Solve a linear programming problem with non-negativity constraints.

Question: Solve a linear programming problem with non-negativity constraints.

Answer: Linear programming is a mathematical technique used to optimize a linear objective function subject to linear constraints. Non-negativity constraints require that the decision variables must be greater than or equal to zero.

Let's consider an example of a company that produces two products, A and B. The company has limited resources of 100 units of labour and 80 units of raw material. Product A requires 2 units of labour and 1 unit of raw material, while product B requires 1 unit of labour and 2 units of raw material. The profit per unit of product A is £5 and for product B is £4. The company wants to maximize its profit.

Let x be the number of units of product A produced and y be the number of units of product B produced. Then the objective function is:

Maximize Z = 5x + 4y

Subject to the constraints:

2x + y ≤ 100 (Labour constraint)
x + 2y ≤ 80 (Raw material constraint)
x ≥ 0, y ≥ 0 (Non-negativity constraints)

We can graph these constraints on a coordinate plane and find the feasible region, which is the region that satisfies all the constraints. The feasible region is the shaded area in the graph below:


The optimal solution occurs at the corner point of the feasible region that maximizes the objective function. We can find this point by calculating the value of the objective function at each corner point:

At (0, 0): Z = 5(0) + 4(0) = 0
At (0, 40): Z = 5(0) + 4(40) = 160
At (50, 0): Z = 5(50) + 4(0) = 250
At (20, 30): Z = 5(20) + 4(30) = 190

To understand further how linear programming can be applied in various fields, explore these additional resources on optimization problems, the theory behind applications of linear programming, and how it can be applied in real-world applications.

A-Level Maths Tutor Summary: In this example, we use linear programming to maximise profit for a company making products A and B with limited labour and materials. By plotting constraints and non-negativity on a graph, we find the best solution is making 50 units of A and none of B, yielding £250 profit. This process demonstrates optimizing decisions within given limits.