Posted by LIZ on Saturday, January 29, 2011 at 12:26am.
graph 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)
There is a name for that - "linear programming"
It 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