0
$\begingroup$

Consider the following problem:

A factory has to produce some amount of a certain mixture, which is made of three substances, A, B and C. This mixture should contain at least 30 kg of substance A, at least 20 kg of substance B and at most 25 kg of substance C. The factory is going to use two intermediate products P1 and P2. Intermediate product P1 is 70% substance A, 10% substance B and 20% substance C, whereas intermediate product P2 -- 30% of A, 30% of B and of C. The price of one kilogramme of these intermediate products is 25 dollars and 15 dollars, respectively.

The question is: how much of intermediate products should the factory buy to fulfill the requirements and minimize the cost?

My approach is as follows: let's denote the amount of intermediate products P1 and P2 by $X$ and $Y$ respectively. The cost of the purchase is then: $$ 25\cdot X+15\cdot Y$$ We have to find the minimum of this function with the following constraints: $$0.7\cdot X+0.3\cdot Y\ge 30$$ $$0.1\cdot X+0.3\cdot Y\ge 20$$ $$0.2\cdot X+0.3\cdot Y\le 25$$

My question is: how does one determine the minimum of such function knowing the constraints? Is there any method other than ''plot the inequalities and try to fit the line''?

  • 1
    Certainly -- [Linear programming: Algorithms](https://en.wikipedia.org/wiki/Linear_programming#Algorithms). Start with the classic [simplex method](https://en.wikipedia.org/wiki/Simplex_algorithm). – 2017-01-10
  • 0
    Well, keep in mind it has to be done on paper. No acces to computer. – 2017-01-10
  • 2
    Anything that can be done by computer can be done on paper, it just takes a little longer. There is no closed-form formula for solving linear programming problems, if that's what you're expecting. You have to use an iterative algorithm such as the simplex method. – 2017-01-10
  • 1
    Actually a plot in this case is the simplest way - draw the polygon boundary for the feasible region, check the corners. – 2017-01-10

1 Answers 1

2

Let's scale the numbers to avoid fractions: \begin{align} \min\quad &25x + 15y\\ \mathrm{s.t.}\quad &7x+3y\geqslant30 \tag 1\\ \quad &x + 3y \geqslant20 \tag 2\\ \quad &2x+3y\leqslant 25 \tag 3\\ \quad &x,y\geqslant0. \end{align} If $x=0$, from constraint $(1)$ we see that $y\geqslant 10$, which violates constraint $(3)$. If $y=0$, then from constraint $(2)$ we see that $x\geqslant 20$, which again violates constraint $(3)$. Because there are two variables, there must be at least two binding constraints. There are three possibilities - $(1)$ and $(2)$, $(1)$ and $(3)$, and $(2)$ and $(3)$. Only the first is a feasible solution, and hence we solve \begin{align} 7x + 3y&=30\\ x+3y&=20 \end{align} which yields $x=\frac53$, $y=\frac{55}9$ for an objective value of $\frac{400}3$.