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SIR, i have to calculate maximum profit for month 3
in question it has been solved by iso – contribution line methodbut for my practice i want to solve this by simultaneous equation method
seabach & herdorf required 5kg and 7kg material = total material available 35000kg
seabach & herdorf required 2hr and 3hr labour hr = total labour hr 24000
seabach & herdorf required 3hr and 2hr machine hr = total machine hr 12000
max demand for seebach = 4000unit and for herdorf = 3000 unitacc to graph given in question it is easily identifiable that s need 2000 unit and h need 3000 unit ( for max contribution)
only two equation are constraints material and machinebut acc to simultaneous method my solution is wrong if its possible can you show me the solution in this method.?
thankyou
If you move the iso cont line out……… 250S + 315H
The simultaneous equations that you solve are:
3S + 2H < 12000
And H = 3000 (The maximum demand for H)Thus 12000 – (2* 3000) = 6000
So S is 6000 / 3 = 2000
S and H
250 and 315 Cont
2,000 and 3,000 Units
Gives a total of
500,000 and 945,000Total Cont = 1,445,000 – (FC) 300,000 = 1,145,000
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