Question 1092374: maximize profit P = 40x + 30y
constraints:
1. x + y <= 240
2. 2x + y <= 320
x >= 0, y >= 0
I need the answer by Using Big M Method..??
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13206)   (Show Source):
You can put this solution on YOUR website!
The term "Big M Method" is not a standard term; it must be something your teacher or textbook uses.
So I will go ahead and outline the process I use to solve problems like this.
The intercepts of the first constraint line are (0,240) and (240,0); its slope is -1.
The intercepts of the second constraint line are (0,320) and (160,0); its slope is -2.
The two constraint lines intersect at (80,160). (I will assume you know how to determine that....)
So the possible points where we get maximum profit are (0,240), (80,160, and (160,0).
If the slope of the constraint line is greater than the slope of the first constraint line (less negative; downward to the right but less steep than the first constraint line), then the maximum profit will be at the y-intercept of the first constraint line.
If the slope of the constraint line is less than the slope of the second constraint line (more negative; steeper than the second constraint line), then the maximum profit will be at the x-intercept of the second constraint line.
If the slope of the constraint line is between the slopes of the two constraint lines, then the maximum profit will be at the intersection of the two constraint lines.
The slope of the constraint line is -4/3, which is between -1 and -2; so the maximum profit will be found at (80,160). That maximum profit is 40x+30y = 3200+4800 = 8000.
Answer by ikleyn(52866)  (Show Source):
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