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1 | 6618-6621 | Then (8 – x – y) units will be transported to depot
at C (Why )
Hence, we have
x ≥ 0, y ≥ 0
and
8 – x – y ≥ 0
i e x ≥ 0, y ≥ 0 and x + y ≤ 8
Now, the weekly requirement of the depot at A is 5 units of the commodity |
1 | 6619-6622 | )
Hence, we have
x ≥ 0, y ≥ 0
and
8 – x – y ≥ 0
i e x ≥ 0, y ≥ 0 and x + y ≤ 8
Now, the weekly requirement of the depot at A is 5 units of the commodity Since
x units are transported from the factory at P, the remaining (5 – x) units need to be
transported from the factory at Q |
1 | 6620-6623 | e x ≥ 0, y ≥ 0 and x + y ≤ 8
Now, the weekly requirement of the depot at A is 5 units of the commodity Since
x units are transported from the factory at P, the remaining (5 – x) units need to be
transported from the factory at Q Obviously, 5 – x ≥ 0, i |
1 | 6621-6624 | x ≥ 0, y ≥ 0 and x + y ≤ 8
Now, the weekly requirement of the depot at A is 5 units of the commodity Since
x units are transported from the factory at P, the remaining (5 – x) units need to be
transported from the factory at Q Obviously, 5 – x ≥ 0, i e |
1 | 6622-6625 | Since
x units are transported from the factory at P, the remaining (5 – x) units need to be
transported from the factory at Q Obviously, 5 – x ≥ 0, i e x ≤ 5 |
1 | 6623-6626 | Obviously, 5 – x ≥ 0, i e x ≤ 5 Similarly, (5 – y) and 6 – (5 – x + 5 – y) = x + y – 4 units are to be transported from
the factory at Q to the depots at B and C respectively |
1 | 6624-6627 | e x ≤ 5 Similarly, (5 – y) and 6 – (5 – x + 5 – y) = x + y – 4 units are to be transported from
the factory at Q to the depots at B and C respectively Thus,
5 – y ≥ 0 , x + y – 4 ≥0
i |
1 | 6625-6628 | x ≤ 5 Similarly, (5 – y) and 6 – (5 – x + 5 – y) = x + y – 4 units are to be transported from
the factory at Q to the depots at B and C respectively Thus,
5 – y ≥ 0 , x + y – 4 ≥0
i e |
1 | 6626-6629 | Similarly, (5 – y) and 6 – (5 – x + 5 – y) = x + y – 4 units are to be transported from
the factory at Q to the depots at B and C respectively Thus,
5 – y ≥ 0 , x + y – 4 ≥0
i e y ≤ 5 , x + y ≥
4
Fig 12 |
1 | 6627-6630 | Thus,
5 – y ≥ 0 , x + y – 4 ≥0
i e y ≤ 5 , x + y ≥
4
Fig 12 12
© NCERT
not to be republished
LINEAR PROGRAMMING 525
Total transportation cost Z is given by
Z = 160 x + 100 y + 100 ( 5 – x) + 120 (5 – y) + 100 (x + y – 4) + 150 (8 – x – y)
= 10 (x – 7 y + 190)
Therefore, the problem reduces to
Minimise Z = 10 (x – 7y + 190)
subject to the constraints:
x ≥ 0, y ≥ 0 |
1 | 6628-6631 | e y ≤ 5 , x + y ≥
4
Fig 12 12
© NCERT
not to be republished
LINEAR PROGRAMMING 525
Total transportation cost Z is given by
Z = 160 x + 100 y + 100 ( 5 – x) + 120 (5 – y) + 100 (x + y – 4) + 150 (8 – x – y)
= 10 (x – 7 y + 190)
Therefore, the problem reduces to
Minimise Z = 10 (x – 7y + 190)
subject to the constraints:
x ≥ 0, y ≥ 0 (1)
x + y ≤ 8 |
1 | 6629-6632 | y ≤ 5 , x + y ≥
4
Fig 12 12
© NCERT
not to be republished
LINEAR PROGRAMMING 525
Total transportation cost Z is given by
Z = 160 x + 100 y + 100 ( 5 – x) + 120 (5 – y) + 100 (x + y – 4) + 150 (8 – x – y)
= 10 (x – 7 y + 190)
Therefore, the problem reduces to
Minimise Z = 10 (x – 7y + 190)
subject to the constraints:
x ≥ 0, y ≥ 0 (1)
x + y ≤ 8 (2)
x ≤ 5 |
1 | 6630-6633 | 12
© NCERT
not to be republished
LINEAR PROGRAMMING 525
Total transportation cost Z is given by
Z = 160 x + 100 y + 100 ( 5 – x) + 120 (5 – y) + 100 (x + y – 4) + 150 (8 – x – y)
= 10 (x – 7 y + 190)
Therefore, the problem reduces to
Minimise Z = 10 (x – 7y + 190)
subject to the constraints:
x ≥ 0, y ≥ 0 (1)
x + y ≤ 8 (2)
x ≤ 5 (3)
y ≤ 5 |
1 | 6631-6634 | (1)
x + y ≤ 8 (2)
x ≤ 5 (3)
y ≤ 5 (4)
and
x + y ≥ 4 |
1 | 6632-6635 | (2)
x ≤ 5 (3)
y ≤ 5 (4)
and
x + y ≥ 4 (5)
The shaded region ABCDEF
represented by the constraints (1) to
(5) is the feasible region (Fig 12 |
1 | 6633-6636 | (3)
y ≤ 5 (4)
and
x + y ≥ 4 (5)
The shaded region ABCDEF
represented by the constraints (1) to
(5) is the feasible region (Fig 12 13) |
1 | 6634-6637 | (4)
and
x + y ≥ 4 (5)
The shaded region ABCDEF
represented by the constraints (1) to
(5) is the feasible region (Fig 12 13) Observe that the feasible region is bounded |
1 | 6635-6638 | (5)
The shaded region ABCDEF
represented by the constraints (1) to
(5) is the feasible region (Fig 12 13) Observe that the feasible region is bounded The coordinates of the corner points
of the feasible region are (0, 4), (0, 5), (3, 5), (5, 3), (5, 0) and (4, 0) |
1 | 6636-6639 | 13) Observe that the feasible region is bounded The coordinates of the corner points
of the feasible region are (0, 4), (0, 5), (3, 5), (5, 3), (5, 0) and (4, 0) Let us evaluate Z at these points |
1 | 6637-6640 | Observe that the feasible region is bounded The coordinates of the corner points
of the feasible region are (0, 4), (0, 5), (3, 5), (5, 3), (5, 0) and (4, 0) Let us evaluate Z at these points Corner Point
Z = 10 (x – 7 y + 190)
(0, 4)
1620
(0, 5)
1550 ←
←
←
←
←
Minimum
(3, 5)
1580
(5, 3)
1740
(5, 0)
1950
(4, 0)
1940
From the table, we see that the minimum value of Z is 1550 at the point (0, 5) |
1 | 6638-6641 | The coordinates of the corner points
of the feasible region are (0, 4), (0, 5), (3, 5), (5, 3), (5, 0) and (4, 0) Let us evaluate Z at these points Corner Point
Z = 10 (x – 7 y + 190)
(0, 4)
1620
(0, 5)
1550 ←
←
←
←
←
Minimum
(3, 5)
1580
(5, 3)
1740
(5, 0)
1950
(4, 0)
1940
From the table, we see that the minimum value of Z is 1550 at the point (0, 5) Hence, the optimal transportation strategy will be to deliver 0, 5 and 3 units from
the factory at P and 5, 0 and 1 units from the factory at Q to the depots at A, B and C
respectively |
1 | 6639-6642 | Let us evaluate Z at these points Corner Point
Z = 10 (x – 7 y + 190)
(0, 4)
1620
(0, 5)
1550 ←
←
←
←
←
Minimum
(3, 5)
1580
(5, 3)
1740
(5, 0)
1950
(4, 0)
1940
From the table, we see that the minimum value of Z is 1550 at the point (0, 5) Hence, the optimal transportation strategy will be to deliver 0, 5 and 3 units from
the factory at P and 5, 0 and 1 units from the factory at Q to the depots at A, B and C
respectively Corresponding to this strategy, the transportation cost would be minimum,
i |
1 | 6640-6643 | Corner Point
Z = 10 (x – 7 y + 190)
(0, 4)
1620
(0, 5)
1550 ←
←
←
←
←
Minimum
(3, 5)
1580
(5, 3)
1740
(5, 0)
1950
(4, 0)
1940
From the table, we see that the minimum value of Z is 1550 at the point (0, 5) Hence, the optimal transportation strategy will be to deliver 0, 5 and 3 units from
the factory at P and 5, 0 and 1 units from the factory at Q to the depots at A, B and C
respectively Corresponding to this strategy, the transportation cost would be minimum,
i e |
1 | 6641-6644 | Hence, the optimal transportation strategy will be to deliver 0, 5 and 3 units from
the factory at P and 5, 0 and 1 units from the factory at Q to the depots at A, B and C
respectively Corresponding to this strategy, the transportation cost would be minimum,
i e , Rs 1550 |
1 | 6642-6645 | Corresponding to this strategy, the transportation cost would be minimum,
i e , Rs 1550 Miscellaneous Exercise on Chapter 12
1 |
1 | 6643-6646 | e , Rs 1550 Miscellaneous Exercise on Chapter 12
1 Refer to Example 9 |
1 | 6644-6647 | , Rs 1550 Miscellaneous Exercise on Chapter 12
1 Refer to Example 9 How many packets of each food should be used to maximise
the amount of vitamin A in the diet |
1 | 6645-6648 | Miscellaneous Exercise on Chapter 12
1 Refer to Example 9 How many packets of each food should be used to maximise
the amount of vitamin A in the diet What is the maximum amount of vitamin A
in the diet |
1 | 6646-6649 | Refer to Example 9 How many packets of each food should be used to maximise
the amount of vitamin A in the diet What is the maximum amount of vitamin A
in the diet Fig 12 |
1 | 6647-6650 | How many packets of each food should be used to maximise
the amount of vitamin A in the diet What is the maximum amount of vitamin A
in the diet Fig 12 13
© NCERT
not to be republished
526
MATHEMATICS
2 |
1 | 6648-6651 | What is the maximum amount of vitamin A
in the diet Fig 12 13
© NCERT
not to be republished
526
MATHEMATICS
2 A farmer mixes two brands P and Q of cattle feed |
1 | 6649-6652 | Fig 12 13
© NCERT
not to be republished
526
MATHEMATICS
2 A farmer mixes two brands P and Q of cattle feed Brand P, costing Rs 250 per
bag, contains 3 units of nutritional element A, 2 |
1 | 6650-6653 | 13
© NCERT
not to be republished
526
MATHEMATICS
2 A farmer mixes two brands P and Q of cattle feed Brand P, costing Rs 250 per
bag, contains 3 units of nutritional element A, 2 5 units of element B and 2 units
of element C |
1 | 6651-6654 | A farmer mixes two brands P and Q of cattle feed Brand P, costing Rs 250 per
bag, contains 3 units of nutritional element A, 2 5 units of element B and 2 units
of element C Brand Q costing Rs 200 per bag contains 1 |
1 | 6652-6655 | Brand P, costing Rs 250 per
bag, contains 3 units of nutritional element A, 2 5 units of element B and 2 units
of element C Brand Q costing Rs 200 per bag contains 1 5 units of nutritional
element A, 11 |
1 | 6653-6656 | 5 units of element B and 2 units
of element C Brand Q costing Rs 200 per bag contains 1 5 units of nutritional
element A, 11 25 units of element B, and 3 units of element C |
1 | 6654-6657 | Brand Q costing Rs 200 per bag contains 1 5 units of nutritional
element A, 11 25 units of element B, and 3 units of element C The minimum
requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively |
1 | 6655-6658 | 5 units of nutritional
element A, 11 25 units of element B, and 3 units of element C The minimum
requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively Determine the number of bags of each brand which should be mixed in order to
produce a mixture having a minimum cost per bag |
1 | 6656-6659 | 25 units of element B, and 3 units of element C The minimum
requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively Determine the number of bags of each brand which should be mixed in order to
produce a mixture having a minimum cost per bag What is the minimum cost of
the mixture per bag |
1 | 6657-6660 | The minimum
requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively Determine the number of bags of each brand which should be mixed in order to
produce a mixture having a minimum cost per bag What is the minimum cost of
the mixture per bag 3 |
1 | 6658-6661 | Determine the number of bags of each brand which should be mixed in order to
produce a mixture having a minimum cost per bag What is the minimum cost of
the mixture per bag 3 A dietician wishes to mix together two kinds of food X and Y in such a way that
the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and
8 units of vitamin C |
1 | 6659-6662 | What is the minimum cost of
the mixture per bag 3 A dietician wishes to mix together two kinds of food X and Y in such a way that
the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and
8 units of vitamin C The vitamin contents of one kg food is given below:
Food
Vitamin A Vitamin B
Vitamin C
X
1
2
3
Y
2
2
1
One kg of food X costs Rs 16 and one kg of food Y costs Rs 20 |
1 | 6660-6663 | 3 A dietician wishes to mix together two kinds of food X and Y in such a way that
the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and
8 units of vitamin C The vitamin contents of one kg food is given below:
Food
Vitamin A Vitamin B
Vitamin C
X
1
2
3
Y
2
2
1
One kg of food X costs Rs 16 and one kg of food Y costs Rs 20 Find the least
cost of the mixture which will produce the required diet |
1 | 6661-6664 | A dietician wishes to mix together two kinds of food X and Y in such a way that
the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and
8 units of vitamin C The vitamin contents of one kg food is given below:
Food
Vitamin A Vitamin B
Vitamin C
X
1
2
3
Y
2
2
1
One kg of food X costs Rs 16 and one kg of food Y costs Rs 20 Find the least
cost of the mixture which will produce the required diet 4 |
1 | 6662-6665 | The vitamin contents of one kg food is given below:
Food
Vitamin A Vitamin B
Vitamin C
X
1
2
3
Y
2
2
1
One kg of food X costs Rs 16 and one kg of food Y costs Rs 20 Find the least
cost of the mixture which will produce the required diet 4 A manufacturer makes two types of toys A and B |
1 | 6663-6666 | Find the least
cost of the mixture which will produce the required diet 4 A manufacturer makes two types of toys A and B Three machines are needed
for this purpose and the time (in minutes) required for each toy on the machines
is given below:
Types of Toys
Machines
I
II
III
A
12
18
6
B
6
0
9
Each machine is available for a maximum of 6 hours per day |
1 | 6664-6667 | 4 A manufacturer makes two types of toys A and B Three machines are needed
for this purpose and the time (in minutes) required for each toy on the machines
is given below:
Types of Toys
Machines
I
II
III
A
12
18
6
B
6
0
9
Each machine is available for a maximum of 6 hours per day If the profit on
each toy of type A is Rs 7 |
1 | 6665-6668 | A manufacturer makes two types of toys A and B Three machines are needed
for this purpose and the time (in minutes) required for each toy on the machines
is given below:
Types of Toys
Machines
I
II
III
A
12
18
6
B
6
0
9
Each machine is available for a maximum of 6 hours per day If the profit on
each toy of type A is Rs 7 50 and that on each toy of type B is Rs 5, show that 15
toys of type A and 30 of type B should be manufactured in a day to get maximum
profit |
1 | 6666-6669 | Three machines are needed
for this purpose and the time (in minutes) required for each toy on the machines
is given below:
Types of Toys
Machines
I
II
III
A
12
18
6
B
6
0
9
Each machine is available for a maximum of 6 hours per day If the profit on
each toy of type A is Rs 7 50 and that on each toy of type B is Rs 5, show that 15
toys of type A and 30 of type B should be manufactured in a day to get maximum
profit 5 |
1 | 6667-6670 | If the profit on
each toy of type A is Rs 7 50 and that on each toy of type B is Rs 5, show that 15
toys of type A and 30 of type B should be manufactured in a day to get maximum
profit 5 An aeroplane can carry a maximum of 200 passengers |
1 | 6668-6671 | 50 and that on each toy of type B is Rs 5, show that 15
toys of type A and 30 of type B should be manufactured in a day to get maximum
profit 5 An aeroplane can carry a maximum of 200 passengers A profit of Rs 1000 is
made on each executive class ticket and a profit of Rs 600 is made on each
economy class ticket |
1 | 6669-6672 | 5 An aeroplane can carry a maximum of 200 passengers A profit of Rs 1000 is
made on each executive class ticket and a profit of Rs 600 is made on each
economy class ticket The airline reserves at least 20 seats for executive class |
1 | 6670-6673 | An aeroplane can carry a maximum of 200 passengers A profit of Rs 1000 is
made on each executive class ticket and a profit of Rs 600 is made on each
economy class ticket The airline reserves at least 20 seats for executive class However, at least 4 times as many passengers prefer to travel by economy class
than by the executive class |
1 | 6671-6674 | A profit of Rs 1000 is
made on each executive class ticket and a profit of Rs 600 is made on each
economy class ticket The airline reserves at least 20 seats for executive class However, at least 4 times as many passengers prefer to travel by economy class
than by the executive class Determine how many tickets of each type must be
sold in order to maximise the profit for the airline |
1 | 6672-6675 | The airline reserves at least 20 seats for executive class However, at least 4 times as many passengers prefer to travel by economy class
than by the executive class Determine how many tickets of each type must be
sold in order to maximise the profit for the airline What is the maximum profit |
1 | 6673-6676 | However, at least 4 times as many passengers prefer to travel by economy class
than by the executive class Determine how many tickets of each type must be
sold in order to maximise the profit for the airline What is the maximum profit © NCERT
not to be republished
LINEAR PROGRAMMING 527
6 |
1 | 6674-6677 | Determine how many tickets of each type must be
sold in order to maximise the profit for the airline What is the maximum profit © NCERT
not to be republished
LINEAR PROGRAMMING 527
6 Two godowns A and B have grain capacity of 100 quintals and 50 quintals
respectively |
1 | 6675-6678 | What is the maximum profit © NCERT
not to be republished
LINEAR PROGRAMMING 527
6 Two godowns A and B have grain capacity of 100 quintals and 50 quintals
respectively They supply to 3 ration shops, D, E and F whose requirements are
60, 50 and 40 quintals respectively |
1 | 6676-6679 | © NCERT
not to be republished
LINEAR PROGRAMMING 527
6 Two godowns A and B have grain capacity of 100 quintals and 50 quintals
respectively They supply to 3 ration shops, D, E and F whose requirements are
60, 50 and 40 quintals respectively The cost of transportation per quintal from
the godowns to the shops are given in the following table:
Transportation cost per quintal (in Rs)
From/To
A
B
D
6
4
E
3
2
F
2 |
1 | 6677-6680 | Two godowns A and B have grain capacity of 100 quintals and 50 quintals
respectively They supply to 3 ration shops, D, E and F whose requirements are
60, 50 and 40 quintals respectively The cost of transportation per quintal from
the godowns to the shops are given in the following table:
Transportation cost per quintal (in Rs)
From/To
A
B
D
6
4
E
3
2
F
2 50
3
How should the supplies be transported in order that the transportation cost is
minimum |
1 | 6678-6681 | They supply to 3 ration shops, D, E and F whose requirements are
60, 50 and 40 quintals respectively The cost of transportation per quintal from
the godowns to the shops are given in the following table:
Transportation cost per quintal (in Rs)
From/To
A
B
D
6
4
E
3
2
F
2 50
3
How should the supplies be transported in order that the transportation cost is
minimum What is the minimum cost |
1 | 6679-6682 | The cost of transportation per quintal from
the godowns to the shops are given in the following table:
Transportation cost per quintal (in Rs)
From/To
A
B
D
6
4
E
3
2
F
2 50
3
How should the supplies be transported in order that the transportation cost is
minimum What is the minimum cost 7 |
1 | 6680-6683 | 50
3
How should the supplies be transported in order that the transportation cost is
minimum What is the minimum cost 7 An oil company has two depots A and B with capacities of 7000 L and 4000 L
respectively |
1 | 6681-6684 | What is the minimum cost 7 An oil company has two depots A and B with capacities of 7000 L and 4000 L
respectively The company is to supply oil to three petrol pumps, D, E and F
whose requirements are 4500L, 3000L and 3500L respectively |
1 | 6682-6685 | 7 An oil company has two depots A and B with capacities of 7000 L and 4000 L
respectively The company is to supply oil to three petrol pumps, D, E and F
whose requirements are 4500L, 3000L and 3500L respectively The distances
(in km) between the depots and the petrol pumps is given in the following table:
Distance in (km |
1 | 6683-6686 | An oil company has two depots A and B with capacities of 7000 L and 4000 L
respectively The company is to supply oil to three petrol pumps, D, E and F
whose requirements are 4500L, 3000L and 3500L respectively The distances
(in km) between the depots and the petrol pumps is given in the following table:
Distance in (km )
From / To
A
B
D
7
3
E
6
4
F
3
2
Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how
should the delivery be scheduled in order that the transportation cost is minimum |
1 | 6684-6687 | The company is to supply oil to three petrol pumps, D, E and F
whose requirements are 4500L, 3000L and 3500L respectively The distances
(in km) between the depots and the petrol pumps is given in the following table:
Distance in (km )
From / To
A
B
D
7
3
E
6
4
F
3
2
Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how
should the delivery be scheduled in order that the transportation cost is minimum What is the minimum cost |
1 | 6685-6688 | The distances
(in km) between the depots and the petrol pumps is given in the following table:
Distance in (km )
From / To
A
B
D
7
3
E
6
4
F
3
2
Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how
should the delivery be scheduled in order that the transportation cost is minimum What is the minimum cost 8 |
1 | 6686-6689 | )
From / To
A
B
D
7
3
E
6
4
F
3
2
Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how
should the delivery be scheduled in order that the transportation cost is minimum What is the minimum cost 8 A fruit grower can use two types of fertilizer in his garden, brand P and brand Q |
1 | 6687-6690 | What is the minimum cost 8 A fruit grower can use two types of fertilizer in his garden, brand P and brand Q The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of
each brand are given in the table |
1 | 6688-6691 | 8 A fruit grower can use two types of fertilizer in his garden, brand P and brand Q The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of
each brand are given in the table Tests indicate that the garden needs at least
240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of
chlorine |
1 | 6689-6692 | A fruit grower can use two types of fertilizer in his garden, brand P and brand Q The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of
each brand are given in the table Tests indicate that the garden needs at least
240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of
chlorine If the grower wants to minimise the amount of nitrogen added to the garden,
how many bags of each brand should be used |
1 | 6690-6693 | The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of
each brand are given in the table Tests indicate that the garden needs at least
240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of
chlorine If the grower wants to minimise the amount of nitrogen added to the garden,
how many bags of each brand should be used What is the minimum amount of
nitrogen added in the garden |
1 | 6691-6694 | Tests indicate that the garden needs at least
240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of
chlorine If the grower wants to minimise the amount of nitrogen added to the garden,
how many bags of each brand should be used What is the minimum amount of
nitrogen added in the garden © NCERT
not to be republished
528
MATHEMATICS
kg per bag
Brand P
Brand Q
Nitrogen
3
3 |
1 | 6692-6695 | If the grower wants to minimise the amount of nitrogen added to the garden,
how many bags of each brand should be used What is the minimum amount of
nitrogen added in the garden © NCERT
not to be republished
528
MATHEMATICS
kg per bag
Brand P
Brand Q
Nitrogen
3
3 5
Phosphoric acid
1
2
Potash
3
1 |
1 | 6693-6696 | What is the minimum amount of
nitrogen added in the garden © NCERT
not to be republished
528
MATHEMATICS
kg per bag
Brand P
Brand Q
Nitrogen
3
3 5
Phosphoric acid
1
2
Potash
3
1 5
Chlorine
1 |
1 | 6694-6697 | © NCERT
not to be republished
528
MATHEMATICS
kg per bag
Brand P
Brand Q
Nitrogen
3
3 5
Phosphoric acid
1
2
Potash
3
1 5
Chlorine
1 5
2
9 |
1 | 6695-6698 | 5
Phosphoric acid
1
2
Potash
3
1 5
Chlorine
1 5
2
9 Refer to Question 8 |
1 | 6696-6699 | 5
Chlorine
1 5
2
9 Refer to Question 8 If the grower wants to maximise the amount of nitrogen
added to the garden, how many bags of each brand should be added |
1 | 6697-6700 | 5
2
9 Refer to Question 8 If the grower wants to maximise the amount of nitrogen
added to the garden, how many bags of each brand should be added What is
the maximum amount of nitrogen added |
1 | 6698-6701 | Refer to Question 8 If the grower wants to maximise the amount of nitrogen
added to the garden, how many bags of each brand should be added What is
the maximum amount of nitrogen added 10 |
1 | 6699-6702 | If the grower wants to maximise the amount of nitrogen
added to the garden, how many bags of each brand should be added What is
the maximum amount of nitrogen added 10 A toy company manufactures two types of dolls, A and B |
1 | 6700-6703 | What is
the maximum amount of nitrogen added 10 A toy company manufactures two types of dolls, A and B Market tests and available
resources have indicated that the combined production level should not exceed 1200
dolls per week and the demand for dolls of type B is at most half of that for dolls of
type A |
1 | 6701-6704 | 10 A toy company manufactures two types of dolls, A and B Market tests and available
resources have indicated that the combined production level should not exceed 1200
dolls per week and the demand for dolls of type B is at most half of that for dolls of
type A Further, the production level of dolls of type A can exceed three times the
production of dolls of other type by at most 600 units |
1 | 6702-6705 | A toy company manufactures two types of dolls, A and B Market tests and available
resources have indicated that the combined production level should not exceed 1200
dolls per week and the demand for dolls of type B is at most half of that for dolls of
type A Further, the production level of dolls of type A can exceed three times the
production of dolls of other type by at most 600 units If the company makes profit of
Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be
produced weekly in order to maximise the profit |
1 | 6703-6706 | Market tests and available
resources have indicated that the combined production level should not exceed 1200
dolls per week and the demand for dolls of type B is at most half of that for dolls of
type A Further, the production level of dolls of type A can exceed three times the
production of dolls of other type by at most 600 units If the company makes profit of
Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be
produced weekly in order to maximise the profit Summary
� A linear programming problem is one that is concerned with finding the optimal
value (maximum or minimum) of a linear function of several variables (called
objective function) subject to the conditions that the variables are
non-negative and satisfy a set of linear inequalities (called linear constraints) |
1 | 6704-6707 | Further, the production level of dolls of type A can exceed three times the
production of dolls of other type by at most 600 units If the company makes profit of
Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be
produced weekly in order to maximise the profit Summary
� A linear programming problem is one that is concerned with finding the optimal
value (maximum or minimum) of a linear function of several variables (called
objective function) subject to the conditions that the variables are
non-negative and satisfy a set of linear inequalities (called linear constraints) Variables are sometimes called decision variables and are non-negative |
1 | 6705-6708 | If the company makes profit of
Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be
produced weekly in order to maximise the profit Summary
� A linear programming problem is one that is concerned with finding the optimal
value (maximum or minimum) of a linear function of several variables (called
objective function) subject to the conditions that the variables are
non-negative and satisfy a set of linear inequalities (called linear constraints) Variables are sometimes called decision variables and are non-negative � A few important linear programming problems are:
(i) Diet problems
(ii) Manufacturing problems
(iii) Transportation problems
� The common region determined by all the constraints including the non-negative
constraints x ≥ 0, y ≥ 0 of a linear programming problem is called the feasible
region (or solution region) for the problem |
1 | 6706-6709 | Summary
� A linear programming problem is one that is concerned with finding the optimal
value (maximum or minimum) of a linear function of several variables (called
objective function) subject to the conditions that the variables are
non-negative and satisfy a set of linear inequalities (called linear constraints) Variables are sometimes called decision variables and are non-negative � A few important linear programming problems are:
(i) Diet problems
(ii) Manufacturing problems
(iii) Transportation problems
� The common region determined by all the constraints including the non-negative
constraints x ≥ 0, y ≥ 0 of a linear programming problem is called the feasible
region (or solution region) for the problem � Points within and on the boundary of the feasible region represent feasible
solutions of the constraints |
1 | 6707-6710 | Variables are sometimes called decision variables and are non-negative � A few important linear programming problems are:
(i) Diet problems
(ii) Manufacturing problems
(iii) Transportation problems
� The common region determined by all the constraints including the non-negative
constraints x ≥ 0, y ≥ 0 of a linear programming problem is called the feasible
region (or solution region) for the problem � Points within and on the boundary of the feasible region represent feasible
solutions of the constraints Any point outside the feasible region is an infeasible solution |
1 | 6708-6711 | � A few important linear programming problems are:
(i) Diet problems
(ii) Manufacturing problems
(iii) Transportation problems
� The common region determined by all the constraints including the non-negative
constraints x ≥ 0, y ≥ 0 of a linear programming problem is called the feasible
region (or solution region) for the problem � Points within and on the boundary of the feasible region represent feasible
solutions of the constraints Any point outside the feasible region is an infeasible solution © NCERT
not to be republished
LINEAR PROGRAMMING 529
� Any point in the feasible region that gives the optimal value (maximum or
minimum) of the objective function is called an optimal solution |
1 | 6709-6712 | � Points within and on the boundary of the feasible region represent feasible
solutions of the constraints Any point outside the feasible region is an infeasible solution © NCERT
not to be republished
LINEAR PROGRAMMING 529
� Any point in the feasible region that gives the optimal value (maximum or
minimum) of the objective function is called an optimal solution � The following Theorems are fundamental in solving linear programming
problems:
Theorem 1 Let R be the feasible region (convex polygon) for a linear
programming problem and let Z = ax + by be the objective function |
1 | 6710-6713 | Any point outside the feasible region is an infeasible solution © NCERT
not to be republished
LINEAR PROGRAMMING 529
� Any point in the feasible region that gives the optimal value (maximum or
minimum) of the objective function is called an optimal solution � The following Theorems are fundamental in solving linear programming
problems:
Theorem 1 Let R be the feasible region (convex polygon) for a linear
programming problem and let Z = ax + by be the objective function When Z
has an optimal value (maximum or minimum), where the variables x and y
are subject to constraints described by linear inequalities, this optimal value
must occur at a corner point (vertex) of the feasible region |
1 | 6711-6714 | © NCERT
not to be republished
LINEAR PROGRAMMING 529
� Any point in the feasible region that gives the optimal value (maximum or
minimum) of the objective function is called an optimal solution � The following Theorems are fundamental in solving linear programming
problems:
Theorem 1 Let R be the feasible region (convex polygon) for a linear
programming problem and let Z = ax + by be the objective function When Z
has an optimal value (maximum or minimum), where the variables x and y
are subject to constraints described by linear inequalities, this optimal value
must occur at a corner point (vertex) of the feasible region Theorem 2 Let R be the feasible region for a linear programming problem,
and let Z = ax + by be the objective function |
1 | 6712-6715 | � The following Theorems are fundamental in solving linear programming
problems:
Theorem 1 Let R be the feasible region (convex polygon) for a linear
programming problem and let Z = ax + by be the objective function When Z
has an optimal value (maximum or minimum), where the variables x and y
are subject to constraints described by linear inequalities, this optimal value
must occur at a corner point (vertex) of the feasible region Theorem 2 Let R be the feasible region for a linear programming problem,
and let Z = ax + by be the objective function 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 (vertex) of R |
1 | 6713-6716 | When Z
has an optimal value (maximum or minimum), where the variables x and y
are subject to constraints described by linear inequalities, this optimal value
must occur at a corner point (vertex) of the feasible region Theorem 2 Let R be the feasible region for a linear programming problem,
and let Z = ax + by be the objective function 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 (vertex) of R � If the feasible region is unbounded, then a maximum or a minimum may not
exist |
1 | 6714-6717 | Theorem 2 Let R be the feasible region for a linear programming problem,
and let Z = ax + by be the objective function 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 (vertex) of R � If the feasible region is unbounded, then a maximum or a minimum may not
exist However, if it exists, it must occur at a corner point of R |
1 | 6715-6718 | 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 (vertex) of R � If the feasible region is unbounded, then a maximum or a minimum may not
exist However, if it exists, it must occur at a corner point of R � Corner point method for solving a linear programming problem |
1 | 6716-6719 | � If the feasible region is unbounded, then a maximum or a minimum may not
exist However, if it exists, it must occur at a corner point of R � Corner point method for solving a linear programming problem The method
comprises of the following steps:
(i) Find the feasible region of the linear programming problem and determine
its corner points (vertices) |
1 | 6717-6720 | However, if it exists, it must occur at a corner point of R � Corner point method for solving a linear programming problem The method
comprises of the following steps:
(i) Find the feasible region of the linear programming problem and determine
its corner points (vertices) (ii) Evaluate the objective function Z = ax + by at each corner point |
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