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Question 1085580: Min: C=19x+15y
Subject to: x+2y_>2
X+2y_<10
2x+y=10
X_>0 ; y_>0
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i believe you mean:
minimize c = 19x + 15y
subject to:
x + 2y >= 2
2 + 2y <= 10
2x + y = 10 *****
x >= 0
y >= 0
you can graph these equations as is using the desmos.com calculator by reversing the inequalities and then looking for the unshaded region of the graph.
the graph will look like this:
the constraint inequalities and equations are:
x + 2y >= 2
2 + 2y <= 10
2x + y = 10 ***** this is an equation and not an inequality.
x >= 0
y >= 0
the corner points of the graph have to be on the line 2x + y = 10.
those corner points are (3 1/3, 3 1/3), (5,0)
the minimum cost point is (5,0), where the cost is 5 * 19 + 0 * 15 = 95
all constraints are met at those corner points.
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if it was 2x + y >= 10, then it would look like this:
x + 2y >= 2
2 + 2y <= 10
2x + y >= 10 ***** testing greater than or equal to 10 in this graph.
x >= 0
y >= 0
the corner points are at (3 1/3, 3 1/3), (5,0), (10,0)
the minimum cost point is (5,0), where the cost is 5 * 19 + 0 * 15 = 95
all constraints are met at that corner point.
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if it was 2x + y <= 10, then it would look like this:
x + 2y >= 2
2 + 2y <= 10
2x + y <= 10 ***** testing smaller than or equal to 10 in this graph.
x >= 0
y >= 0
the corner points are (0,5), (3 1/3, 3 1/3), (0,1), (2,0), (5,0)
the minimum cost point is (0,1), where the cost is 19 * 0 + 15 * 1 = 15.
all constraints are met at those corner points.
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keep in mind that the inequalities on the graphs are reversed.
when you want 2x + y <= 10, you graph 2x + y >= 10
when you want 2x + y >= 10, you graph 2x + y <= 10
what this does is shade the regions you DON'T want and leaves the regions that you DO want unshaded.
it's much easier to see the region you are looking for this way.
with the software, shading the regions you DO want makes locating that region more difficult.
try it yourself and see what i mean.
the corner points of the feasible region are where you will find your solution.
you would evaluate your objective function at those corner points.
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if i understood your problem correctly, then the first graph shows you what the graph would look like if you shaded the regions that do NOT satisfy the inequalities.
in that graph, the solution has to be on the line 2x + y = 10, therefore this restrictions only allows two corner points where that line intersects with the inequalities.
if you were doing this manually, you would shade the regions that DO satisfy all inequalities and equations.
in that case this graph would be shaded where it is now unshaded and the solution would still have to be on the line where the line meets the limits of the feasible region.
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