## Linear Optimization

The task of optimization is to find the optimal value for a given objective, while satisfying given constraints.
The program determines the optimal solution for a two-variable objective function with linear inequalities as constraints.

### Example 1 (Maximization)

A factory produces two different types of cell phones. X units of type A and y units of type B are to be completed daily.

Constraints:
The individual departments have the following daily production capacities:
1. The assembly line for type A can produce a maximum of 600 units.
x ≤ 600
2. The assembly line for type B can produce a maximum of 700 units.
y ≤ 700
3. The plastic department can produce a maximum of 750 units.
4. x + y ≤ 750
5. The electrical department can produce a maximum of 400 type A units or 1200 type B units or a combination of both. This means that 1/400 of the total time is required for each of type A and 1/1200 for each of type B.
1/400·x + 1/1200·y ≤ 1   or   3·x + 1·y ≤ 1200;
Objective function:
How many units of each type must be produced per day in order to obtain maximum profit if the profit for type A is \$140 per unit 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```

Therefor, the maximum profit of \$73500 is achieved if 225 type A and 525 type B units are produced each day.

### 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```