Question 468975: How do I minimize: z=13x+9y
subject to: 3x+4y>=21; 8x+4y>=32; x>=0; y>=0 ?
Answer by ccs2011(207)   (Show Source):
You can put this solution on YOUR website! First graph your bounds which are defined as:
3x+4y>=21; 8x+4y>=32; x>=0; y>=0
Each of these represents an area above a line.
Determine slope and y_intercept of each line by rewriting in slope-intercept form.
y = mx + b
x = 0 and y =0 are just the axis of the graph
Rewrite 3x+4y>=21 by solving for y:
Subtract 3x on both sides
Divide by 4 on both sides
Rewrite 8x+4y>=32 by solving for y:
Subtract 8x on both sides
Divide by 4 on both sides
Now graph these lines
All the possible (x,y) points are in the region that is on or above both lines
Lets look at z=13x+9y
The goal is to minimize z, there is a direct variation with z to x and y so to decrease z you must decrease x and/or y.
If you pick any point in the possible region and move it straight down until you hit one of the 2 lines, then you kept the same x but decreased y thus decreasing z.
From that logic the smallest z values will be on the line.
Typically the smallest z values will be on the endpoints.
There are 3 end-points, x-intercept (7,0) y-intercept (0,8) and point where lines intersect.
To find where they intersect, set 2 equations equal to each other.
Eliminate fractions by multiplying everything by 4
Add 8x on both sides
Subtract 21 on both sides
Divide by 5 on both sides
Substitute back into equation to find y
Finally evaluate z at each end-point, smallest value will be minimum
For (7,0): z = 13*7 = 91
For (0,8): z = 9*8 = 72
For (11/5,18/5): z = 13*(11/5) + 9*(18/5) = 305/5 = 61
Therefore z has minimum of 61 at point(11/5,18/5)
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