Problem: A company produces and sells two different products. The demand for each product is unlimited, but the company is constrained by cash avaliable and machine capacity.
Each unit of the first and second product requires 3 and 4 machine hours, respectively. There are 20.000 machine hours avaliable in the current production period. The production costs are $\$$3 and $\$$2 per unit of the first and second product, respectively; The selling prices of the first and second product are $\$$6 and $\$$5,40 per unit, respectively. The avaliable cash is $\$$4.000; furthermore, 45% of the sales revenues from the first product and 30% of the sales revelues from the second product will be made avaliable to finance operations durint the current period.
Formulate a linear programming problem that aims at maximizing net income subject to the cash availability and machine capacity limitations.
My attempt:
Let $x_1, x_2$ be the number of units of the product 1 and 2 produced, respectively. Our goal is to maximize the function
$$ (6-3)x_1 + (5.4-2)x_2 $$
respect to the following constraints:
1) The machine hour avaliable: $3x_1 + 4 x_2 \leq 20.000$
2) The cash avaliable: $3 x_1 + 2 x_2 \leq 4.000 + (0.45)6 x_1 + (0.3)5.4 x_2$
3) $x_1, x_2 \geq 0$
I Would like to plot the feasible region (I mean, the region in $R^2$ that contains all the possible solutions), but I am having trouble with those constraints.
Thanks in advance.
Best Answer
If you solve $3x_1+4_2\leq 20,000$ for $x_2$, then you get $x_2 \leq 5,000-\frac{3}{4}x_1$
And if you solve $3x_1+3.8x_2\leq 40.000$ for $x_2$, then you get $x_2\leq \frac{40,000}{3}-\frac{3}{3.8}x_1$
If you calculate the intersection of $\frac{40,000}{3}-\frac{3}{3.8}x_1$ and $5,000-\frac{3}{4}x_1$, you will see, there is no intersection for $x_1 \geq 0$.
Thus one of the constraints is not neccessary. It is the second one. You do not have to regard this constraint for further calculations. A sketch might helpful for understanding the situation.