NCERT Class 12 Maths Linear Programming
NCERT Class 12 Maths Linear Programming
Exercise 12.1
Exercise 12.2
Miscellaneous Exercise
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Exercise 12.1
Chapter 12 Ex.12.1 Question 1
Maximize \(Z = 3x + 4y\)
Subject to the constraints: \(x + y \le 4,\;x \ge 0,\;y \ge 0\)
Solution
The feasible region determined by the constraints, \(x + y \le 4,\;x \ge 0\) and \(y \ge 0\) is given by
Since, the corner points of the feasible region are \(O\left( {0,0} \right),A\left( {4,0} \right)\) and \(B\left( {0,4} \right)\). The value of \(Z\) at these points are as follows:
Corner point 
\(Z = 3x + 4y\) 

\(O\left( {0,0} \right)\) 
0 

\(A\left( {4,0} \right)\) 
12 

\(B\left( {0,4} \right)\) 
16 
\( \to \)Maximum 
Thus, the maximum value of \(Z\) is \(16\) at the point \({\rm{B}}\left( {0,4} \right)\).
Chapter 12 Ex.12.1 Question 2
Minimize \(Z =  3x + 4y\)
Subject to \(x + 2y \le 8,\;3x + 2y \le 12,\;x \ge 0,\;y \ge 0\)
Solution
The feasible region determined by the system of constraints, \(x + 2y \le 8,\;3x + 2y \le 12,\;x \ge 0\) and \(y \ge 0\), is given by
Since, the corner points of the feasible region are \(O\left( {0,0} \right),A\left( {4,0} \right),B\left( {2,3} \right)\) and \(C\left( {0,4} \right)\). The value of \(Z\) at these corner points are as follows:
Corner point 
\(Z =  3x + 4y\) 

\({\rm{O}}\left( {0,0} \right)\) 
0 

\({\rm{A}}\left( {4,0} \right)\) 
\(  12\) 
\( \to \)Minimum 
\({\rm{B}}\left( {2,3} \right)\) 
6 

\({\rm{C}}\left( {0,4} \right)\) 
16 
Thus, the minimum value of \(Z\) is \(  12\) at the point \({\rm{A}}\left( {4,0} \right)\).
Chapter 12 Ex.12.1 Question 3
Maximize \(Z = 5x + 3y\)
Subject to \(3x + 5y \le 15,\;5x + 2y \le 10,\;x \ge 0,\;y \ge 0\)
Solution
The feasible region determined by the system of constraints \(3x + 5y \le 15,\;5x + 2y \le 10,\;x \ge 0\) and \(y \ge 0\), is given by
Since, the corner points of the feasible region are \(O\left( {0,0} \right),A\left( {2,0} \right),B\left( {0,3} \right)\) and \(C\left( {\frac{{20}}{{19}},\frac{{45}}{{19}}} \right)\). The value of \(Z\) at these corner points are as follows:
Corner point 
\(Z = 5x + 3y\) 

\({\rm{O}}\left( {0,0} \right)\) 
0 

\({\rm{A}}\left( {2,0} \right)\) 
10 

\({\rm{B}}\left( {0,3} \right)\) 
9 

\({\rm{C}}\left( {\frac{{20}}{{19}},\frac{{45}}{{19}}} \right)\) 
\(\frac{{235}}{{19}}\) 
\( \to \)Maximum 
Thus, the maximum value of \(Z\) is \(\left( {\frac{{235}}{{19}}} \right)\) at the point \({\rm{C}}\left( {\frac{{20}}{{19}},\frac{{45}}{{19}}} \right)\).
Chapter 12 Ex.12.1 Question 4
Minimize \(Z = 3x + 5y\)
Subject to \(x + 3y \ge 3,\;x + y \ge 2,\;x,\;y \ge 0\)
Solution
The feasible region determined by the system of constraints, \(x + 3y \ge 3,\;x + y \ge 2\) and \(x,\;y \ge 0\), is given by
Since, the feasible region is unbounded, the corner points of the feasible region are \(A\left( {3,0} \right),B\left( {\frac{3}{2},\frac{1}{2}} \right)\) and \(C\left( {0,2} \right)\). The values of \(Z\) at these corner points are as follows:
Corner point 
\(Z = 3x + 5y\) 

\({\rm{A}}\left( {3,0} \right)\) 
9 

\({\rm{B}}\left( {\frac{3}{2},\frac{1}{2}} \right)\) 
7 
\( \to \)Minimum 
\({\rm{C}}\left( {0,2} \right)\) 
10 
As the feasible region is unbounded, therefore, \(7\) may or may not be the minimum value of \(Z.\)
For this, we draw the graph of the inequality, \(3x + 5y < 7\), and check whether the resulting half plane has points in common with the feasible region or not.
Since, feasible region has no common point with \(3x + 5y < 7\)
Thus, the minimum value of \(Z\) is \(7\) at \(\left( {\frac{3}{2},\frac{1}{2}} \right)\).
Chapter 12 Ex.12.1 Question 5
Maximize \(Z = 3x + 2y\)
Subject to \(x + 2y \le 10,\;3x + y \le 15,\;x,\;y \ge 0\)
Solution
The feasible region determined by the constraints, \(x + 2y \le 10,\;3x + y \le 15\) and \(x,\;y \ge 0\), is given by
Since the corner points of the feasible region are \(A\left( {5,0} \right),\;B\left( {4,3} \right)\) and \(C\left( {0,5} \right)\).
The values of \(Z\) at these corner points are as follows:
Corner point 
\(Z = 3x + 2y\) 

\({\rm{A}}\left( {5,0} \right)\) 
15 

\({\rm{B}}\left( {4,3} \right)\) 
18 
\( \to \) Maximum 
\({\rm{C}}\left( {0,5} \right)\) 
10 
Thus, the maximum value of \(Z\) is \(18\) at the point \(B\left( {4,3} \right)\).
Chapter 12 Ex.12.1 Question 6
Minimize \(Z = x + 2y\)
Subject to \(2x + y \ge 3,\;x + 2y \ge 6,\;x,\;y \ge 0\)
Solution
The feasible region determined by the constraints, \(2x + y \ge 3,\;x + 2y \ge 6\) and \(x,\;y \ge 0\), is given by
Since the corner points of the feasible region are \(A\left( {6,0} \right)\) and \(B\left( {0,3} \right)\).
The values of \(Z\) at these corner points are as follows:
Corner point 
\(Z = x + 2y\) 
\({\rm{A}}\left( {6,0} \right)\) 
6 
\({\rm{B}}\left( {0,3} \right)\) 
6 
Since the values of \(Z\) at points \(A\) and \(B\) is same. If we take any other point such as \(\left( {2,2} \right)\)on line \(x + 2y = 6\), then \(Z = 6\).
Thus, the minimum value of \(Z\) occurs for more than 2 points.
Thus, the value of \(Z\) is minimum at every point on the line, \(x + 2y = 6\).
Chapter 12 Ex.12.1 Question 7
Minimize and Maximize \(Z = 5x + 10y\)
subject to \(x + 2y \le 120,\;x + y \ge 60,\;x  2y \ge 0,\;x,\;y \ge 0\)
Solution
The feasible region determined by the constraints,\(x + 2y \le 120,\;x + y \ge 60,\;x  2y \ge 0\) and \(x,\;y \ge 0\) is given by
Since the corner points of the feasible region are \(A\left( {60,0} \right),{\rm{ }}B\left( {120,0} \right),C\left( {60,30} \right)\) and \(D\left( {40,20} \right)\).
The values of \(Z\) at these corner points are as follows:
Corner point 
\(Z = 5x + 10y\) 

\({\rm{A}}\left( {60,0} \right)\) 
300 
\( \to \)Minimum 
\({\rm{B}}\left( {120,0} \right)\) 
600 
\( \to \)Maximum 
\({\rm{C}}\left( {60,30} \right)\) 
600 
\( \to \)Maximum 
\({\rm{D}}\left( {40,20} \right)\) 
400 
The minimum value of \(Z\) is \(300\) at \(A\left( {60,0} \right)\) and the maximum value of \(Z\) is \(600\) at all the points on the line segment joining \(B\left( {120,0} \right)\) and \(C\left( {60,30} \right)\).
Chapter 12 Ex.12.1 Question 8
Minimize and Maximize \(Z = x + 2y\)
Subject to \(x + 2y \ge 100,\;2x  y \le 0,\;2x + y \le 200,\;x,\;y \ge 0\)
Solution
The feasible region determined by the constraints, \(x + 2y \ge 100,\;2x  y \le 0,\;2x + y \le 200\) and \(x,\;y \ge 0\) is given by
The corner points of the feasible region are \({\rm{A}}\left( {0,50} \right){\rm{, B}}\left( {20,40} \right),\;{\rm{C}}\left( {50,100} \right)\;{\rm{and}}\;{\rm{D}}\left( {0,200} \right)\).
The values of \(Z\) at these corner points are as follows:
Corner point 
\(Z = x + 2y\) 

\({\rm{A}}\left( {0,50} \right)\) 
100 
\( \to \)Minimum 
\({\rm{B}}\left( {20,40} \right)\) 
100 
\( \to \)Minimum 
\({\rm{C}}\left( {50,100} \right)\) 
250 

\({\rm{D}}\left( {0,200} \right)\) 
400 
\( \to \)Maximum 
The maximum value of \(Z\) is \(400\) at \(D\left( {0,200} \right)\) and the minimum value of \(Z\) is \(100\) at all the points on the line segment joining \(A\left( {0,50} \right)\) and \(B\left( {20,40} \right)\).
Chapter 12 Ex.12.1 Question 9
Maximize \(Z =  x + 2y\)
subject to the constraints: \(x \ge 3,\;x + y \ge 5,\;x + 2y \ge 6,\;y \ge 0\)
Solution
The feasible region determined by the constraints, \(x \ge 3,\;x + y \ge 5,\;x + 2y \ge 6\) and \(y \ge 0\), is given by
Since, the feasible region is unbounded, the values of \(Z\) at corner points \(A\left( {6,0} \right),B\left( {4,1} \right)\) and \(C\left( {3,2} \right)\) are as follows:
Corner point 
\(Z =  x + 2y\) 
\({\rm{A}}\left( {6,0} \right)\) 
\(Z =  6\) 
\({\rm{B}}\left( {4,1} \right)\) 
\(Z =  2\) 
\({\rm{C}}\left( {3,2} \right)\) 
\(Z = 1\) 
As the feasible region is unbounded, therefore, \(Z = 1\) may or may not be the maximum value.
For this, we graph the inequality, \(  x + 2y > 1\), and check whether the resulting half plane has points in common with the feasible region or not.
The resulting feasible region has points in common with the feasible region.
Thus, \(Z = 1\) is not the maximum value.
Therefore, \(Z\) has no maximum value.
Chapter 12 Ex.12.1 Question 10
Maximize \(Z = x + y\)
Subject to \(x  y \le  1,\;  x + y \le 0,\;x,\;y \ge 0\)
Solution
The feasible region determined by the constraints,\(x  y \le  1,\;  x + y \le 0\) and \(x,\;y \ge 0\), is given by
There is no feasible region and thus, \(Z\) has no maximum value.