacturer produces two types of dolls;
a basic version doll A and a deluxe version doll
B. Each doll of type B takes twice as long to
produce as one doll of type A.The company
have time to make a maximum of 2000 dolls of
type A per day, the supply of plastic is sufficient
to produce 1500 dolls per day and each type
requires equal amount of it. The deluxe version,
i.e. type B requires a fancy dress of which there
are only 600 per day available. If the company
makes a profit of Rs. 3 and Rs.5 per doll,
respectively, on doll A and B, then the number of
each should be produced per day in order to
maximise profit, is
a. 800, 500 b. 500, 600
c. 450,450 d. 1000,500
Direction (Q. Nos. 20-23) Consider the graph of
constraints stated as linear inequalities as below:
5x + y < 100 ... (i)
x+ y < 60 ... (ii)
x > 0 ... (iii)
y > 0 ... (iv)
where, x and yare number of tables and chairs on
which a furniture dealer wants to make his profit.
20. The shaded region in the graph is called
a. feasible region b. critical region
c. non-feasible region d. None of these
21. Every point in the feasible region is called
a. no solution b. unbounded solution
c. feasible solution d. non-feasible solution
22. There is/are ... D. .. point(s) in the feasible region
OABC that satisfy all the constraints given in
(i) to (ii). Here, D refers to
a. Only 1 b. Only 2
c. finitely many d. infinitely many
23. Any point in the feasible region that gives the
optimal value (maximum or minimum) of the
objective function is called alan
a. optimal solution b. no solution
c. finite solution d. None of these
24. Let R be the feasible region (convex polygon) for
a linear programming problem and Z = ax + by be
the objective function.
Then, which of the following statements is false?
a. When Z has an optimal value, where the
variables x and yare subject to constraints
described by linear inequalities, this optimal
value must occur at a corner point (vertex) of
the feasible region
b. If R is bounded, then the objective function
Z has both a maximum and a minimum value
on R and each of these occurs at a corner
point of R
c. If R is unbounded, then a maximum or a
minimum value of the objective function may
not exist
d. If R is unbounded and a maximum or a
minimum value of the objective function, then
it does not occur at a corner point
25. The minimum value of the objective function
Z = x + 2y
subject to the constraints,
2x + y > 3,
x + 2y > 6
x, y > 0 occurs
a. at every point on the line x + 2y = 6
b. at every point on the line 2x + y = 3
c. at every point on the line x + 2y = 3
d. at every point on the line 2x + y = 6
26. Let the feasible region of the linear programming
problem with the objective function Z = ax + by
is unbounded and let Mand m be the maximum
and minimum value of Z, respectively.
Now, consider the following statements
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I. M is the maximum value of Z, if the open half
plane determined by ax + by > M has no point
in common with the feasible region. Otherwise,
Z has no maximum value.
II. m is the minimum value of Z, if the open half
plane determined by ax + by < m has no point
in common with the feasible region. Otherwise,
Z has no minimum value.
Choose the correct option.
a. Only I is true
b. Only II is true
c. Both I and II are true
d. Neither I nor II is true
27. The minimum and maximum values of the
objective function, Z = 5x + 10y
subject to the constraints
x + 2 < s 120, x + y > 60
x – 2y > 0, x, y > 0
are respectively
a. 300 and 500 b. 300 and 600
c. 600 and 700 d. 300 and 400
28. Consider the following statements
I. If the feasible region of an LPP is unbounded,
then maximum or minimum value of the
objective function Z = ax + by mayor may not
exist.
II. Maximum value of the objective function Z =
ax + by in an LPP always occurs at only one
corner point of the feasible region.
III.In an LPP, the minimum value of the objective
function Z = ax + by is always 0, if origin is
one of the corner point of the feasible region.
IV. In an LPP, the maximum value of the objective
function Z = ax + by is always finite.
Which of the following statements are true?
a. I and IV b. II and III
c. I and III d. II and IV
29. The corner points of the feasible region determined
by the system of linear constraints are (0, 10),
(5, 5) (15, 15), (0, 20). Let Z = px + qy , where
p, q > 0.
Then, the condition on p and q so that the
maximum of Z occurs at both the points (15, 15)
and (0, 20), is
a. p = q
b. p = 2q
c. q = 2p
d. q = 3 P
30. A cooperative society of farmers has 50 hectare
of land to grow two crops X and Y. The profit
from crops X and Y per hectare are estimated as
Rs. 10500 and Rs. 9000, respectively. To control
weeds, a liquid herbicide has to be used for crops
X and Yat rates of 20 Land 10 L per hectare.
Further no more than Rs.800 L of herbicide should
be used in order to protect fish and wildlife using
a pond which collects drainage from his lan
To maximise the total profit of the society, the
land should be allocated to each crop is
a. 20 hec for crop X and 30 hec for crop Y
b. 20 hec each for both crop X and crop Y
c. 30 hec for crop X and 20 hec for crop Y
d. 30 hec each for both crop X and crop Y
31. Anil wants to invest atmost Rs. 12000 in bonds
A and B. According to the rules, he has to invest
atieast Rs. 2000 in bond A and atleast Rs. 4000
in bond B. If the rate of interest in bond A is 8%
per annum and on bond B is 10% per annum,
then to maximise the interest, the investment in
bond A and B are respectively
a. Rs. 10000 and Rs. 2000
b. Rs.2000 and Rs.10000
c. Rs.6000 and Rs. 6000
d. None of these
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MHT-CET Corner
1. The constraints – x1 + x
2 < 1,–x
1 + 3x
2 < 9,
x1,x
2 > 0 defines on
a. bounded feasible space
b. unbounded feasible space
c. both bounded and unbounded feasible space
d. None of the above
2. The maximum value of the objective function
Z = 3x + 2y for linear constraints x + y < 7,
2x + 3y < 16, x > 0, y > 0 is
a. 16 b 21
c. 25 d. 28
3. The maximum value of z = 9 x + 13 y subject to
constraints 2x + 3 y – 18, 2x + y < 10, x > 0,
y > 0 is
a. 130 b. 81
c. 79 d. 99
4. For the LPP Min Z = x1 + x
2 such that inequalities
5x1 + 10x
2 > 0, x
1+ x
2 < 1, x
2 < 4 and x
1 x
2 > 0
a. There is a bounded solution 2008
b. There is no solution
c. There are infinite solutions
d. None of the above
5. The region represented by the inequation system
x, y > 0, y < 6, x + y < 3, is
a. unbounded in first quadrant
b. unbounded in first and second quadrants
c. bounded in first quadrant
d. None of the above
6. A wholesale merchant wants to start the business
of cereal with Rs. 24000. Wheat is Rs.400 per
quintal and rice is Rs. 600 per quintal. He has
capacity to store 200 quintal cereal. He earns the
profit Rs. 25 per quintal on wheat and Rs. 40 per
quintal on rice. If he stores x quintal rice and y
quintal wheat, then for maximum profit the
objective function is
a. 25x + 40y b. 40x + 25y
c. 400x + 600y d.400 600
x40 25
y
7. Which of the term is not used in a linear
programming problem?
a. Optimal solution b. Feasible solution
c. Concave region d. Objective function
8. If given constraints are 5x + 4y > 2, x < 6, y < 7,
then the maximum value of the function
Z = x + 2y is
a. 13 b. 14
c. 15 d. 20
9. Z = 30x + 20y, x + y < 8,x + 2y > 4, 6x + 4y > 12,
x > 0, y > 0 has
a. unique solution b. infinitely many solution
c. minimumat (4, 0) d. minimum60 at point (0, 3)
10. Minimize Z = 3x + y, subject to constraints
2x + 3y < 6, x + y > 1, x > 0, y > 0. Then
a. x = 1,y = 1 b. x = 0, y = 1
c. x = 1,y = 0 d. x = –1, y = –1
11. The shaded region for the inequality x + 5y < 6 is
a. to the non-origin side of x + 5y = 6
b. to the either side of x + 5y = 6
c. to the origin side of x + 5y = 6
d. to the neither side of x + 5y = 6
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Answers
Exercise 1
1. (b) 2. (a) 3. (c) 4. (d) 5. (c) 6. (a 7. (b) 8. (c) 9. (d 10. (c)
11. (c) 12. (c) 13. (a) 14. (a) 15. (c) 16. (a 17. (b) 18. (b) 19. (d 20. (b)
21. (d) 22. (c) 23. (b) 24. (c) 25. (c) 26. (a 27. (a 28. (b) 29. (a 30. (c)
31. (a) 32. (a) 33. (b) 34. (d) 35. (a) 36. (b) 37. (c) 38. (c)) 39. (d 40. (b)
41. (d) 42. (d) 43. (b) 44. (b) 45. (a) 46. (a 47. (d 48. (c)) 49. (c)) 50. (d)
Exercise 2
1. (b) 2. (c) 3. (c)) 4. (b) 5. (d) 6. (b) 7. (b) 8. (b) 9. (a 10. (b)
11. (b) 12. (a) 13. (b) 14. (c) 15. (a) 16. (a 17. (c)) 18. (b) 19. (d 20. (a)
21. (c) 22. (d) 23. (a 24. (d 25. (a) 26. (c)) 27. (b) 28. (a 29. (d 30. (c)
31. (c)
MHT- CET Corner
1. (b) 2. (b) 3. (c) 4. (a) 5. (c) 6. (b) 7. (c)) 8. (d 9. (d 10. (b)
11. (c)
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