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August 4, 2015

August 4, 2015

Posted by **LIZ** on Saturday, January 29, 2011 at 12:26am.

x + 5y ≥ 6

5x + y ≥ 6

x ≥ 0, y ≥ 0.

- math -
**Damon**, Saturday, January 29, 2011 at 9:32amgraph the two lines

x + 5y ≥ 6 (intercepts at (0,6/5)(6,0))

5x + y ≥ 6 (intercepts at ((6/5,0)(0,6)

x ≥ 0, y ≥ 0.

find intersection

5 x + 25 y = 30

5 x + y = 6

------------------

24 y = 24

y = 1

x = 6-5 = 1

so at (1,1)

evaluate c = x+y at three points

(0,6) (1,1) (6,0)

minimum is obviously 2 at (1,1)

- linear programming -
**Damon**, Saturday, January 29, 2011 at 9:33amThere is a name for that - "linear programming"

- math -
**MathMate**, Saturday, January 29, 2011 at 9:41amIt is similar to the previous problems. You could plot them on a graph, find the intersection(s) and evaluate the objective function Z=x+y.

The lines to plot are:

for x+5y≥6, put the inequality in the standard form: y≥(6-x)/5.

Similarly, for 5x+y≥6, plot it as y≥(6-5x).

There are lines x≥0 and y≥0, which come into play whne x≥6 or y≥6.

Plot the graph and find all possible intersections (there are three). The feasibility region is an open region in the first quadrant (towards the top-right).

Now evaluate the objective function Z=x+y at all three intersections and find the point that gives the*smallest*value of Z.

See:

http://img9.imageshack.us/img9/7284/1296278783.png