Solving integer programming problem using the graphical method I have an integer programming problem I need to solve using the graphical method.

Maximize $55x_1 + 500x_2$
   such that
  $$\begin{align}
4x_1 + 5x_2 &\le 2000\\
2.5x_1 + 7x_2 &\le 1750\\
5x_1 + 4x_2 &\le 2200
\end{align}$$
  $$x_1,x_2 \ge 0$$

The optimal solution is known and it's $(0,250)$. The problem is that I need to draw the graph by hand and I don't know how to do it properly when the numbers are quite big.
Thanks!
 A: Here's how I usually solve these kinds of problems graphically:


*

*Start out by drawing a cartesian 2-dimensional coordinate system ($(x_1, x_2)$ in your case).

*Draw your limits as lines; i.e. by isolating $x_2 = \ldots$ and plotting the lines (scale the $x_1$- and $x_2$-axises according to your needs).

*These lines will confine an area, this is your solution space; i.e. the area which contains your of your valid solutions.
Now for finding the optimal solution;
The maximization function can be seen as just another 'line', i.e. by isolating $x_2 = \ldots$


*

*Draw this line into the 'limits' graph, as relationship between $x_1$ and $x_2$ of the maximization function is 'encoded' in the slope of the line. The optimum value can be found by simply 'shoving' the line 'up' until only a single point of the line is within the area of valid solutions.


Example:
Isolating your constraints for $x_2$ yields:
$$\begin{align}
4x_1 + 5x_2 \le 2000 \quad\Rightarrow\quad x_2 \leq 400 - \frac{4x_1}{5} \\
2.5x_1 + 7x_2 \le 1750 \quad\Rightarrow\quad x_2 \leq 250 - \frac{5x_1}{14} \\
5x_1 + 4x_2 \le 2200 \quad\Rightarrow\quad x_2 \leq 550 - \frac{5x_1}{4}
\end{align}$$
Plotting these yields;

Where the area below the graph, but with $x_1, x_2 > 0$ is the solution space.
We can see that the blue line ($x_2 \leq 400 - \frac{4x_1}{5}$) is superfluous for defining the solution space, and thus leave it out.
Your maximization function isolated for $x_2$ yields:
$$
55x_1 + 500x_2 = 0 \\
\Downarrow \\
x_2 = -\frac{11x_1}{100}
$$
Adding this to the plot, yields the following graph (new blue line = maximization function):

Now 'shoving' this maximization function line 'up' yields the following;

At this point the line cannot be 'shoved' further 'up', without entirely leaving the solution space.
