Algebraic Simplification in Linear Programming

[Guest]

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!

[Guest]

One of your equations is [tex].07x+ .5y= .3(x+ y)[/tex], a boundary of the region satisfying [tex].07x+ .5y\ge .3(x+ y)[/tex].

If you just want the x and y intercepts, yes, you can set x= 0 to get .5y= .3y. Subtract .3y from both sides to get .2y= 0 so y= 0 and set y= 0 to get .07x= .3x so x= 0. Unfortunately that only gives the one point (0, 0). To draw the line you need a second point. You can get that by setting either x or y to any number other than 0, For example, if you set x= 1, you get -07+ .5y= .3(1+ y)= .3+ .3y. Subtract .3y from both sides to get .07+ .2y= .3. Now subtract .07 from both sides of the equation to get .2y= .23. Divide both sides by .2 to get y= 1.15. The line goes through (0, 0) and (1, 1.15). Mark those points and draw the line through them.

If it really bothers you that there are "x" and "y" on both sides of the equation, first do the indicated multiplication on the right: .07x+ .5y= .3x+ .3y. Now subtract .3x and .3y from both sides of the equation: -.23x+ .2y= 0,

[Guest]

[tex]\int[/tex]Thank you for the feedback. I think the vector spaces and bases need to be covered as well but I just put the first 10 topics first and continue later.