Let the manufacturer produce x packages of screw A and y packages of screw B.
Let Z be the profit.
Table
Type of Screw |
Number of packages |
Time on Automatic Machine (minutes) |
Time on Hand Machine (minutes) |
Profit (Rs.) |
A |
x |
4x |
6x |
7x |
B |
y |
6y |
3y |
10y |
Total |
4x + 6y |
6x + 3y |
7x + 10y |
We are to maximise
P = 7x + 10y
subject to constraints
4x + 6y ≤ 240 or 2x + 3y ≤ 120
6x + 3y ≤ 240 or 2x + y ≤ 80
x ≥ 0. y ≥ 0
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.
We draw the graph of 2x + 3y = 120
For a = 0, 3y = 120 or y = 40
For y = 0, 2x = 120 or x = 60
∴ line meets OX in A(60, 0) and OY in L(0, 40)
Again we draw the graph of 2x + y = 80
For x = 0, y = 80
For y = 0, 2 x = 80 or x = 40
∴ line meets OX in B(40, 0) and OY in M(0, 80),
Since feasible region satisfies all the constraints.
∴ OBCL is the feasible region.
The corner points are O(0, 0), B(40, 0), C(30, 20), L(0, 40)
At O(0, 0), P = 7 × 0 + 10 × 0 = 0 + 0 = 0
At B(40, 0), P = 7 × 40 + 10 × 0 = 280 + 0 = 280
At C(30, 20), P = 7 × 30 + 10 × 20 = 210 + 200 = 410
At L(0, 40), P = 7 × 0 + 10 × 40 = 0 + 400 = 400
∴ maximum value = 410 at (30, 20)
∴ 30 packages of screws A and 20 packages of screws B arc produced for maximum profit of Rs. 410.
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs Rs 4 per unit food and F2 costs Rs 6 per unit. One unit of food F2 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.