Graphical Method of Solution | Mathematics Optional Notes for UPSC PDF Download

 Page 1


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