Let the company manufacture x Souvenirs of type A and y Souvenirs of type B.
Let P be the profit.
Table
Type |
Number |
Time of cutting (minutes) |
Time of assembling (minutes) |
Profit (Rs.) |
A |
x |
5x |
10x |
5x |
B |
y |
8y |
8y |
6y |
Total |
5x + 8y |
10x + 8y |
5x + 6 |
We are to maximise
P = 5x + 6y
subject to constraints
5x + 8y ≤ 200
10x + 8y ≤ 240 or 5x + 4y ≤ 120
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.
First we draw the graph of 5x + 8y = 200
For x = 0, 8y = 200 or y = 25
For y = 0, 5 x = 200 or x = 40
∴ line meets OX in A(40, 0) and OY in L(0, 25)
Again we draw the graph of 5x + 4y = 120
For x = 0, 4y = 120 or y = 30
For y = 0, 5x = 120 or x = 24
∴ line meets OX in B(24, 0) and OY in M(0. 30).
Since feasible region satisfies all the constraints.
∴ OBCL is the feasible region.
The comer points are O(0, 0), B(24, 0), C(8, 20), L(0, 25)
At O(0, 0), P = 5 × 0 + 6 × 0 = 0 + 0 = 0
At B(24, 0), P = 5 × 24 + 6 × 0 = 120 + 0 = 120
At C(8, 20), P = 5 × 8 + 6 × 20 = 40 + 120 = 160
At L(0, 25) P = 5 × 0 + 6 × 25 = 0 + 150 = 150
∴ maximum value = 160 at (8, 20)
∴ 8 souvenirs of type A and 20 souvenirs of type B are manufactured for maximum profit of Rs. 160.
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.