Maximize z = 6x + 3y subject to:
2x + y < 11
3x + 4y < 24
x > 0 y > 0
2006-09-23 09:27:07 · 5 answers · asked by nileri_1 2 in Science & Mathematics ➔ Mathematics
for more complex equations, the simplex method could be used... this one is fairly simple...
solution 1
to solve it graphically, plot the four equations 2x+y=11 and 3x+4y=24, x=0, y=0.... we get four vertices of a ploygon the inside of which would fulfill the given conditions... but the theorem says that the optimal solution will be at the vertices.. hence by solving the two equations we could obtain the intersection as (4,3) which must be the solution...
the inequalities ( "<" instead of the usual "<=") present a problem here.. if this problem requires integer solution, then (4,2) must be a solution....(close to the vertex... parallel to the line 6x+3y)... if it is not an integer programming problem, then an infinitesmally close point to (4,3) is an acceptable solution... something like (3.999..., 3)
2006-09-23 09:51:34 · answer #1 · answered by m s 3 · 0⤊ 0⤋
unless you are in a very advanced math class (use of "simplex" method for instance), you have to use a graph. each of you in-equations represent a region of space. For instance x>0 means all point in your space above the line x=0 (horizontal axis). Same for y>0, which means all point on the right side of the line y=0 (vertical axis). Combining the two here means that the point you are looking for is in the top right corner.
For the other two, you have to draw the two lines 2x+y=11 and 3x+4y=24.
The point (0,0) verify 2x+y<11, so the valid region for the first line is below that line. Now combining the three condition, you have only a triangle that remains.
The point (0,0) also verify your in-equation 3x+4y<24. So the final region of your space you have to work with is contained in the top right corner below the two lines.
Now you take you ruler to represent the z = 6x+3y equation (try to have you ruler with a slope of -3/6=-1/2, higher on the left than on the right). You start with z=0 and you slowly move it up. The z value is going to be given by your ruler when you can only see a little triangle remaining from your valid region.
Your probably going to find that the ruler will reach the intersection of the two lines 2x+y=11 and 3x+4y=24 - which is the point (x=4;y=3), which make z = 6.4+3.3 = 33.
Note however that because you have strict inequality, the point you are looking cannot be on these line. so z will be close to 33 but not exactly 33.
2006-09-23 16:54:50 · answer #2 · answered by sebourban 4 · 0⤊ 0⤋
8x+4y<44
3x+4y<24
5x<20
0
0 < z < 35
2006-09-23 17:04:24 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋
You graph each boundary line. The maximum (and minimum for that matter) will be one of the vertices of the polygon formed by the intersection. Just find all these coordinates, then try them in your z equation and see which is biggest.
2006-09-23 16:32:44 · answer #4 · answered by hayharbr 7 · 0⤊ 0⤋
you have to graph and then find the solution for Zmax
2006-09-23 16:29:56 · answer #5 · answered by raj 7 · 0⤊ 0⤋