# Linear Programming Examples

A store has requested a manufacturer to produce pants and sports jackets.

For materials, the manufacturer has 750 m^{2} of cotton textile and 1,000 m^{2} of polyester. Every pair of pants (1 unit) needs 1 m^{2} of cotton and 2 m^{2} of polyester. Every jacket needs 1.5 m^{2} of cotton and 1 m^{2} of polyester.

The price of the pants is fixed at $50 and the jacket, $40.

What is the number of pants and jackets that the manufacturer must give to the stores so that these items obtain a maximum sale?

1**Choose the unknowns.**

**x = number of pants**

**y = number of jackets**

2**Write the objective function**.

**f(x,y)= 50x + 40y**

3Write the constraints as a system of inequalities.

To write the constraints, use a table:

pants | jackets | available | |
---|---|---|---|

cotton | 1 | 1,5 | 750 |

polyester | 2 | 1 | 1,000 |

x + 1.5y ≤ 750 **2x+3y ≤ 1500**

**2x + y ≤ 1000**

As the number of pants and jackets are natural numbers, there are two more constraints:

**x ≥ 0**

**y ≥ 0**

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

Represent the constraints graphically.

As x ≥ 0 and y ≥ 0, work in the first quadrant.

Represent the straight lines from their points of intersection with the axes.

Solve the inequation graphically: 2x +3y ≤ 1500, and take a point on the plane, for example (0,0).

2 · 0 + 3 · 0 ≤ 1,500

Since 0 ≤ 1,500 then the point (0,0) is in the half plane where the inequality is satisfied.

Similarly, solve 2x + y ≤ 1,000.

2 · 0 + 0 ≤ 1,000

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**.

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

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

2x + 3y = 1,500; x = 0 (0, 500)

2x + y = 1,000; y = 0 (500, 0)

2x + 3y =1,500; 2x + y = 1,000 (375, 250)

6 **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.

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

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

f(0, 500) = 50·0 + 40·500 = $20,000

f(500, 0) = 50·500 + 40·0 = $25,000

f(375, 250) = 50·375 + 40·250 = $28,750 **Maximum**

The optimum solution is to make 375 pants and 250 jackets to obtain a benefit of $28,750.

The solution is not always unique, so we can also find other solutions.

#### Example

If the objective function of the previous exercise had been:

f(x,y) = 20x + 30y

f(0,500) = 20·0 + 30 · 500 = $15,000 ** Maximum**

f(500, 0) = 20·500 + 30·0 = $10,000

f(375, 250) = 20·375 + 30·250 = $15,000 **Maximum**

In this case, all the pairs with integer solutions of the segment drawn in black would be the maximum.

f(300, 300)= 20·300 + 30·300 = $15,000 **Maximum**