Q 3177180986
(Diet problem): A dietician wishes to mix two types of foods in such a
way that vitamin contents of the mixture contain atleast 8 units of vitamin A and 10
units of vitamin C. Food ‘I’ contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C.
Food ‘II’ contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs
Rs 50 per kg to purchase Food ‘I’ and Rs 70 per kg to purchase Food ‘II’. Formulate
this problem as a linear programming problem to minimise the cost of such a mixture.
Class 12 Chapter 12 Example 6
way that vitamin contents of the mixture contain atleast 8 units of vitamin A and 10
units of vitamin C. Food ‘I’ contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C.
Food ‘II’ contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs
Rs 50 per kg to purchase Food ‘I’ and Rs 70 per kg to purchase Food ‘II’. Formulate
this problem as a linear programming problem to minimise the cost of such a mixture.
Class 12 Chapter 12 Example 6
Solution:
Let the mixture contain x kg of Food ‘I’ and y kg of Food ‘II’. Clearly, x ≥ 0,
y ≥ 0. We make the following table from the given data:
Since the mixture must contain at least 8 units of vitamin A and 10 units of
vitamin C, we have the constraints:
2x + y ≥ 8
x + 2y ≥ 10
Total cost Z of purchasing x kg of food ‘I’ and y kg of Food ‘II’ is
Z = 50x + 70y
Hence, the mathematical formulation of the problem is:
Minimise Z = 50x + 70y ... (1)
subject to the constraints:
2x + y ≥ 8 ... (2)
x + 2y ≥ 10 ... (3)
x, y ≥ 0 ... (4)
Let us graph the inequalities (2) to (4). The feasible region determined by the
system is shown in the Fig 12.7. Here again, observe that the feasible region is
unbounded.
Let us evaluate Z at the corner points A(0,8), B(2,4) and C(10,0).
In the table, we find that smallest value of Z is 380 at the point (2,4). Can we say
that the minimum value of Z is 380? Remember that the feasible region is unbounded.
Therefore, we have to draw the graph of the inequality
50x + 70y < 380 i.e., 5x + 7y < 38
to check whether the resulting open half plane has any point common with the feasible
region. From the Fig 12.7, we see that it has no points in common.
Thus, the minimum value of Z is 380 attained at the point (2, 4). Hence, the optimal
mixing strategy for the dietician would be to mix 2 kg of Food ‘I’ and 4 kg of Food ‘II’,
and with this strategy, the minimum cost of the mixture will be Rs 380.
Let the mixture contain x kg of Food ‘I’ and y kg of Food ‘II’. Clearly, x ≥ 0,
y ≥ 0. We make the following table from the given data:
Since the mixture must contain at least 8 units of vitamin A and 10 units of
vitamin C, we have the constraints:
2x + y ≥ 8
x + 2y ≥ 10
Total cost Z of purchasing x kg of food ‘I’ and y kg of Food ‘II’ is
Z = 50x + 70y
Hence, the mathematical formulation of the problem is:
Minimise Z = 50x + 70y ... (1)
subject to the constraints:
2x + y ≥ 8 ... (2)
x + 2y ≥ 10 ... (3)
x, y ≥ 0 ... (4)
Let us graph the inequalities (2) to (4). The feasible region determined by the
system is shown in the Fig 12.7. Here again, observe that the feasible region is
unbounded.
Let us evaluate Z at the corner points A(0,8), B(2,4) and C(10,0).
In the table, we find that smallest value of Z is 380 at the point (2,4). Can we say
that the minimum value of Z is 380? Remember that the feasible region is unbounded.
Therefore, we have to draw the graph of the inequality
50x + 70y < 380 i.e., 5x + 7y < 38
to check whether the resulting open half plane has any point common with the feasible
region. From the Fig 12.7, we see that it has no points in common.
Thus, the minimum value of Z is 380 attained at the point (2, 4). Hence, the optimal
mixing strategy for the dietician would be to mix 2 kg of Food ‘I’ and 4 kg of Food ‘II’,
and with this strategy, the minimum cost of the mixture will be Rs 380.
