Answer:
z (max) = 1250 $
x₁ = 25 x₂ = 0 x₃ = 25
Step-by-step explanation:
Profit $ mach. 1 mach. 2
Product 1 ( x₁ ) 30 0.5 1
Product 2 ( x₂ ) 50 2 1
Product 3 ( x₃ ) 20 0.75 0.5
Machinne 1 require 2 operators
Machine 2 require 1 operator
Amaximum of 100 hours of labor available
Then Objective Function:
z = 30*x₁ + 50*x₂ + 20*x₃ to maximize
Constraints:
1.-Machine 1 hours available 40
In machine 1 L-H we will need
0.5*x₁ + 2*x₂ + 0.75*x₃ ≤ 40
2.-Machine 2 hours available 40
1*x₁ + 1*x₂ + 0.5*x₃ ≤ 40
3.-Labor-hours available 100
Machine 1 2*( 0.5*x₁ + 2*x₂ + 0.75*x₃ )
Machine 2 x₁ + x₂ + 0.5*x₃
Total labor-hours :
2*x₁ + 5*x₂ + 2*x₃ ≤ 100
4.- Production requirement:
x₁ ≤ 0.5 *( x₁ + x₂ + x₃ ) or 0.5*x₁ - 0.5*x₂ - 0.5*x₃ ≤ 0
5.-Production requirement:
x₃ ≥ 0,2 * ( x₁ + x₂ + x₃ ) or -0.2*x₁ - 0.2*x₂ + 0.8*x₃ ≥ 0
General constraints:
x₁ ≥ 0 x₂ ≥ 0 x₃ ≥ 0 all integers
The model is:
z = 30*x₁ + 50*x₂ + 20*x₃ to maximize
Subject to:
0.5*x₁ + 2*x₂ + 0.75*x₃ ≤ 40
1*x₁ + 1*x₂ + 0.5*x₃ ≤ 40
2*x₁ + 5*x₂ + 2*x₃ ≤ 100
0.5*x₁ - 0.5*x₂ - 0.5*x₃ ≤ 0
-0.2*x₁ - 0.2*x₂ + 0.8*x₃ ≥ 0
x₁ ≥ 0 x₂ ≥ 0 x₃ ≥ 0 all integers
After 6 iterations with the help of the on-line solver AtomZmaths we find
z (max) = 1250 $
x₁ = 25 x₂ = 0 x₃ = 25
* 13 / 3 + π *