# Math Help - help linear programming

1. ## help linear programming

Find minimum and maximum for these,objective
1.C = 10x + 7y,constraints:0<=x<=60,0<=y<=45,5x+6y<=420;
2.C=4x+6y,constraints:-x+y<=11,x+y<=27,2x+5y<=90;
3.C=4x+3y,constraints:x=>0,2x+3y=>6,3x-2y<=9,x+5y<=20
4.C=10x+3y,contraints:x=>0,Y=>0,-x+y=>0,2x+y=>4,2x+y<=13 using Linear programing(optimization)

2. Originally Posted by zasi
Find minimum and maximum for these,objective
1.C = 10x + 7y,constraints:0<=x<=60,0<=y<=45,5x+6y<=420;
2.C=4x+6y,constraints:-x+y<=11,x+y<=27,2x+5y<=90;
3.C=4x+3y,constraints:x=>0,2x+3y=>6,3x-2y<=9,x+5y<=20
4.C=10x+3y,contraints:x=>0,Y=>0,-x+y=>0,2x+y=>4,2x+y<=13 using Linear programing(optimization)
Now these are all essentialy the same so I will do the first one and leave
the rest to you.

First sketch the feasible region defined by the constraints (see attachment).

The objective function takes its maximum and minimum at one of the
corners or vetices of the feasible region, so we evaluate it at each
at the maximum value it takes is the maximum on the feasible region,
and similarly the minimum value at a vertex in the minimum on the feasible
region.

The vertices are (0,45), (30,45), (60,20), (60,0), (0,0), and the objective
takes values : 315, 615, 740, 600, 0.

So the maximum value of the objective is 740, which is achieved at (60,20),
and the minimum is 0, which is achived at (0,0).

RonL