A furniture company produces three types of couches. The first type uses 1 foot of framing wood and 3 feet of cabinet wood. The second type uses 2 feet of framing wood and 2 feet of cabinet wood. The third type uses 2 feet of framing wood and 1 foot of cabinet wood. The profit of the three types of couches are $10, $8, and $5, respectively. The factory produces 500 couches each month of the first type, 300 of the second type, and 200 of the third type. However, this month there is a shortage of cabinet wood and the supply to the factory will have to be reduced by 600 feet, but the supply of framing wood is increased by 100 feet. How should the production of the three types of couches be adjusted to minimize the decrease in profit?

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

5c=d for c
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