A company manufactures three kinds of calculators: A, B, and C in its two factories I and II. The company has got an order for manufacturing at least 6400 calculators of kind A, 4000 of kind B, and 4800 of kind C. The daily output of factory I is 50 calculators of kind A, 50 calculators of kind B, and 30 calculators of kind C. The daily output of factory II is 40 calculators of kind A, 20 of kind B, and 40 of kind C. The cost per day to run factory I is Rs. 12,000 and of factory II is Rs. 15,000. How many days do the two factories have to be in operation to produce the order with the minimum cost?

Let factories I and II should be operated for x and y number of days respectively. Then the problem can be formulated as in L.P.P. as:
Minimise Z = 12000x + 15000y
Subject to constraints
50x + 40y ≥ 6400 i.e., 5x + 4y ≥ 640
50x + 20y ≥ 4000 i.e., 5x + 2y ≥ 400
30x + 40y ≥ 4800 i.e., 3x + 4y ≥ 480
x ≥ 0, y ≥ 0
We draw the lines 5x + 4y ≥ 640, 5x + 2y ≥ 400, 3x + 4y ≥ 480 and obtain the feasible region (unbounded and convex) shown shaded in the adjoining figure. The corner points are A(0, 200), B(32, 120), C(80, 60) and D(160, 0).
The values of Z at these points are 3000000, 2184000, 1860000 and 1920000 respectively. As the feasible region is unbounded, we draw the graph of the half plane.
12000x + 15000y < 1860000
i.e., 12x + 15y < 1860
and note that there is no point common with the feasible region, therefore Z has minimum and minimum value is Rs. 1860000. It occurs at the point (80, 60) i.e., Factory I should be operated for 80 days and factory II should be operated for 60 days to minimise the cost.
Running Cost per day (in Rs )
ABC
I 12000 50 50 30
II 15000 40 20 40