maximize Z = 30x+50y

subject to
2x+4y=100 and x+5y=80

1 answer

I assume you are supposed to do a graphical solution.
first, change each constraint equation to slope/inercept form.

y=-1/2 x+25 and
y=-1/5 x + 16

Now, on an x/y graph, plot those lines. Now what is neat about this solution, the max or min is always at an intersection.

Here, if you assume x>0, and y>0, then you have 4 corners (intersections).
0,0
16,0
0,25 and finally where the lines cross at 30,10

check all those points, I did them in my head.

Then, take your objective function Z, and calculate the value at each of these corners: One will be the max solution.

Now to make a genius of yourself, try a point x,y inside the area, and somewhere along any line, just to prove to yourself the maximum and minimum will always be on a corner.
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