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प्रश्न
A chemical company produces two compounds, A and B. The following table gives the units of ingredients, C and D per kg of compounds A and B as well as minimum requirements of C and D and costs per kg of A and B. Find the quantities of A and B which would give a supply of C and D at a minimum cost.
| Compound | Minimum requirement | ||
| A | B | ||
| Ingredient C Ingredient D |
1 3 |
2 1 |
80 75 |
| Cost (in Rs) per kg | 4 | 6 | - |
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उत्तर
Let x kg of compound A and y kg of compound B were produced.
Quantity cannot be negative.
Therefore, \[x, y \geq 0\]
| Compound | Minimum requirement | ||
| A | B | ||
| Ingredient C Ingredient D |
1 3 |
2 1 |
80 75 |
| Cost (in Rs) per kg | 4 | 6 | - |
According to question, the constraints are
\[x + 2y \geq 80\]
\[3x + y \geq 75\]
Cost (in Rs) per kg of
compound A and compound B is Rs 4 and Rs 6 respectively.Therefore, cost of x kg of compound A and y kg of compound B is 4x and 6y respectively.
Total cost = Z = \[4x + 6y\]
Thus, the mathematical formulation of the given linear programming problem is
Min Z = \[4x + 6y\]
subject to
\[x + 2y \geq 80\]
\[3x + y \geq 75\]
x + 2y = 80, 3x + y =75, x = 0 and y = 0
Region represented by x + 2y ≥ 80:
The line x + 2y = 80 meets the coordinate axes at A1(80, 0) and B1(0, 40) respectively. By joining these points we obtain the line x + 2y = 80. Clearly (0,0) does not satisfies the x + 2y = 80. So, the region which does not contain the origin represents the solution set of the inequation x + 2y ≥ 80.
Region represented by 3x + y ≥ 75:
The line 3x + y =75 meets the coordinate axes at C1(25, 0) and D1(0, 75) respectively. By joining these points we obtain the line 3x + y =75. Clearly (0,0) does not satisfies the inequation 3x + y ≥ 75. So,the region which does not contain the origin represents the solution set of the inequation 3x + y ≥ 75.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + 2y ≥ 80, 3x + y ≥ 75, x ≥ 0, and y ≥ 0 are as follows.
The corner points are D1(0, 75), E1(14, 33) and A1(80, 0).
The values of Z at these corner points are as follows
| Corner point | Z= 4x + 6y |
| D1 | 450 |
| E1 | 254 |
| A1 | 320 |
The minimum value of Z is 254 which is attained at E1 \[\left( 14, 33 \right)\]
Thus, the minimum cost is Rs 254 obtained when 14 units of compound A and 33 units of compound B were produced.
संबंधित प्रश्न
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| Calcium Protein Calories |
10 5 2 |
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| kg per bag | ||
| Brand P | Brand P | |
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A B |
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|
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| Machines | |||
| I | II | III | |
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| From \ To | Cost (in ₹) | ||
| A | B | C | |
| P | 160 | 100 | 150 |
| Q | 100 | 120 | 100 |
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It is being given that at least one of each must be produced.
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| Transportation cost per quintal(in Rs.) | ||
| From-> | A | B |
| To | ||
| D | 6.00 | 4.00 |
| E | 3.00 | 2.00 |
| F | 2.50 | 3.00 |
How should the supplies be transported in order that the transportation cost is minimum?
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| Transportation Cost per packet(in Rs.) | ||
| From-> | A | B |
| To | ||
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A linear programming problem is given by Z = px + qy, where p, q > 0 subject to the constraints x + y ≤ 60, 5x + y ≤ 100, x ≥ 0 and y ≥ 0.
- Solve graphically to find the corner points of the feasible region.
- If Z = px + qy is maximum at (0, 60) and (10, 50), find the relation of p and q. Also mention the number of optimal solution(s) in this case.
