Let x units of model A and y units of model B is ordered.
Then from the given condition we must have selling atleast 100 units.
So x+y ≥100 ... i
Next is to maximise profit. Profit is 40 for A and 60 for B.
40x+60y is the total profit
From the condition profit ≥4800
i.e. 40x+60y≥4800
Divide by 20 to get
2x+3y≥240 ... ii
The objective function is Cost function
Z = 300x+400y and to minimize Z
Solving constraints we get 2x+3y≥240 and
2x+2y≥200
y≥40 and x≥60.
So possibilities are (120,0) (0,100) or (40,60)
Let us calculate Z for each of these points.
Z(120,0) = 36000
z(0.100) = 40000
Z(60,40) = 18000+16000 = 34000
Since cost is minimum for (60,40)
60 units of model A and 40 units of model B should be ordered.