Solve the linear programming problem. Maximize

Question

P=40x + 50y

subject to

2x+y<=10

x+y<=7

x+2y<=12

x,y<=0

Answer 1

line 1 vs line 2: 2x+y=10 vs x+y=7

subtracting them: x = 3 --> y = 4, so they intersect at (3,4)

line 1 vs line 3: 2x+y=10 vs x + 2y = 12

Then y = 10 - 2x.

Substitution: x + 2(10-2x) = 12

x + 20 - 4x = 12

-3x + 20 = 12

-3x = -8

x = 8/3

then y = 10 - 2(8/3) = 10 - 16/3 = (30 - 16)/3 = 14/3

(8/3, 14/3)

line 2 vs line 3:

x+y = 7 vs x+2y = 12

subtracting them: y = 5 ---> x=2 so they intersect at (2,5)

(8/3,14/3) lies outside the feasibility region.

The critical points are (0,6) . (2,5), (3,4) and (5,0), and of course (0,0) which does nothing

P(x,y) = 40x+50y is the objective function

P(0,6) = 300

P(2,5) = 80+250 = 330 <---max

P(3,4) = 120 + 200 = 320

P(5,0) = 200 <-- min