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Exercise 1

A company manufactures and sells two models of lamps, L1 and L2. To manufacture each lamp, the manual work involved in model L1 is 20 minutes and for L2, 30 minutes. The mechanical (machine) work involved for L1 is 20 minutes and for L2, 10 minutes. The manual work available per month is 100 hours and the machine is limited to only 80 hours per month. Knowing that the profit per unit is 10 for L1 and L2, respectively, determine the quantities of each lamp that should be manufactured to obtain the maximum benefit.

Exercise 2

With the start of school approaching, a store is planning on having a sale on school materials. They have 600 notebooks, 500 folders and 400 pens in stock, and they plan on packing it in two different forms. In the first package, there will be 2 notebooks, 1 folder and 2 pens, and in the second one, 3 notebooks, 1 folder and 1 pen. The price of each package will be 7.00 respectively. How many packages should they put together of each type to obtain the maximum benefit?

Exercise 3

On a chicken farm, the poultry is given a healthy diet to gain weight. The chickens have to consume a minimum of 15 units of Substance A and another 15 units of Substance B. In the market there are only two classes of compounds: Type X, with a composition of one unit of A to five units of B, and another type, Y, with a composition of five units of A to one of B. The price of Type X is 30. What are the quantities of each type of compound that have to be purchased to cover the needs of the diet with a minimal cost?

Exercise 4

There is only 600 milograms of a certain drug that is needed to make both large and small pills for small scale pharmaceutical distribution. The large tablets weigh 40 milograms and the small ones, 30 milograms. Consumer research determines that at least twice the amount of the smaller tablets are needed than the large ones and there needs to be least three large tablets made. Each large tablet is sold for a profit of 1. How many tablets of each type have to be prepared to obtain the maximum profit?

 

Solution of exercise 1

A company manufactures and sells two models of lamps, L1 and L2. To manufacture each lamp, the manual work involved in model L1 is 20 minutes and for L2, 30 minutes. The mechanical (machine) work involved for L1 is 20 minutes and for L2, 10 minutes. The manual work available per month is 100 hours and the machine is limited to only 80 hours per month. Knowing that the profit per unit is 10 for L1 and L2, respectively, determine the quantities of each lamp that should be manufactured to obtain the maximum benefit.

Step 1 - Identify the decision variables

x = number of lamps L1

y = number of lamps L2

 

Step 2 - Write the objective function

Write the objective function.

 

Step 3 - Write the set of constraints

Write the constraints as a system of inequalities.

Convert the time from minutes to hours.

20 min = 1/3 h

30 min = 1/2 h

10 min = 1/6 h

As the number of lamps are natural numbers, there are two more constraints:

 

Step 4 - Choose the method to solve the problem

We will solve this problem graphically.

 

Step 5 - Construct the graph

Find the set of feasible solutions that graphically represent the constraints.

Represent the constraints graphically.

As and , work in the first quadrant.

Solve the inequality graphically:

Take a point on the plane, for example (0,0).

Example 1 - Graph

 

Step 6 - Find the feasibility region of the graph

The area of intersection of the solutions of the inequalities would be the solution to the system of inequalities, which is the set of feasible solutions.

Feasibility region of the graph

 

Step 7 - Find the optimum point

Calculate the coordinates of the vertices from the compound of feasible solutions.

The optimal solution, if unique, is a vertex. These are the solutions to systems:

                 

                 

                   

Finding the optimum point from the coordinates of the vertices

Calculate the value of the objective function at each of the vertices to determine which of them has the maximum or minimum values.

In the objective function, place each of the vertices that were determined in the previous step.

    Maximum

The optimum solution is to manufacture 210 units of model L1 and 60 units of model L1 to obtain a benefit of 6.50 and f(x, y) = 6.5x + 7yx + y leq 5002x + 3y leq 6002x + y leq 400x geq 0y geq 0f(x,y) = 6.5 cdot 200 + 7 cdot 0 = 1300f(x,y) = 6.5 cdot 0 + 7 cdot 200 = 1400f(x,y) = 6.5 cdot 150 + 7 cdot 100 = 16751675.

 

Solution of exercise 3

On a chicken farm, the poultry is given a healthy diet to gain weight. The chickens have to consume a minimum of 15 units of Substance A and another 15 units of Substance B. In the market there are only two classes of compounds: Type X, with a composition of one unit of A to five units of B, and another type, Y, with a composition of five units of A to one of B. The price of Type X is 30. What are the quantities of each type of compound that have to be purchased to cover the needs of the diet with a minimal cost?

Step 1 - Identify the decision variables

x = X

y = Y

 

Step 2 - Write the objective function.

Step 3 - Identify the constraints

Write the constraints as a system of inequalities.


 

Step 4 - Choose the method to solve the problem

We will solve this problem graphically.

 

Step 5 - Construct the graph

Graph - Example 3

 

Step 6 - Find the feasibility region

Find the set of feasible solutions that graphically represent the constraints.

Feasibility region of the graph

The grey highlighted area represents the feasibility region.

 

Step 7 - Find the optimum point

Calculate the coordinates of the vertices from the compound of feasible solutions.

Optimal point from the coordinates of the vertices

Calculate the value of the objective function at each of the vertices to determine which of them has the maximum or minimum values. It must be taken into account the possible non-existence of a solution if the compound is not bounded.

      Minimum

The minimum cost is X = frac {5} {2} Y = frac{5}{2}2 and the small tablet, f(x, y) = 2x + y40x + 30y leq 600x geq 3y geq 2xx geq 0y geq 0f(x,y) = 2 cdot 3 + 16 = 22f(x,y) = 2 cdot 3 + 6 = 12f(x,y) = 2 cdot 6 + 12 = 2424, and is obtained by making 6 units of the large tablets and 12 units of the small tablets.

 

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Emma

Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.