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 1.5 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 IS 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?
Let the farmer mix x bags of brand P and y bags of brand Q.
Let Z be the cost.
Table,
We are to minimise
subject to the constraints
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of 2 x + y = 12
For x = 0, y = 12
For y = 0, 2 x = 12 or x = 6
∴ line meets OX in A(6, 0) and OY in L(0, 12).
Also we draw the graph of 2x + 9y = 36
For x = 0, 9 y = 36 or y = 4
For y = 0, 2 x = 36 or x = 18
∴ line meets OX in B(18, 0) and OY in M(0, 4).
Again we draw the graph of 2x + 3y = 24
For x = 0, 3y = 24 or y = 8
For y = 0, 2x = 24 or x = 12
∴ line meets OX in C(12, 0) and OY in N(0, 8).
Since feasible region satisfies all the constraints.
∴ shaded region is the feasible region which is unbounded and comer points are B(18, 0), D(9, 2), E(3, 6), L(0, 12).
At B(18, 0), Z = 250 × 18 + 200 × 0 = 4500 + 0 = 4500
At D(9, 2), Z = 250 × 9 + 200 × 2 = 2250 + 400 = 2650
At E(3, 6), Z = 250 × 3 + 200 × 6 = 750 + 1200 = 1950
At L(0, 12), Z = 250 × 0 + 200 × 12 = 0 + 2400 = 2400
∴ smallest value = 1950 at (3, 6)
Since feasible region is unbounded.
∴ we are to check w hether this value is minimum.
For this we draw the graph of
250x + 200y < 1950 or 5x + 4y < 39 ...(1)
Since (1) has no common point with feasible region.
∴ minimum value = 1950 at (3, 6)
∴ minimum cost is Rs. 1950 when 3 bags of brand P and 6 bags of brand Q are mixed.
