Question 227686: Minimize the expression Z= 2x +3y subject to the following constraints.
y <= 5, x <=6, x+y >= 2, x >= 0, y >=0
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! To solve problems like this:- Graphs the constraints. This should form some form of polygon.
- Find the coordinates of the vertices of the polygon.
Unfortunately Algebra.com's graphing software does not handle graphs like this very well. So I will have to describe the graph of the contraints. Each contraint will form a side of the polygon.- y <= 5: A horizontal line through 5 on the y-axis
- x <= 6: a vertical line through 6 on the x-axis
- x+y >= 2. In slope-intercept form this is y >= -x + 2. This is a line with a slope of -1 through 2 on the y-axis
- x >=0: the y=axis
- y >=0: the x-axis
Here's a partial graph (without the x <=6 line):
Remembering the missing vertical line through 6 on the x-axis, we end up with a pentagon (5-sided polygon). Any points that are within the pentagon or on the sides of the pentagon are possible solutions. The minimum (or maximum) value for Z will come from one of the vertices. So we need to find the vertices. By inspection (or with a little calculation) we can find the following vertices: (0, 2), (0, 5), (6, 5), (6, 0) and (2, 0).
Now all we need to do is use these vertices to find the minimum value for Z. Take each vertex, one at a time, and put its coordinates into the formula for Z. This will give you 5 different values for Z. The lowest one is the minimum.
I'll do one Z and leave the rest for you. The Z for the vertex (0, 2):
Z = 2(0) + 3(2) = 0 + 6 = 6
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