A construction company produces two products (doors and windows) that require two plants (A and B). Management wants to determine how many units of each product should be produced to maximize profit. For each unit of Product 1 (i. E. Door), 2 workers are required in Plant A and 4 workers are required in Plant B. For each unit of Product 2 (i. E. Window), 3 workers are required in Plant A and 4 workers are required in plant B. The company has 40 workers in Plant A and 60 workers in Plant B. Each unit of Product 1 gives a profit of $100. Each unit of Product 2, up to 10 units, gives a profit of $80. Any excess over 10 units of Product 2 brings no profit; so, such excess has to be ruled out.

Z max  = 1500

x₁  = 15

x₂ = 0

Step-by-step explanation:

From problem statement:

                                                      Plant A              Plant B     Profit $/each

Product 1 (Doors )          x₁                  2                          4            100

Product  2 (windows )   x₂                  3                           4              80

Product  2 (windows )   x₂   above 10 is scrap

Objective function to maximize

z = 100*x₁   +  80*x₂

Subject to:

Plant A capacity

2*x₁ + 3*x₂   ≤  40

Plant B capacity

4*x₁  + 4*x₂  ≤ 60

Windows condition:

0*x₁  + x₂  ≤  10

Model:

Max      z  =  100*x₁   +  80*x₂

Subject to:

2*x₁ + 3*x₂   ≤  40

4*x₁  + 4*x₂  ≤ 60

              x₂ ≤ 10

x₁ , x₂ ≥  0    and integers

Afther 6 iterations using Simplex Solver AtomZmath

Z max  = 1500

x₁  = 15

x₂ = 0

-2x-4

step-by-step explanation:

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