## Linear Optimization

The task of optimization is to determine the optimal value for a given goal, whereby given conditions must be observed.

The program determines the optimal solution for a two-variable objective function with linear inequalities as boundary conditions.

### Example 1 (Maximization)

A factory produces two different mobile phones. X devices of type A and y devices of type B are to be completed daily.

- Boundary conditions:
- The individual departments have the following production capacities per day:

- The assembly line for Type A can produce a maximum of 600 devices.

x ≤ 600 - The assembly line for type B can produce a maximum of 700 devices.

y ≤ 700 - The plastics department produces a maximum of 750 cases in total. x + y ≤ 750
- The electrical department produces a maximum of 400 type A devices or 1200 type B devices or a combination thereof. This means that 1/400 of the total time is required per device of type A and 1/1200 per device of type B.

1/400·x + 1/1200·y ≤ 1 or 3·x + 1·y ≤ 1200;

- The assembly line for Type A can produce a maximum of 600 devices.

- Objective function:
- How many devices of each must be produced per day in order to achieve maximum profit if the profit for type A is $140 per device and for type B $80.

Objective function: ƒ(x,y) = 140·x + 80·y → Maximum Constraints: x ≥ 0 y ≥ 0 x ≤ 600 y ≤ 700 x + y ≤ 750 3·x + y ≤ 1200 Maximum x = 225 y = 525 ƒ(x,y) = 73500

The maximum profit of $73500 is therefore achieved if 225 type A and 525 type B devices are manufactured daily.

### Example 2 (Minimization)

Objective function: ƒ(x,y) = 140·x + 80·y → Minimum Constraints: x ≥ 0 y ≥ 0 x ≥ 160 y ≥ 80 x + y ≥ 750 3 x + y ≥ 1200 Minimum x = 225 y = 525 ƒ(x,y) = 73500