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1 | 6318-6321 | We will now illustrate these steps of Corner Point Method by considering some
examples:
Example 1 Solve the following linear programming problem graphically:
Maximise Z = 4x + y (1)
subject to the constraints:
x + y ≤ 50 (2)
3x + y ≤ 90 (3)
x ≥ 0, y ≥ 0 |
1 | 6319-6322 | (1)
subject to the constraints:
x + y ≤ 50 (2)
3x + y ≤ 90 (3)
x ≥ 0, y ≥ 0 (4)
Solution The shaded region in Fig 12 |
1 | 6320-6323 | (2)
3x + y ≤ 90 (3)
x ≥ 0, y ≥ 0 (4)
Solution The shaded region in Fig 12 2 is the feasible region determined by the system
of constraints (2) to (4) |
1 | 6321-6324 | (3)
x ≥ 0, y ≥ 0 (4)
Solution The shaded region in Fig 12 2 is the feasible region determined by the system
of constraints (2) to (4) We observe that the feasible region OABC is bounded |
1 | 6322-6325 | (4)
Solution The shaded region in Fig 12 2 is the feasible region determined by the system
of constraints (2) to (4) We observe that the feasible region OABC is bounded So,
we now use Corner Point Method to determine the maximum value of Z |
1 | 6323-6326 | 2 is the feasible region determined by the system
of constraints (2) to (4) We observe that the feasible region OABC is bounded So,
we now use Corner Point Method to determine the maximum value of Z The coordinates of the corner points O, A, B and C are (0, 0), (30, 0), (20, 30) and
(0, 50) respectively |
1 | 6324-6327 | We observe that the feasible region OABC is bounded So,
we now use Corner Point Method to determine the maximum value of Z The coordinates of the corner points O, A, B and C are (0, 0), (30, 0), (20, 30) and
(0, 50) respectively Now we evaluate Z at each corner point |
1 | 6325-6328 | So,
we now use Corner Point Method to determine the maximum value of Z The coordinates of the corner points O, A, B and C are (0, 0), (30, 0), (20, 30) and
(0, 50) respectively Now we evaluate Z at each corner point © NCERT
not to be republished
510
MATHEMATICS
Fig 12 |
1 | 6326-6329 | The coordinates of the corner points O, A, B and C are (0, 0), (30, 0), (20, 30) and
(0, 50) respectively Now we evaluate Z at each corner point © NCERT
not to be republished
510
MATHEMATICS
Fig 12 2
Hence, maximum value of Z is 120 at the point (30, 0) |
1 | 6327-6330 | Now we evaluate Z at each corner point © NCERT
not to be republished
510
MATHEMATICS
Fig 12 2
Hence, maximum value of Z is 120 at the point (30, 0) Example 2 Solve the following linear programming problem graphically:
Minimise Z = 200 x + 500 y |
1 | 6328-6331 | © NCERT
not to be republished
510
MATHEMATICS
Fig 12 2
Hence, maximum value of Z is 120 at the point (30, 0) Example 2 Solve the following linear programming problem graphically:
Minimise Z = 200 x + 500 y (1)
subject to the constraints:
x + 2y ≥ 10 |
1 | 6329-6332 | 2
Hence, maximum value of Z is 120 at the point (30, 0) Example 2 Solve the following linear programming problem graphically:
Minimise Z = 200 x + 500 y (1)
subject to the constraints:
x + 2y ≥ 10 (2)
3x + 4y ≤ 24 |
1 | 6330-6333 | Example 2 Solve the following linear programming problem graphically:
Minimise Z = 200 x + 500 y (1)
subject to the constraints:
x + 2y ≥ 10 (2)
3x + 4y ≤ 24 (3)
x ≥ 0, y ≥ 0 |
1 | 6331-6334 | (1)
subject to the constraints:
x + 2y ≥ 10 (2)
3x + 4y ≤ 24 (3)
x ≥ 0, y ≥ 0 (4)
Solution The shaded region in Fig 12 |
1 | 6332-6335 | (2)
3x + 4y ≤ 24 (3)
x ≥ 0, y ≥ 0 (4)
Solution The shaded region in Fig 12 3 is the feasible region ABC determined by the
system of constraints (2) to (4), which is bounded |
1 | 6333-6336 | (3)
x ≥ 0, y ≥ 0 (4)
Solution The shaded region in Fig 12 3 is the feasible region ABC determined by the
system of constraints (2) to (4), which is bounded The coordinates of corner points
Corner Point Corresponding value
of Z
(0, 0)
0
(30, 0)
120 ←
Maximum
(20, 30)
110
(0, 50)
50
Corner Point Corresponding value
of Z
(0, 5)
2500
(4, 3)
2300
(0, 6)
3000
Minimum
←
Fig 12 |
1 | 6334-6337 | (4)
Solution The shaded region in Fig 12 3 is the feasible region ABC determined by the
system of constraints (2) to (4), which is bounded The coordinates of corner points
Corner Point Corresponding value
of Z
(0, 0)
0
(30, 0)
120 ←
Maximum
(20, 30)
110
(0, 50)
50
Corner Point Corresponding value
of Z
(0, 5)
2500
(4, 3)
2300
(0, 6)
3000
Minimum
←
Fig 12 3
© NCERT
not to be republished
LINEAR PROGRAMMING 511
A, B and C are (0,5), (4,3) and (0,6) respectively |
1 | 6335-6338 | 3 is the feasible region ABC determined by the
system of constraints (2) to (4), which is bounded The coordinates of corner points
Corner Point Corresponding value
of Z
(0, 0)
0
(30, 0)
120 ←
Maximum
(20, 30)
110
(0, 50)
50
Corner Point Corresponding value
of Z
(0, 5)
2500
(4, 3)
2300
(0, 6)
3000
Minimum
←
Fig 12 3
© NCERT
not to be republished
LINEAR PROGRAMMING 511
A, B and C are (0,5), (4,3) and (0,6) respectively Now we evaluate Z = 200x + 500y
at these points |
1 | 6336-6339 | The coordinates of corner points
Corner Point Corresponding value
of Z
(0, 0)
0
(30, 0)
120 ←
Maximum
(20, 30)
110
(0, 50)
50
Corner Point Corresponding value
of Z
(0, 5)
2500
(4, 3)
2300
(0, 6)
3000
Minimum
←
Fig 12 3
© NCERT
not to be republished
LINEAR PROGRAMMING 511
A, B and C are (0,5), (4,3) and (0,6) respectively Now we evaluate Z = 200x + 500y
at these points Hence, minimum value of Z is 2300 attained at the point (4, 3)
Example 3 Solve the following problem graphically:
Minimise and Maximise Z = 3x + 9y |
1 | 6337-6340 | 3
© NCERT
not to be republished
LINEAR PROGRAMMING 511
A, B and C are (0,5), (4,3) and (0,6) respectively Now we evaluate Z = 200x + 500y
at these points Hence, minimum value of Z is 2300 attained at the point (4, 3)
Example 3 Solve the following problem graphically:
Minimise and Maximise Z = 3x + 9y (1)
subject to the constraints:
x + 3y ≤ 60 |
1 | 6338-6341 | Now we evaluate Z = 200x + 500y
at these points Hence, minimum value of Z is 2300 attained at the point (4, 3)
Example 3 Solve the following problem graphically:
Minimise and Maximise Z = 3x + 9y (1)
subject to the constraints:
x + 3y ≤ 60 (2)
x + y ≥ 10 |
1 | 6339-6342 | Hence, minimum value of Z is 2300 attained at the point (4, 3)
Example 3 Solve the following problem graphically:
Minimise and Maximise Z = 3x + 9y (1)
subject to the constraints:
x + 3y ≤ 60 (2)
x + y ≥ 10 (3)
x ≤ y |
1 | 6340-6343 | (1)
subject to the constraints:
x + 3y ≤ 60 (2)
x + y ≥ 10 (3)
x ≤ y (4)
x ≥ 0, y ≥ 0 |
1 | 6341-6344 | (2)
x + y ≥ 10 (3)
x ≤ y (4)
x ≥ 0, y ≥ 0 (5)
Solution First of all, let us graph the feasible region of the system of linear inequalities
(2) to (5) |
1 | 6342-6345 | (3)
x ≤ y (4)
x ≥ 0, y ≥ 0 (5)
Solution First of all, let us graph the feasible region of the system of linear inequalities
(2) to (5) The feasible region ABCD is shown in the Fig 12 |
1 | 6343-6346 | (4)
x ≥ 0, y ≥ 0 (5)
Solution First of all, let us graph the feasible region of the system of linear inequalities
(2) to (5) The feasible region ABCD is shown in the Fig 12 4 |
1 | 6344-6347 | (5)
Solution First of all, let us graph the feasible region of the system of linear inequalities
(2) to (5) The feasible region ABCD is shown in the Fig 12 4 Note that the region is
bounded |
1 | 6345-6348 | The feasible region ABCD is shown in the Fig 12 4 Note that the region is
bounded The coordinates of the corner points A, B, C and D are (0, 10), (5, 5), (15,15)
and (0, 20) respectively |
1 | 6346-6349 | 4 Note that the region is
bounded The coordinates of the corner points A, B, C and D are (0, 10), (5, 5), (15,15)
and (0, 20) respectively Fig 12 |
1 | 6347-6350 | Note that the region is
bounded The coordinates of the corner points A, B, C and D are (0, 10), (5, 5), (15,15)
and (0, 20) respectively Fig 12 4
Corner
Corresponding value of
Point
Z = 3x + 9y
A (0, 10)
90
B (5, 5)
60
C (15, 15)
180
D (0, 20)
180
Minimum
Maximum
(Multiple
optimal
solutions)
←
}←
We now find the minimum and maximum value of Z |
1 | 6348-6351 | The coordinates of the corner points A, B, C and D are (0, 10), (5, 5), (15,15)
and (0, 20) respectively Fig 12 4
Corner
Corresponding value of
Point
Z = 3x + 9y
A (0, 10)
90
B (5, 5)
60
C (15, 15)
180
D (0, 20)
180
Minimum
Maximum
(Multiple
optimal
solutions)
←
}←
We now find the minimum and maximum value of Z From the table, we find that
the minimum value of Z is 60 at the point B (5, 5) of the feasible region |
1 | 6349-6352 | Fig 12 4
Corner
Corresponding value of
Point
Z = 3x + 9y
A (0, 10)
90
B (5, 5)
60
C (15, 15)
180
D (0, 20)
180
Minimum
Maximum
(Multiple
optimal
solutions)
←
}←
We now find the minimum and maximum value of Z From the table, we find that
the minimum value of Z is 60 at the point B (5, 5) of the feasible region The maximum value of Z on the feasible region occurs at the two corner points
C (15, 15) and D (0, 20) and it is 180 in each case |
1 | 6350-6353 | 4
Corner
Corresponding value of
Point
Z = 3x + 9y
A (0, 10)
90
B (5, 5)
60
C (15, 15)
180
D (0, 20)
180
Minimum
Maximum
(Multiple
optimal
solutions)
←
}←
We now find the minimum and maximum value of Z From the table, we find that
the minimum value of Z is 60 at the point B (5, 5) of the feasible region The maximum value of Z on the feasible region occurs at the two corner points
C (15, 15) and D (0, 20) and it is 180 in each case Remark Observe that in the above example, the problem has multiple optimal solutions
at the corner points C and D, i |
1 | 6351-6354 | From the table, we find that
the minimum value of Z is 60 at the point B (5, 5) of the feasible region The maximum value of Z on the feasible region occurs at the two corner points
C (15, 15) and D (0, 20) and it is 180 in each case Remark Observe that in the above example, the problem has multiple optimal solutions
at the corner points C and D, i e |
1 | 6352-6355 | The maximum value of Z on the feasible region occurs at the two corner points
C (15, 15) and D (0, 20) and it is 180 in each case Remark Observe that in the above example, the problem has multiple optimal solutions
at the corner points C and D, i e the both points produce same maximum value 180 |
1 | 6353-6356 | Remark Observe that in the above example, the problem has multiple optimal solutions
at the corner points C and D, i e the both points produce same maximum value 180 In
such cases, you can see that every point on the line segment CD joining the two corner
points C and D also give the same maximum value |
1 | 6354-6357 | e the both points produce same maximum value 180 In
such cases, you can see that every point on the line segment CD joining the two corner
points C and D also give the same maximum value Same is also true in the case if the
two points produce same minimum value |
1 | 6355-6358 | the both points produce same maximum value 180 In
such cases, you can see that every point on the line segment CD joining the two corner
points C and D also give the same maximum value Same is also true in the case if the
two points produce same minimum value © NCERT
not to be republished
512
MATHEMATICS
Example 4 Determine graphically the minimum value of the objective function
Z = – 50x + 20y |
1 | 6356-6359 | In
such cases, you can see that every point on the line segment CD joining the two corner
points C and D also give the same maximum value Same is also true in the case if the
two points produce same minimum value © NCERT
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512
MATHEMATICS
Example 4 Determine graphically the minimum value of the objective function
Z = – 50x + 20y (1)
subject to the constraints:
2x – y ≥ – 5 |
1 | 6357-6360 | Same is also true in the case if the
two points produce same minimum value © NCERT
not to be republished
512
MATHEMATICS
Example 4 Determine graphically the minimum value of the objective function
Z = – 50x + 20y (1)
subject to the constraints:
2x – y ≥ – 5 (2)
3x + y ≥ 3 |
1 | 6358-6361 | © NCERT
not to be republished
512
MATHEMATICS
Example 4 Determine graphically the minimum value of the objective function
Z = – 50x + 20y (1)
subject to the constraints:
2x – y ≥ – 5 (2)
3x + y ≥ 3 (3)
2x – 3y ≤ 12 |
1 | 6359-6362 | (1)
subject to the constraints:
2x – y ≥ – 5 (2)
3x + y ≥ 3 (3)
2x – 3y ≤ 12 (4)
x ≥ 0, y ≥ 0 |
1 | 6360-6363 | (2)
3x + y ≥ 3 (3)
2x – 3y ≤ 12 (4)
x ≥ 0, y ≥ 0 (5)
Solution First of all, let us graph the feasible region of the system of inequalities (2) to
(5) |
1 | 6361-6364 | (3)
2x – 3y ≤ 12 (4)
x ≥ 0, y ≥ 0 (5)
Solution First of all, let us graph the feasible region of the system of inequalities (2) to
(5) The feasible region (shaded) is shown in the Fig 12 |
1 | 6362-6365 | (4)
x ≥ 0, y ≥ 0 (5)
Solution First of all, let us graph the feasible region of the system of inequalities (2) to
(5) The feasible region (shaded) is shown in the Fig 12 5 |
1 | 6363-6366 | (5)
Solution First of all, let us graph the feasible region of the system of inequalities (2) to
(5) The feasible region (shaded) is shown in the Fig 12 5 Observe that the feasible
region is unbounded |
1 | 6364-6367 | The feasible region (shaded) is shown in the Fig 12 5 Observe that the feasible
region is unbounded We now evaluate Z at the corner points |
1 | 6365-6368 | 5 Observe that the feasible
region is unbounded We now evaluate Z at the corner points From this table, we find that – 300 is the smallest value of Z at the corner point
(6, 0) |
1 | 6366-6369 | Observe that the feasible
region is unbounded We now evaluate Z at the corner points From this table, we find that – 300 is the smallest value of Z at the corner point
(6, 0) Can we say that minimum value of Z is – 300 |
1 | 6367-6370 | We now evaluate Z at the corner points From this table, we find that – 300 is the smallest value of Z at the corner point
(6, 0) Can we say that minimum value of Z is – 300 Note that if the region would
have been bounded, this smallest value of Z is the minimum value of Z (Theorem 2) |
1 | 6368-6371 | From this table, we find that – 300 is the smallest value of Z at the corner point
(6, 0) Can we say that minimum value of Z is – 300 Note that if the region would
have been bounded, this smallest value of Z is the minimum value of Z (Theorem 2) But here we see that the feasible region is unbounded |
1 | 6369-6372 | Can we say that minimum value of Z is – 300 Note that if the region would
have been bounded, this smallest value of Z is the minimum value of Z (Theorem 2) But here we see that the feasible region is unbounded Therefore, – 300 may or may
not be the minimum value of Z |
1 | 6370-6373 | Note that if the region would
have been bounded, this smallest value of Z is the minimum value of Z (Theorem 2) But here we see that the feasible region is unbounded Therefore, – 300 may or may
not be the minimum value of Z To decide this issue, we graph the inequality
– 50x + 20y < – 300 (see Step 3(ii) of corner Point Method |
1 | 6371-6374 | But here we see that the feasible region is unbounded Therefore, – 300 may or may
not be the minimum value of Z To decide this issue, we graph the inequality
– 50x + 20y < – 300 (see Step 3(ii) of corner Point Method )
i |
1 | 6372-6375 | Therefore, – 300 may or may
not be the minimum value of Z To decide this issue, we graph the inequality
– 50x + 20y < – 300 (see Step 3(ii) of corner Point Method )
i e |
1 | 6373-6376 | To decide this issue, we graph the inequality
– 50x + 20y < – 300 (see Step 3(ii) of corner Point Method )
i e ,
– 5x + 2y < – 30
and check whether the resulting open half plane has points in common with feasible
region or not |
1 | 6374-6377 | )
i e ,
– 5x + 2y < – 30
and check whether the resulting open half plane has points in common with feasible
region or not If it has common points, then –300 will not be the minimum value of Z |
1 | 6375-6378 | e ,
– 5x + 2y < – 30
and check whether the resulting open half plane has points in common with feasible
region or not If it has common points, then –300 will not be the minimum value of Z Otherwise, –300 will be the minimum value of Z |
1 | 6376-6379 | ,
– 5x + 2y < – 30
and check whether the resulting open half plane has points in common with feasible
region or not If it has common points, then –300 will not be the minimum value of Z Otherwise, –300 will be the minimum value of Z Fig 12 |
1 | 6377-6380 | If it has common points, then –300 will not be the minimum value of Z Otherwise, –300 will be the minimum value of Z Fig 12 5
Corner Point
Z = – 50x + 20y
(0, 5)
100
(0, 3)
60
(1, 0)
–50
(6, 0)
– 300
smallest
←
© NCERT
not to be republished
LINEAR PROGRAMMING 513
As shown in the Fig 12 |
1 | 6378-6381 | Otherwise, –300 will be the minimum value of Z Fig 12 5
Corner Point
Z = – 50x + 20y
(0, 5)
100
(0, 3)
60
(1, 0)
–50
(6, 0)
– 300
smallest
←
© NCERT
not to be republished
LINEAR PROGRAMMING 513
As shown in the Fig 12 5, it has common points |
1 | 6379-6382 | Fig 12 5
Corner Point
Z = – 50x + 20y
(0, 5)
100
(0, 3)
60
(1, 0)
–50
(6, 0)
– 300
smallest
←
© NCERT
not to be republished
LINEAR PROGRAMMING 513
As shown in the Fig 12 5, it has common points Therefore, Z = –50 x + 20 y
has no minimum value subject to the given constraints |
1 | 6380-6383 | 5
Corner Point
Z = – 50x + 20y
(0, 5)
100
(0, 3)
60
(1, 0)
–50
(6, 0)
– 300
smallest
←
© NCERT
not to be republished
LINEAR PROGRAMMING 513
As shown in the Fig 12 5, it has common points Therefore, Z = –50 x + 20 y
has no minimum value subject to the given constraints In the above example, can you say whether z = – 50 x + 20 y has the maximum
value 100 at (0,5) |
1 | 6381-6384 | 5, it has common points Therefore, Z = –50 x + 20 y
has no minimum value subject to the given constraints In the above example, can you say whether z = – 50 x + 20 y has the maximum
value 100 at (0,5) For this, check whether the graph of – 50 x + 20 y > 100 has points
in common with the feasible region |
1 | 6382-6385 | Therefore, Z = –50 x + 20 y
has no minimum value subject to the given constraints In the above example, can you say whether z = – 50 x + 20 y has the maximum
value 100 at (0,5) For this, check whether the graph of – 50 x + 20 y > 100 has points
in common with the feasible region (Why |
1 | 6383-6386 | In the above example, can you say whether z = – 50 x + 20 y has the maximum
value 100 at (0,5) For this, check whether the graph of – 50 x + 20 y > 100 has points
in common with the feasible region (Why )
Example 5 Minimise Z = 3x + 2y
subject to the constraints:
x + y ≥ 8 |
1 | 6384-6387 | For this, check whether the graph of – 50 x + 20 y > 100 has points
in common with the feasible region (Why )
Example 5 Minimise Z = 3x + 2y
subject to the constraints:
x + y ≥ 8 (1)
3x + 5y ≤ 15 |
1 | 6385-6388 | (Why )
Example 5 Minimise Z = 3x + 2y
subject to the constraints:
x + y ≥ 8 (1)
3x + 5y ≤ 15 (2)
x ≥ 0, y ≥ 0 |
1 | 6386-6389 | )
Example 5 Minimise Z = 3x + 2y
subject to the constraints:
x + y ≥ 8 (1)
3x + 5y ≤ 15 (2)
x ≥ 0, y ≥ 0 (3)
Solution Let us graph the inequalities (1) to (3) (Fig 12 |
1 | 6387-6390 | (1)
3x + 5y ≤ 15 (2)
x ≥ 0, y ≥ 0 (3)
Solution Let us graph the inequalities (1) to (3) (Fig 12 6) |
1 | 6388-6391 | (2)
x ≥ 0, y ≥ 0 (3)
Solution Let us graph the inequalities (1) to (3) (Fig 12 6) Is there any feasible region |
1 | 6389-6392 | (3)
Solution Let us graph the inequalities (1) to (3) (Fig 12 6) Is there any feasible region Why is so |
1 | 6390-6393 | 6) Is there any feasible region Why is so From Fig 12 |
1 | 6391-6394 | Is there any feasible region Why is so From Fig 12 6, you can see that
there is no point satisfying all the
constraints simultaneously |
1 | 6392-6395 | Why is so From Fig 12 6, you can see that
there is no point satisfying all the
constraints simultaneously Thus, the
problem is having no feasible region and
hence no feasible solution |
1 | 6393-6396 | From Fig 12 6, you can see that
there is no point satisfying all the
constraints simultaneously Thus, the
problem is having no feasible region and
hence no feasible solution Remarks From the examples which we
have discussed so far, we notice some
general features of linear programming
problems:
(i)
The feasible region is always a
convex region |
1 | 6394-6397 | 6, you can see that
there is no point satisfying all the
constraints simultaneously Thus, the
problem is having no feasible region and
hence no feasible solution Remarks From the examples which we
have discussed so far, we notice some
general features of linear programming
problems:
(i)
The feasible region is always a
convex region (ii)
The maximum (or minimum)
solution of the objective function occurs at the vertex (corner) of the feasible
region |
1 | 6395-6398 | Thus, the
problem is having no feasible region and
hence no feasible solution Remarks From the examples which we
have discussed so far, we notice some
general features of linear programming
problems:
(i)
The feasible region is always a
convex region (ii)
The maximum (or minimum)
solution of the objective function occurs at the vertex (corner) of the feasible
region 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) value |
1 | 6396-6399 | Remarks From the examples which we
have discussed so far, we notice some
general features of linear programming
problems:
(i)
The feasible region is always a
convex region (ii)
The maximum (or minimum)
solution of the objective function occurs at the vertex (corner) of the feasible
region 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) value EXERCISE 12 |
1 | 6397-6400 | (ii)
The maximum (or minimum)
solution of the objective function occurs at the vertex (corner) of the feasible
region 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) value EXERCISE 12 1
Solve the following Linear Programming Problems graphically:
1 |
1 | 6398-6401 | 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) value EXERCISE 12 1
Solve the following Linear Programming Problems graphically:
1 Maximise Z = 3x + 4y
subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0 |
1 | 6399-6402 | EXERCISE 12 1
Solve the following Linear Programming Problems graphically:
1 Maximise Z = 3x + 4y
subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0 Fig 12 |
1 | 6400-6403 | 1
Solve the following Linear Programming Problems graphically:
1 Maximise Z = 3x + 4y
subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0 Fig 12 6
© NCERT
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514
MATHEMATICS
2 |
1 | 6401-6404 | Maximise Z = 3x + 4y
subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0 Fig 12 6
© NCERT
not to be republished
514
MATHEMATICS
2 Minimise Z = – 3x + 4 y
subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0 |
1 | 6402-6405 | Fig 12 6
© NCERT
not to be republished
514
MATHEMATICS
2 Minimise Z = – 3x + 4 y
subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0 3 |
1 | 6403-6406 | 6
© NCERT
not to be republished
514
MATHEMATICS
2 Minimise Z = – 3x + 4 y
subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0 3 Maximise Z = 5x + 3y
subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0 |
1 | 6404-6407 | Minimise Z = – 3x + 4 y
subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0 3 Maximise Z = 5x + 3y
subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0 4 |
1 | 6405-6408 | 3 Maximise Z = 5x + 3y
subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0 4 Minimise Z = 3x + 5y
such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0 |
1 | 6406-6409 | Maximise Z = 5x + 3y
subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0 4 Minimise Z = 3x + 5y
such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0 5 |
1 | 6407-6410 | 4 Minimise Z = 3x + 5y
such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0 5 Maximise Z = 3x + 2y
subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0 |
1 | 6408-6411 | Minimise Z = 3x + 5y
such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0 5 Maximise Z = 3x + 2y
subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0 6 |
1 | 6409-6412 | 5 Maximise Z = 3x + 2y
subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0 6 Minimise Z = x + 2y
subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0 |
1 | 6410-6413 | Maximise Z = 3x + 2y
subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0 6 Minimise Z = x + 2y
subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0 Show that the minimum of Z occurs at more than two points |
1 | 6411-6414 | 6 Minimise Z = x + 2y
subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0 Show that the minimum of Z occurs at more than two points 7 |
1 | 6412-6415 | Minimise Z = x + 2y
subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0 Show that the minimum of Z occurs at more than two points 7 Minimise and Maximise Z = 5x + 10 y
subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0 |
1 | 6413-6416 | Show that the minimum of Z occurs at more than two points 7 Minimise and Maximise Z = 5x + 10 y
subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0 8 |
1 | 6414-6417 | 7 Minimise and Maximise Z = 5x + 10 y
subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0 8 Minimise and Maximise Z = x + 2y
subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0 |
1 | 6415-6418 | Minimise and Maximise Z = 5x + 10 y
subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0 8 Minimise and Maximise Z = x + 2y
subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0 9 |
1 | 6416-6419 | 8 Minimise and Maximise Z = x + 2y
subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0 9 Maximise Z = – x + 2y, subject to the constraints:
x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0 |
1 | 6417-6420 | Minimise and Maximise Z = x + 2y
subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0 9 Maximise Z = – x + 2y, subject to the constraints:
x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0 10 |
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