Question

LP 3. A manufacturer wants to maximize the profit of two products. Product X yields a profit of $ 1.50 per unit, and Product Y yields a profit of $2.00 per unit. Market tests and available resources have indicated the following constraints: The combined production level should not exceed 1200 units per month. The demand for product Y is no more than half the demand for product X. The production level of product X is less than or equal to 600 units plus three times the production level of product Y.

Please provide step by step of how do you solve the constrains. I have been reviewing other posts on the same question, but they do not specify how constraints were solved. I want to learn how to solve it. no just to copy your answer.

Solve using the graphing the intercept with the elimination method. Do not use excel.

Answer 1

Let the no. of product X be x and Product Y be y

Total Profit = 1.5*x + 2*y

We have to maximize this profit

Subject to constraints:

x + y <= 1200

y <= 0.5*x

x <= 600 + 3*y

x - 3y >= 600

We graph the constraints as shown below:

The feasible region is shown hatched in orange. We find the corner points:

Green line = x - 3y = 600, Put y = 0, x = 600

Hence, Point D = (600, 0)

We solve equations for green line and red line above two lines simultaneously

x - 3y = 600...........(1)

x + y = 1200...........(2)

Subtract equations (1) from (2)

(x + y) - (x - 3y) = 1200 - 600

4y = 600

y = 150

Substituting in equation (2)

x + 150 = 1200

x = 1050

Hence, Point C = (1050, 150)

Point A = 0

We solve equations for blue line and red line above two lines simultaneously

y = 0.5x...........(3)

x + y = 1200...........(4)

Substitute equation 3 in 4

x + 0.5x = 1200

x = 800

Substitue in (4)

800 + y = 1200

y = 400

Hence, Point B = (800, 400)

Hence, we get 4 corner points as:

A = (0, 0), B = (800, 400), C = (1050, 150) and D = (600, 0)

We find the objective value as shown below for each point

A = (0, 0); objective = 1.5*x + 2*y = 1.5 * 0 + 2 * 0 = 0

B = (800, 400); objective = 1.5*800 + 2 * 400 = 1200 + 800 = 2000

C = (1050, 150); objective = 1.5*1050 + 2*150 = 1875

D = (600, 0); objective = 1.5*600 + 2*0 = 900

Maximum value = 2000

Hence, the optimum solution = X = 800, Y = 400

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