A company produces two different products. They require two types of ingredients: M and N. The first product require 90 grams of the ingredient M and 10 grams of the ingredient N. The second product require 20 grams of both ingredients. The company has a maximum of 2,100 grams of the ingredient M and because of a contract with suppliers has to maintain in stock 500 grams of the ingredient N. Also, the company has agreed with a buyer to produce at least 10 items of the first product. The cost of production of the first product is 10 dollars and 20 dollars for the second. There is a fixed cost of production of $100.

  1. How many units of both products should be produced so the company minimizes its cost?
  2. What is the minimum cost?

Ok, this is what I have done. I've identified the decision variables as

  • x: the first product
  • y: the second product

The objective function es $$C = 10x+20y+100$$

The restrictions:

$$x\geq0$$ $$y\geq0$$ $$90x+20y\leq2100$$ $$10x+20y\geq500$$ $$x\geq10$$

When I do the graph I get these vertexes (see the graph below) of the feasible region but the problem is that when I evaluate them in the objective function I get "two optimal solutions" when I think should be one. These are the results when I evaluate the objective function with the vertex:

  • (10,60) -> 1400
  • (10,20) -> 600
  • (20,15) -> 600

Could someone help me!!! So is everything right with the whole process? or do I have something wrong that I'm getting two optimal solutions? I need some lighting.

This is the graph:

linear programming graph

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  • $\begingroup$ The slope of your objective function is equal to the slope of the line connecting the two points. Any point on that line segment between the black dot and the red dot will give the same value. $\endgroup$ – John Douma Mar 7 '19 at 1:32
  • $\begingroup$ so what would the optimal solution then? is the whole process right according to the established problem? $\endgroup$ – gi2302 Mar 7 '19 at 1:39
  • $\begingroup$ Your process is correct. In this case, the optimal solution is any one of those points. $\endgroup$ – John Douma Mar 7 '19 at 1:41
  • $\begingroup$ But is weird. So a linear programming problem could have then two feasible solutions? I've always thought that should be just one. $\endgroup$ – gi2302 Mar 7 '19 at 2:09
  • 2
    $\begingroup$ It is important that you understand how Linear Programming works. As you increase $C$, the graph of your objective function moves in the positive $y$ direction. Normally, it will encounter one of the vertices of the feasible region first which will give you your minimum. In your case, the graph of the objective function is parallel to the bottom of your feasibility region so you get multiple solutions. $\endgroup$ – John Douma Mar 7 '19 at 2:25

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