Only that x1+ x2>0 0.The constraints, I think, would be: 1/3x1 + 1/2x2 <= 60 (maximum amount of peanuts)
2/3x1 + 1/2x2 <= 90 (maximum amount of raisins)
x1 + x2 <= 150 (maximum amount of mix packages)
Yes, subtract off the costs. If you use (1/3)x1 kg of peanuts in RP and (1/2)x2 kg of peanuts in Pr, and each kilogram of peanuts cost £0.6 then you need to subtract .6((1/3)x_1+ (1/2)x_2) from the profit. Do the same for the raisins and packages.The optimization equation is what I can't quite figure out, though. I figure I could just write it F = 2.9x1 + 2.55x2 --> max. But this formula would calculate revenue maximisation, not profit. So any way to make the formula take costs into account as well?
Great. But "point of maximization" of what. The point you want should be the vertex where the profit is largest and you are saying you do not know how to find that.I did draw the graph (the feasibility region) and point of maximization does in fact show (90;60), which should be the correct answer. Point is, how to use the optimization equation to calculate the comparative values of all the corner points of the feasibility region?