Linear problem: maximizing net income

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$that contains all the possible solutions), but I am having trouble with those constraints. Thanks in advance. • Satisfying both inequalities simultaneously would accept negative values of$x_2$, which is why the third inequality is needed. But it doesn't require negative values of$x_2$. For example, the solution$x_1 = x_2 = 1$trivially satisfies all of the given inequalities. – COTO Aug 17 '14 at 16:15 • As for your amended question: your solution is restricted to being: i) to the right of$x_2 = 0$, ii) above$x_1 = 0$, iii) to the bottom left of the line$3x_1 + 4x_2 = 20,000$, and iv) to the bottom left of the line$3x_1 + 3.8x_2 = 40,000$. This forms a nice convex tetragon. – COTO Aug 17 '14 at 16:19 • My apologies. The intersection is a triangle. – COTO Aug 17 '14 at 16:28 • Assuming that$x_1,x_2\ge 0$we can conclude that the second constraint is always satisfied if the first one is satisfied, so just forget about the second constraint. – Alexander Vigodner Aug 17 '14 at 16:33 1 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.

• It was, thanks. I made a sketch, so I could see my mistake – Giiovanna Aug 17 '14 at 17:12
• @Giiovanna Nice to read. You are welcome. – callculus Aug 17 '14 at 17:19