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I have following question that I kindly need assistance:

2 Products; $x$ and $y$

For every $x$ units sold profit$= 20$

For every $y$ units sold profit$= 50$

Therefore profit function$= (20*x) + (50* y)$

$2$ machines; $1$ and $2$

Machine $1= 100$ hrs

Machine $2= 80$ hrs

To produce unit $x$ requires $1$ hr in Machine $1$

To produce unit $x$ requires $2$ hrs in Machine $2$

To produce unit $y$ requires $3$ hrs in Machine $1$

To produce unit y requires $5$ hrs in Machine $2$

$1X + 3Y< 100,000$

$2X + 5Y < 80,000$

So, how many units of $x$ and $y$ do I need to produce to maximize profit.

Your help will be highly appreciated.

Maurice
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1 Answers1

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\begin{align} \max_{x,y}\quad&20x+50y\\ x+3y&\leq 100000\\ 2x+5y&\leq 80000 \end{align} You also have the constraints that $x,y\geq 0$

There are many ways you can solve this. Like:

  • Graphically
  • Simplex Method
  • Interior Point Method (Super Overkill)
  • $\vdots$

Let's solve it graphically. If you plot the equations out and look at the points of intersection of the constraints, you get a shape that looks like this.

Your points of interest are: $(0,0), (0,16000),(40000,0)$. This is because for a Linear Programming Problem, the optimal solution always lies on the corner points.

If you write out the objective value at each of those points, you will find that the objective value is maximized when you choose the point $(40000,0)$ which means you should produce $40000$ of $x$ and no $y$.

$\color{red}{\mbox{Warning!}}$ This problem has multiple solutions. The entire line connecting $(0,16000)$ and $(40000,0)$ in the first quadrant has same objective value and hence is optimal.

Nitish
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  • The first constraint is redundant, and the second constraint takes the form of the objective value function (2x + 5y) < c i.e. (20x + 50y) < 10c. So the whole of the line bordering the second constraint maximises the objective value function. – oks Feb 21 '14 at 23:16
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    Yup. When you say "the first constraint" is redundant. You are implicitly doing an "active-set" or "primal-dual" approach which is completely valid as well. Graphically, it translated to the first constraint not being involved in determining the critical points. – Nitish Feb 21 '14 at 23:18