Linear Programming - Setting variables equal to zero to draw line

[rtsmith555]

Hi. I have an LP problem in my class which has generated the following answer:

min 8.5x + 9.7y

subject to the following constraints:

C1: x + y >= 30
C2: .07x + .5y >= .3 (x + y)
C3: .03 x + .05y <= .05 (x + y)
C4: x >= 0, y > = 0

This is the final answer as confirmed by the online class. (x + y) is included on the RHS of the constraints in red (C2 and C3). The class instructions state that we can graph it on our own to see the binding lines of the feasible region. We are supposed to do this by setting a variable in each equation (x or y) equal to zero and solving for the other variable in each equation to generate points on the lines. Then we reverse procedure for the other variable.

I've had no problem with this for simpler equations with both variables only on the LHS and a non-variable value on the RHS, but for these equations I cannot arrive at a result that makes sense and class support is virtually nonexistent.

When you sub x or y out with zero in each equation and solve for the other variable, the result is zero. This gives me ordered pairs of 0,0 and 0,0 for each equation. That's not a line to draw.

Can anyone spot my error? Apparently, I am supposed to be able to graph these equations to illustrate a feasible region. Help is appreciated!

 

[Otis]

Hi RT. For C2, substituting x=0 does result in y=0. That means the line passes through the Origin point (0,0). Because both intercepts are zero, we need to substitute a different value for x, and then solve for y. That will provide a second point. (Try x=20 or x=40).

For C3, the graphing method you were taught isn't good. (Are you sure that you've posted C3 as given?) If we use the Distributive Property to multiply the right-hand side, we get:

0.03x + 0.05y <= 0.05x + 0.05y

Subtracting 0.05y from each side yields:

0.03x <= 0.05x

That's true for infinite values of x, and the disappearance of variable y tells us (for any solution x) that y can be any value. In other words, the graphic solution is not a region on one side of a single line; the solution region will be one or more quadrants.

Can you solve 0.03x <= 0.05x ? Hint: Think about negative numbers, positive numbers, and zero.

?

 

[Dr.Peterson]

The class instructions state that we can graph it on our own to see the binding lines of the feasible region. We are supposed to do this by setting a variable in each equation (x or y) equal to zero and solving for the other variable in each equation to generate points on the lines. Then we reverse procedure for the other variable.

This method involves finding the x- and y-intercepts, and plotting those two points in order to draw the line.

It doesn't work when the line passes through the origin, or when the line is horizontal or vertical, as there is only one intercept.

So you have to fall back on another method, such as solving for y and using the slope and the one intercept (at the origin).