Question

Bokomo Inc., produces corn-flakes and bran-flakes. The manufacturing process is highly mechanised; both products are produced by the same machinery by using different settings.

For the coming year, 200 000 machine hours are available. Management is trying to decide on the quantities of each product to produce. The following data are available:

Corn-flakes
Bran-flakes

Machine hours per unit
1.00
0.50

Unit selling price
R2.50
R3.00

Unit variable cost
R1.50
R2.25

Because of market conditions, the company can sell no more than 150 000 boxes of corn-flakes and 300 000 boxes of bran-flakes.

Required:

5.1 Use the graphical method to determine the optimal production mix.            (20)

Answer 1

Here let the no. of corn flakes sold be “x” and bran flakes be “y”

Objective function = maximum profit = (2.5-1.5)*x + (3-2.25)*y

= 1x + 0.75y. (Note that profit = revenue – costs)

Constraints:

(i): 1x+0.5y<=200,000 (machine hours)

(ii): x <= 150,000

(iii): y <= 300,000

Now we can solve this graphically:

The first constraint is 1x+0.5y<=200,000. Now let us put y as nil and we get x = 200,000. Here if we put x as nil then y = 200,000/0.5 = 400,000. Thus points for this constraint line is (200000, 400000)

Point for 2^nd constraint line is x = 150,000 and point for 3^rd constraint line is y = 300,000. We can now make the graph (which is attached below).

The feasible regions is shaded in the graph and the corner points are A (0, 300000); B (50000, 300000); C (0,0) and D (150000, 0).

Values of objective function: at A = 0.75*300000 = 225,000. At B = (50,000*1)+(300,000+0.75) = 275,000. At C = 0 and at D = (150,000*1)+0 = 150,000.

Thus profit is getting maximized when units of corn flakes are 50,000 and units of bran-flakes are 300,000. Maximized profit = $275,000