# linear programming with two unknowns

Nowadays the programs of linear programming are very complex because there are several variables. The resolution of these complex programs is done by the "simplex method" with the help of a computer.

We are going to solve simple problems with only two variables, following these steps:

–Make a table, determine the variables, write constraints and find the objective function.

–Represent the feasible region and find its vertices

–Calculate the value of the function in each vertex and determine the optimum solution

Examples:

1) A small business enterprise makes dresses and trousers. Making a dress requires ½ hour's cutting work and 20 minutes' stitching. Making a pair of trousers requires 15 minutes' cuttings and ½ hour's stitching. The profit on a dress is 40 € and on a pair of trousers 50 € respectively. The business operates for a maximum of 8 hours per day. Determine how many dresses and trousers should be made to maximize profit and what the maximum profit is.

Let x the number of dresses and y the pairs of trousers.

The objective function (profit) is

P(x,y) = 40x + 50y

Constraints are:

Feasible region is:

Solving the systems of this pair of equations, we obtain the vertices of the feasible region: (0,0), (0,16), (16,0) and (12,8). Then:

P(0,0) = 0

P(0,16) = 800

P(16,0) = 640

P(12,8) = 880

12 dresses and 8 pairs of trousers should be made and the maximum profit is 880 €

2) A farmer has 80 hectares of his farm available for planting maize and cabbages. He must grow at least 10 hectares of maize and 20 of cabbages to meet demands. He prefers to plant more cabbages than maize but his work force and equipment will only allow him to cultivate a maximum of three times more maize than cabbages. If the profit of maize is 800 €/ha and on cabbages 500 €/ha, how should the farmer plant the two crops to make a maximum profit and what is this profit.

Let x the number of hectares of maize and y the number of hectares of cabbages.

The objective function (profit) is

P(x,y) = 800x + 500y

Constraints are:

Feasible region is:

Solving the systems of pair of equations we obtain the vertices of the feasible region: (20,60), (10,20), (10,30) and (60,20). Then:

P(20,60) = 46000

P(10,20) = 18000

P(10,30) = 23000

P(60,20) = 58000

The farmer should plant 60 hectares of maize and 20 hectares of cabbages and the maximum profit is 58000 €

3) A school is preparing a trip for 400 students. The company who is providing the transportation has 10 buses of 50 seats each and 8 buses of 40 seats, but it only has 9 drivers available. The rental cost for a large bus is 800 € and 600 € for the small bus. Calculate how many buses of each type should be used for the trip for the least possible cost.

Let x the number of small buses and y the number of big buses.

The objective function (cost) is

C(x,y) = 600x + 800y

Constraints are:

Feasible region is:

P(0,8) = 6400

P(0,9) = 7200

P(5,4) = 6200

5 small buses and 4 big ones should be used for the trip for the least possible cost (6200 €)

You can use the graph method too. Examples:

This is very useful with unbounded feasible regions:

And you can find these situations:

**Exercises**:

1) A transport company has two types of trucks, Type A and Type B. Type A has a refrigerated capacity of 20 m^{3} and a non-refrigerated capacity of 40 m^{3} while Type B has the same overall volume with equal sections for refrigerated and non-refrigerated stock. A grocer needs to hire trucks for the transport of 3,000 m^{3} of refrigerated stock and 4,000 m^{3} of non-refrigerated stock. The cost per kilometer of a Type A is 30 €, and 40 € for Type B. How many trucks of each type should the grocer rent to achieve the minimum total cost?

2) A store wants to liquidate 200 of its shirts and 100 pairs of pants from last season. They have decided to put together two offers, A and B. Offer A is a package of one shirt and a pair of pants which will sell for 30 €. Offer B is a package of three shirts and a pair of pants, which will sell for 50 €. The store does not want to sell less than 20 packages of Offer A and less than 10 of Offer B. How many packages of each do they have to sell to maximize the money generated from the promotion?

Solutions: 1) 50 Type A trucks and 67 Type B trucks. 2) 50 packages of each offer

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