Bag sewing time=4 mins, jacket sewing time=2 mins; bag pressing time=1 min, jacket pressing time=3 mins. Available sewing time=2,000 mins per week, available pressing time=1,500 mins per week. Profit=5RM per bag, 6RM per jacket. Call B=number of bags per week and J=number of jackets per week. Plot two lines on the same graph if available time for pressing and sewing is fully utilised. The first equation is based on the sewing time: 4B+2J=2000, and the second on the pressing time: B+3J=1500. Let B be the vertical axis and J the horizontal axis. Plot the first equation: draw a line between 1500 on B axis and 500 on J axis. These are the points where no jackets are made, just bags (B=1500 and J=0) and where no bags are made, just jackets (B=0 and J=500). Plot the second equation similarly with a line joining 500 on B to 1000 on J. Where they cross shows how many bags and jackets can be made to optimise the time available and maximise profit.
The two equations can be solved by substituting B=1500-3J into 4B+2J=2000, giving us 6000-12J+2J=2000. So J=400, therefore B=300. The intersection point is (400,300) for (J,B). The profit from this optimisation would be 300*5+400*6 RM = 3,900 RM.