Find solution to this system of inequalities such that x+y is minimal

1. Jun 30, 2006

Ajoo

I propsed myself a problem but i'm having some trouble solving it. I've narrowed it down to this but i need some help on this final part:

I have this 2 inequations that must be obeyed:

(17/15)x + y > 50000 (1)
(2/25)x + (3/10)y > 5000 (2)

and I want to find values for x and y that respect those 2 conditions and so that (x+y) is as low as possible.

I don't even know if there is a single value of x+y or multiple ones.
Is the solution the intersection between the equations (1) and (2)?
Plz, tell me how to do this.

PS: I'm in high school so try to keep it simple.

2. Jun 30, 2006

0rthodontist

Actually the inequalities have to be "greater than or equal" not just greater than. To give you the idea why (unrelated to this problem), if you want to minimize x such that x > 2, then there is no solution because whatever x0 you pick that is greater than 2, you can always pick a smaller x1 that twice as close to 2 as x0 is.

Maybe the best way to solve this for you is to draw a picture of the intersection of the two inequalities and intuitively look for the point where x + y is smallest.

It also happens to be true that for a system of inequalities like that, the minimum value of the function (x+y) (if there is one) will always occur at the intersection of the equations, i.e. 17/15x + y = 50000 and 1/25x + 3/10y = 5000.

3. Jul 1, 2006

Ajoo

Yeah. It should be >= where it is >. I just typed this in a hurry.
I had already tried drawing the 2 graphs in my calculator but how can i be shure the intersection is the smallest value for (x+y)?
However, it's either that a point in equation (1) past the intersection point or a point in equation (2) be4 the intersection point.

4. Jul 1, 2006

gnomedt

The optimal solution in linear programming always falls at one of the vertexes of the feasable region (the convex polygon defined by your linear constraints). This should be fairly intuitive, but many Linear Programming books will have proofs.