The WorldLight Company produces two light fixtures (Products 1 and 2) that require both metal frame parts and electrical components. Management wants to determine how many units of each product to produce per week so as to maximize profit. For each unit of Product 1, one unit of frame parts and two units of electrical components are required. For each unit of Product 2, three units of frame parts and two units of electrical components are required. The company has a weekly supply of 3,000 units of frame parts and 4,500 units of electrical components. Each unit of Product 1 gives a profit of $13, and each unit of Product 2, up to 900 units, gives a profit of $26. Any excess over 900 units of Product 2 brings no profit, so such an excess has been ruled out. Formulate this same model algebraically.

Question

Emanuel Pitts

Mathematics

8 July, 15:55

The WorldLight Company produces two light fixtures (Products 1 and 2) that require both metal frame parts and electrical components. Management wants to determine how many units of each product to produce per week so as to maximize profit. For each unit of Product 1, one unit of frame parts and two units of electrical components are required. For each unit of Product 2, three units of frame parts and two units of electrical components are required. The company has a weekly supply of 3,000 units of frame parts and 4,500 units of electrical components. Each unit of Product 1 gives a profit of $13, and each unit of Product 2, up to 900 units, gives a profit of $26. Any excess over 900 units of Product 2 brings no profit, so such an excess has been ruled out. Formulate this same model algebraically.

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Answers (1)

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Salas · Answer 1

Step-by-step explanation:

Let call:

x₁ number of units of product 1

x₂ number of units of product 2

We build the next table

metal frame electrical components Profit $/u

Product 1 (x₁) 1 2 13

Product 2 (x₂) 3 2 26 (*)

Total supply (in units) 3000 4500

(*) Up to 900 units. each unit above that number has to be ruled out

From the above table we have:

Objective function z

z = 13*x₁ + 26*x₂ to maximize

1*x₁ + 3*x₂ ≤ 3000

2*x₁ + 2*x₂ ≤ 4500

x₁ ≥ 0; 900 ≥ x₂ ≥ 0

Rearranging

z - 13 * x₁ - 26*x₂ = 0

0 + x₁ + 3*x₂ ≤ 0

0 + 2*x₁ + 2*x₂ ≤ 0

x₁ ≥ 0

0 ≤ x₂ ≤ 900

Or

z - 13 * x₁ - 26*x₂ = 0

0 + x₁ + 3*x₂ + s₁ + 0s₂ = 0

0 + 2*x₁ + 2*x₂ + 0s₁ + s₂ = 0

x₂ + 0s₁ + 0s₂ + s₃ = 900

x₁ ≥ 0 x₂ ≥ 0

And the table is ready for use Simplex Method