Algebra
posted by Monique on .
maximize Z = 30x+50y
subject to
2x+4y=100 and x+5y=80

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.