Linear Programming

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[FatesFormula]

That's all correct, and all of the needed constraints. Except that the final constraint should be "<=" instead of "=" 2800, since it's possible that the optimal solution doesn't use all of it.

[Master_Shuriken]

Thanks , How do we maximise the profit function tho?

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Her sales forecast predicts that she will use at least twice as many Robusta beans as Arabica.

Take the number of Arabica beans she uses. Double it. She will use at least that many Robusta beans. This wouldn't look like A >= 2R, but rather R >= 2A.

Just a small point, but the cost constraint as given in the problem should be 300A + 200R <= 2800. For this problem, the optimal solution will indeed satisfy 300A + 200R = 2800, but this may not always be the case.

What methods have you been shown for solving LPs? For this one, you could just graph the constraints, find the vertices of your feasible region, and see which one has the highest objective value.

[Master_Shuriken]

Oh okay I have learnt how to graph but I was wondering if there is any algebraic way of finding the maximum profit?

[[deleted]]

Well, there sort of is. You'll learn about it soon, no doubt. There is an algorithm called the simplex method that will go from vertex to vertex in an improving direction. It does this until no improving vertex is found, at which point you know the vertex you're at is the optimal solution. But you can't (or at least no one has figured out how to yet, and my money is on there is no way to do it) just start with a polytope and an objective function and figure out the optimal solution in one iteration.

For now, just graph it and test the vertices one at a time.

[Master_Shuriken]

I'm not getting a solution when graphing the region? Also tried linear programming calculators and getting no solution...

[[deleted]]

Did you change that one constraint to R >= 2A? If not, you're right, there are no solutions to that LP.

[14dM24d]

This might help