Answer:
z = 10x₁ + 8x₂ + 5x₃
Constraints are;
x₁ + 2x₂ + 2x₃ ≤ 1600
3x₁ + 2x₂ + x₃ ≤ 1700
x₁ ≥ 0
x₂ ≥ 0
x₃ ≥ 0
Step-by-step explanation:
There are three types of couches.
Let the first be x₁, second be x₂ and third be x₃.
We are told that the first type uses 1 foot of framing wood, The second type uses 2 feet of framing wood and the third type uses 2 feet of framing wood.
Thus, the first constraint will be;
x₁ + 2x₂ + 2x₃ ≤ A₁
Where A is the amount of framing wood that is available at hand.
Now, we are told that the factory produces 500 couches of the first type, 300 of the second type, and 200 of the third type each month.
Also that the framing wood is increased by 100.
Thus;
A₁ = (1(500) + 2(300) + 2(200)) + 100
A₁ = 1600
Thus,first constraint in full is;
x₁ + 2x₂ + 2x₃ ≤ 1600
For the cabinet wood, we are told that he uses 3 foot of cabinet wood, The second type uses 2 feet of cabinet wood and the third type uses 1 feet of cabinet wood.
Thus, the second constraint will be;
3x₁ + 2x₂ + x₃ ≤ A₂
Now, we are told that the factory produces 500 couches of the first type, 300 of the second type, and 200 of the third type each month.
Also that the cabinet wood is reduced by 600.
Thus;
A₂ = (3(500) + 2(300) + 1(200)) - 600
A₂ = 1700
Thus, full second constraint is;
3x₁ + 2x₂ + x₃ ≤ 1700
Now, for all these to work, x₁, x₂ and x₃ cannot be equal to or less than zero.
Thus, the third constraint is;
x₁ ≥ 0
x₂ ≥ 0
x₃ ≥ 0
We are told that the profit of the three couches are $10, $8, and $5 respectively.
Thus, the function to be maximized can be expressed as;
z = 10x₁ + 8x₂ + 5x₃
