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Edurev123 
3.- Graphical Method of Solution 
3.1 Write down the dual of the following LP problem and hence solve it by 
graphical method: 
Minimize,?????????????????????????????????????????????????????????????????? =?? ?? ?? +?? ?? ?? 
?????????????????????? ?????????????????????????????????????????????????? ?? ?? +?? ?? ?=?? ?? ?? ?? +?? ?? ?? ?=?? .?? ?? ?? ,?? ?? ?=?? 
(2011 : 20 Marks) 
Solution: 
The given linear programming problem is 
Minimize,?????????????????????????????????????????????????????????????????? =6?? 1
+4?? 2
 
Constraints ????????????????????????????????????????????????2?? 1
+?? 2
?=1
3?? 1
+4?? 2
?=1.5
?? 1
,?? 2
?=0
 
The given L.P.P. is in the standard primal form (all the constraints involve the sign = if it 
is a problem of minimization). 
In the matrix form, the above problem can be written as 
 Min. ?? ?=(6,4)[?? 1
,?? 2
]=????
 S.T. ?=[
2 1
3 4
][
?? 1
?? 2
]=[
1
1.5
]
 
or 
???? =?? 
? The dual of the given problem is 
 Max. ?? ?? ?=?? '
?? =(1,1.5)[?? 1
,?? 2
]
?=?? 1
+1.5?? 2
 S.T. ?=?? '
?? =?? 
 i.e., [
2 3
1 4
][
?? 1
?? 2
]=[
6
4
]
? 2?? 1
+3?? 2
=6
?? 1
+4?? 2
=4
?? 1
,?? 2
=0
 
Page 2


Edurev123 
3.- Graphical Method of Solution 
3.1 Write down the dual of the following LP problem and hence solve it by 
graphical method: 
Minimize,?????????????????????????????????????????????????????????????????? =?? ?? ?? +?? ?? ?? 
?????????????????????? ?????????????????????????????????????????????????? ?? ?? +?? ?? ?=?? ?? ?? ?? +?? ?? ?? ?=?? .?? ?? ?? ,?? ?? ?=?? 
(2011 : 20 Marks) 
Solution: 
The given linear programming problem is 
Minimize,?????????????????????????????????????????????????????????????????? =6?? 1
+4?? 2
 
Constraints ????????????????????????????????????????????????2?? 1
+?? 2
?=1
3?? 1
+4?? 2
?=1.5
?? 1
,?? 2
?=0
 
The given L.P.P. is in the standard primal form (all the constraints involve the sign = if it 
is a problem of minimization). 
In the matrix form, the above problem can be written as 
 Min. ?? ?=(6,4)[?? 1
,?? 2
]=????
 S.T. ?=[
2 1
3 4
][
?? 1
?? 2
]=[
1
1.5
]
 
or 
???? =?? 
? The dual of the given problem is 
 Max. ?? ?? ?=?? '
?? =(1,1.5)[?? 1
,?? 2
]
?=?? 1
+1.5?? 2
 S.T. ?=?? '
?? =?? 
 i.e., [
2 3
1 4
][
?? 1
?? 2
]=[
6
4
]
? 2?? 1
+3?? 2
=6
?? 1
+4?? 2
=4
?? 1
,?? 2
=0
 
Solution by Graphical Method : 
Draw the lines 
and 
2?? 1
+3?? 2
?=6
?? 1
+4?? 2
?=4 on the graph paper. 
 
 
The shaded region is the permissible region. We can see that maximum ?? ?? is obtained 
at (
12
5
,
2
5
) and (3,0) . 
??? ?? is maximum at all points of the line joining (
12
5
,
2
5
) and (3,0) . 
3.2 For each hour per day that Ashok studies mathematics, it yields him 10 marks 
and for each hour that he studies physics, it yields him 5 marks. He can study at 
most 14 hours a day and he must get at least 40 marks in each. Determine 
graphically how many hours a day he should study mathematics and physics 
each, in order to maximize his marks? 
(2012 : 12 Marks) 
Solution: 
Let Ashok studies math ?? hours per day and Physics ?? hours per day. 
??                                    Total marks =10?? +5?? 
He can study at the most 14 hours a day. 
??????????????????????????????????????????????????????? +?? =14 
He must get at least 40 marks in each paper. 
? ????????????????????????????????????????10?? =40 and 5?? =40
 i.e., ?? =4 and ?? =8
 
Hence, the linear programming problem is 
Page 3


Edurev123 
3.- Graphical Method of Solution 
3.1 Write down the dual of the following LP problem and hence solve it by 
graphical method: 
Minimize,?????????????????????????????????????????????????????????????????? =?? ?? ?? +?? ?? ?? 
?????????????????????? ?????????????????????????????????????????????????? ?? ?? +?? ?? ?=?? ?? ?? ?? +?? ?? ?? ?=?? .?? ?? ?? ,?? ?? ?=?? 
(2011 : 20 Marks) 
Solution: 
The given linear programming problem is 
Minimize,?????????????????????????????????????????????????????????????????? =6?? 1
+4?? 2
 
Constraints ????????????????????????????????????????????????2?? 1
+?? 2
?=1
3?? 1
+4?? 2
?=1.5
?? 1
,?? 2
?=0
 
The given L.P.P. is in the standard primal form (all the constraints involve the sign = if it 
is a problem of minimization). 
In the matrix form, the above problem can be written as 
 Min. ?? ?=(6,4)[?? 1
,?? 2
]=????
 S.T. ?=[
2 1
3 4
][
?? 1
?? 2
]=[
1
1.5
]
 
or 
???? =?? 
? The dual of the given problem is 
 Max. ?? ?? ?=?? '
?? =(1,1.5)[?? 1
,?? 2
]
?=?? 1
+1.5?? 2
 S.T. ?=?? '
?? =?? 
 i.e., [
2 3
1 4
][
?? 1
?? 2
]=[
6
4
]
? 2?? 1
+3?? 2
=6
?? 1
+4?? 2
=4
?? 1
,?? 2
=0
 
Solution by Graphical Method : 
Draw the lines 
and 
2?? 1
+3?? 2
?=6
?? 1
+4?? 2
?=4 on the graph paper. 
 
 
The shaded region is the permissible region. We can see that maximum ?? ?? is obtained 
at (
12
5
,
2
5
) and (3,0) . 
??? ?? is maximum at all points of the line joining (
12
5
,
2
5
) and (3,0) . 
3.2 For each hour per day that Ashok studies mathematics, it yields him 10 marks 
and for each hour that he studies physics, it yields him 5 marks. He can study at 
most 14 hours a day and he must get at least 40 marks in each. Determine 
graphically how many hours a day he should study mathematics and physics 
each, in order to maximize his marks? 
(2012 : 12 Marks) 
Solution: 
Let Ashok studies math ?? hours per day and Physics ?? hours per day. 
??                                    Total marks =10?? +5?? 
He can study at the most 14 hours a day. 
??????????????????????????????????????????????????????? +?? =14 
He must get at least 40 marks in each paper. 
? ????????????????????????????????????????10?? =40 and 5?? =40
 i.e., ?? =4 and ?? =8
 
Hence, the linear programming problem is 
 Max. ?? =10?? +5?? 
Subject to the constraints 
?? +?? ?=14
?? ?=4
?? ?=8
 
Draw the lines, ?? +?? =14,?? =4,?? =8. 
 
The shaded region ?????? is the permissible region Here, 
?? =(4,10),?? =(4,8),?? =(6,8) 
?? at ?? (4,10)?=40+50=90
?? at ?? (4,8)?=40+40=80
?? at ?? (6,8)?=60+40=100
 
? For maximum ?? , 
?? =6,?? =8 
3.3 Maximize ?? =?? ?? ?? +?? ?? ?? -?? ?? ?? subject to ?? ?? +?? ?? +?? ?? =?? and ?? ?? ?? -?? ?? ?? +?? ?? =
???? ;?? ?? =?? . 
(2013 : 10 Marks) 
Solution: 
Approach : A graphical solution is possible only for two variables. We use the first 
condition of equality to convert it into a problem of 2 variables. 
?? 1
+?? 2
+?? 3
?=7??? 3
=7-?? 1
-?? 2
Again????? 3
=0??????????????????????????????????????????7-?? 1
-?? 2
?=0
?? 1
+?? 2
?=7
 
So, the problem becomes 
Page 4


Edurev123 
3.- Graphical Method of Solution 
3.1 Write down the dual of the following LP problem and hence solve it by 
graphical method: 
Minimize,?????????????????????????????????????????????????????????????????? =?? ?? ?? +?? ?? ?? 
?????????????????????? ?????????????????????????????????????????????????? ?? ?? +?? ?? ?=?? ?? ?? ?? +?? ?? ?? ?=?? .?? ?? ?? ,?? ?? ?=?? 
(2011 : 20 Marks) 
Solution: 
The given linear programming problem is 
Minimize,?????????????????????????????????????????????????????????????????? =6?? 1
+4?? 2
 
Constraints ????????????????????????????????????????????????2?? 1
+?? 2
?=1
3?? 1
+4?? 2
?=1.5
?? 1
,?? 2
?=0
 
The given L.P.P. is in the standard primal form (all the constraints involve the sign = if it 
is a problem of minimization). 
In the matrix form, the above problem can be written as 
 Min. ?? ?=(6,4)[?? 1
,?? 2
]=????
 S.T. ?=[
2 1
3 4
][
?? 1
?? 2
]=[
1
1.5
]
 
or 
???? =?? 
? The dual of the given problem is 
 Max. ?? ?? ?=?? '
?? =(1,1.5)[?? 1
,?? 2
]
?=?? 1
+1.5?? 2
 S.T. ?=?? '
?? =?? 
 i.e., [
2 3
1 4
][
?? 1
?? 2
]=[
6
4
]
? 2?? 1
+3?? 2
=6
?? 1
+4?? 2
=4
?? 1
,?? 2
=0
 
Solution by Graphical Method : 
Draw the lines 
and 
2?? 1
+3?? 2
?=6
?? 1
+4?? 2
?=4 on the graph paper. 
 
 
The shaded region is the permissible region. We can see that maximum ?? ?? is obtained 
at (
12
5
,
2
5
) and (3,0) . 
??? ?? is maximum at all points of the line joining (
12
5
,
2
5
) and (3,0) . 
3.2 For each hour per day that Ashok studies mathematics, it yields him 10 marks 
and for each hour that he studies physics, it yields him 5 marks. He can study at 
most 14 hours a day and he must get at least 40 marks in each. Determine 
graphically how many hours a day he should study mathematics and physics 
each, in order to maximize his marks? 
(2012 : 12 Marks) 
Solution: 
Let Ashok studies math ?? hours per day and Physics ?? hours per day. 
??                                    Total marks =10?? +5?? 
He can study at the most 14 hours a day. 
??????????????????????????????????????????????????????? +?? =14 
He must get at least 40 marks in each paper. 
? ????????????????????????????????????????10?? =40 and 5?? =40
 i.e., ?? =4 and ?? =8
 
Hence, the linear programming problem is 
 Max. ?? =10?? +5?? 
Subject to the constraints 
?? +?? ?=14
?? ?=4
?? ?=8
 
Draw the lines, ?? +?? =14,?? =4,?? =8. 
 
The shaded region ?????? is the permissible region Here, 
?? =(4,10),?? =(4,8),?? =(6,8) 
?? at ?? (4,10)?=40+50=90
?? at ?? (4,8)?=40+40=80
?? at ?? (6,8)?=60+40=100
 
? For maximum ?? , 
?? =6,?? =8 
3.3 Maximize ?? =?? ?? ?? +?? ?? ?? -?? ?? ?? subject to ?? ?? +?? ?? +?? ?? =?? and ?? ?? ?? -?? ?? ?? +?? ?? =
???? ;?? ?? =?? . 
(2013 : 10 Marks) 
Solution: 
Approach : A graphical solution is possible only for two variables. We use the first 
condition of equality to convert it into a problem of 2 variables. 
?? 1
+?? 2
+?? 3
?=7??? 3
=7-?? 1
-?? 2
Again????? 3
=0??????????????????????????????????????????7-?? 1
-?? 2
?=0
?? 1
+?? 2
?=7
 
So, the problem becomes 
 Max. ?? =2?? 1
+3?? 2
-5(7-?? 1
-?? 2
)=7?? 1
+8?? 2
-35 
?? 1
+?? 2
=7 
2?? 1
-5?? 2
+(7-?? 1
-?? 2
)=10 
or 
?? 1
-6?? 2
=3 and ?? 1
,?? 2
=0 
We use the graphical method to solve the given problem. 
 
Maxima will occur at a corner point value of ?? at various corner points : 
Point 
?? =7?? 1
+8?? 2
-35 
(0,7) 21
(0,0) -35
(3,0) -14
(
45
7
,
4
7
) 14
4
7
 
? Maxima is at (0,7) and maximum value is 21 . 
? Maxima is 21 at ?? 1
=0,?? 2
=7,?? 3
=0. 
3.4 Solve graphically: 
Maximize : 
?? =?? ?? ?? +?? ?? ?? 
subject to : 
Page 5


Edurev123 
3.- Graphical Method of Solution 
3.1 Write down the dual of the following LP problem and hence solve it by 
graphical method: 
Minimize,?????????????????????????????????????????????????????????????????? =?? ?? ?? +?? ?? ?? 
?????????????????????? ?????????????????????????????????????????????????? ?? ?? +?? ?? ?=?? ?? ?? ?? +?? ?? ?? ?=?? .?? ?? ?? ,?? ?? ?=?? 
(2011 : 20 Marks) 
Solution: 
The given linear programming problem is 
Minimize,?????????????????????????????????????????????????????????????????? =6?? 1
+4?? 2
 
Constraints ????????????????????????????????????????????????2?? 1
+?? 2
?=1
3?? 1
+4?? 2
?=1.5
?? 1
,?? 2
?=0
 
The given L.P.P. is in the standard primal form (all the constraints involve the sign = if it 
is a problem of minimization). 
In the matrix form, the above problem can be written as 
 Min. ?? ?=(6,4)[?? 1
,?? 2
]=????
 S.T. ?=[
2 1
3 4
][
?? 1
?? 2
]=[
1
1.5
]
 
or 
???? =?? 
? The dual of the given problem is 
 Max. ?? ?? ?=?? '
?? =(1,1.5)[?? 1
,?? 2
]
?=?? 1
+1.5?? 2
 S.T. ?=?? '
?? =?? 
 i.e., [
2 3
1 4
][
?? 1
?? 2
]=[
6
4
]
? 2?? 1
+3?? 2
=6
?? 1
+4?? 2
=4
?? 1
,?? 2
=0
 
Solution by Graphical Method : 
Draw the lines 
and 
2?? 1
+3?? 2
?=6
?? 1
+4?? 2
?=4 on the graph paper. 
 
 
The shaded region is the permissible region. We can see that maximum ?? ?? is obtained 
at (
12
5
,
2
5
) and (3,0) . 
??? ?? is maximum at all points of the line joining (
12
5
,
2
5
) and (3,0) . 
3.2 For each hour per day that Ashok studies mathematics, it yields him 10 marks 
and for each hour that he studies physics, it yields him 5 marks. He can study at 
most 14 hours a day and he must get at least 40 marks in each. Determine 
graphically how many hours a day he should study mathematics and physics 
each, in order to maximize his marks? 
(2012 : 12 Marks) 
Solution: 
Let Ashok studies math ?? hours per day and Physics ?? hours per day. 
??                                    Total marks =10?? +5?? 
He can study at the most 14 hours a day. 
??????????????????????????????????????????????????????? +?? =14 
He must get at least 40 marks in each paper. 
? ????????????????????????????????????????10?? =40 and 5?? =40
 i.e., ?? =4 and ?? =8
 
Hence, the linear programming problem is 
 Max. ?? =10?? +5?? 
Subject to the constraints 
?? +?? ?=14
?? ?=4
?? ?=8
 
Draw the lines, ?? +?? =14,?? =4,?? =8. 
 
The shaded region ?????? is the permissible region Here, 
?? =(4,10),?? =(4,8),?? =(6,8) 
?? at ?? (4,10)?=40+50=90
?? at ?? (4,8)?=40+40=80
?? at ?? (6,8)?=60+40=100
 
? For maximum ?? , 
?? =6,?? =8 
3.3 Maximize ?? =?? ?? ?? +?? ?? ?? -?? ?? ?? subject to ?? ?? +?? ?? +?? ?? =?? and ?? ?? ?? -?? ?? ?? +?? ?? =
???? ;?? ?? =?? . 
(2013 : 10 Marks) 
Solution: 
Approach : A graphical solution is possible only for two variables. We use the first 
condition of equality to convert it into a problem of 2 variables. 
?? 1
+?? 2
+?? 3
?=7??? 3
=7-?? 1
-?? 2
Again????? 3
=0??????????????????????????????????????????7-?? 1
-?? 2
?=0
?? 1
+?? 2
?=7
 
So, the problem becomes 
 Max. ?? =2?? 1
+3?? 2
-5(7-?? 1
-?? 2
)=7?? 1
+8?? 2
-35 
?? 1
+?? 2
=7 
2?? 1
-5?? 2
+(7-?? 1
-?? 2
)=10 
or 
?? 1
-6?? 2
=3 and ?? 1
,?? 2
=0 
We use the graphical method to solve the given problem. 
 
Maxima will occur at a corner point value of ?? at various corner points : 
Point 
?? =7?? 1
+8?? 2
-35 
(0,7) 21
(0,0) -35
(3,0) -14
(
45
7
,
4
7
) 14
4
7
 
? Maxima is at (0,7) and maximum value is 21 . 
? Maxima is 21 at ?? 1
=0,?? 2
=7,?? 3
=0. 
3.4 Solve graphically: 
Maximize : 
?? =?? ?? ?? +?? ?? ?? 
subject to : 
?? ?? ?? +?? ?? ?=????
?? ?? +?? ?? ?=????
?? ?? +?? ?? ?? ?=?? ?? ?? ?? +?? ?? ?? ?=????
?? ?? ,?? ?? ?=?? 
(2014: 10 marks) 
Solution: 
Since every point which satisfies the condition ?? 1
=0,?? 2
=0 lies in the first quadrant 
only. 
? The desired pair (?? 1
,?? 2
) is restricted to the points of the first quadrant only. 
The following are the end points of the straight line in 1st quadrant. 
Let 
2?? 1
+?? 2
=16?(8,0)(0,16)
?? 1
+?? 2
=11?(11,0)(0,11)
?? 1
+2?? 2
=6?(6,0)(0,3)
5?? 1
+6?? 2
=90?(18,0)(0,15)
 
 
The shaded region ?????????? is the feasible region corresponding to the given constraints 
with ?? (0,3),?? (6,0) , ?? (8,0),?? (5,6),?? (0,11) as the extreme points. 
The values of the objective function ?? =6?? 1
+5?? 2
 at these extreme points are: 
?? (0,3)=0+15=15
?? (6,0)=36+0=36
?? (8,0)=48+0=48
?? (5,6)=30+30=60?
?? (0,11)=0+55=55
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Nov 21, 2024 Last updated
Document Description: Graphical Method of Solution for UPSC 2024 is part of Mathematics Optional Notes for UPSC preparation. The notes and questions for Graphical Method of Solution have been prepared according to the UPSC exam syllabus. Information about Graphical Method of Solution covers topics like and Graphical Method of Solution Example, for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Graphical Method of Solution.
Introduction of Graphical Method of Solution in English is available as part of our Mathematics Optional Notes for UPSC for UPSC & Graphical Method of Solution in Hindi for Mathematics Optional Notes for UPSC course. Download more important topics related with notes, lectures and mock test series for UPSC Exam by signing up for free. UPSC: Graphical Method of Solution | Mathematics Optional Notes for UPSC
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