Linear Programming Word Problems

A linear programming word problem describes a real situation, such as a factory choosing how many of two products to make, where you want to maximize a quantity like profit or minimize a quantity like cost. The catch is that resources are limited: hours of labor, amount of material, budget, or space. Linear programming turns those limits into a system of linear inequalities and turns the goal (profit, cost, distance) into a linear expression called the objective function.

Every constraint in a linear program is really just a linear inequality in two (or more) variables. If you need a refresher on writing and solving those inequalities before tackling word problems, review solving one-step linear inequalities and multi step linear inequalities.

1. Define the variables. Let \(x\) and \(y\) represent the quantities you are deciding on, such as the number of two products.
2. Write the objective function. This is the linear expression you want to maximize or minimize, for example profit \(P = 4x + 3y\).
3. Write the constraints. Translate every limit in the problem into a linear inequality involving \(x\) and \(y\).
4. Add the non-negativity restrictions. Since \(x\) and \(y\) usually represent physical quantities, include \(x \ge 0\) and \(y \ge 0\).
5. Graph the feasible region. Shade the area where all the constraints overlap.
6. Apply the method of corners. Find the coordinates of every vertex of the feasible region and evaluate the objective function there.
7. Choose the optimal point. The largest (or smallest) value found at a corner is the solution, and you interpret it back in the context of the word problem.

Before graphing, it helps to write the linear program in standard form: the objective function on its own line, followed by every constraint written as a linear inequality, and finally the non-negativity conditions. For a maximization problem this looks like:

Maximize \(P = 4x + 3y\)
subject to \(x + 2y \le 10\), \(3x + y \le 15\), \(x \ge 0\), \(y \ge 0\).

Writing every problem this way keeps the objective function and constraints organized, and it makes it easy to check that you have not missed a limit stated in the word problem, including any combined restriction that behaves like a compound inequality.

For a linear program in two variables, the maximum or minimum of the objective function always occurs at a corner (vertex) of the feasible region, never in the interior. That is why the graphical method is often called the method of corners: graph the constraints, identify the vertices of the shaded region, and test the objective function at each one.

A furniture workshop makes tables and chairs. Let \(x\) be the number of tables and \(y\) be the number of chairs produced. Each table needs 1 hour of assembly and each chair needs 2 hours, with 10 hours of assembly time available, so \(x + 2y \le 10\). Wood usage gives a second limit: each table uses 3 units of wood and each chair uses 1 unit, with 15 units on hand, so \(3x + y \le 15\). The profit function is \(P = 4x + 3y\).

The standard form of this linear program is:

Maximize \(P = 4x + 3y\)
subject to \(x + 2y \le 10\), \(3x + y \le 15\), \(x \ge 0\), \(y \ge 0\).

The feasible region is the shaded polygon, and its corners are \((0,0)\), \((5,0)\), \((4,3)\), and \((0,5)\). The point \((4,3)\) is where the two constraint lines cross, found by solving \(x + 2y = 10\) and \(3x + y = 15\) together.

The largest value, \(P = 25\), occurs at \((4, 3)\). So the workshop should build 4 tables and 3 chairs to maximize profit, given the available assembly time and wood.

• Forgetting the non-negativity restrictions \(x \ge 0\) and \(y \ge 0\), which trims the feasible region to the first quadrant in most applied problems.
• Testing the objective function only at intercepts instead of at every true vertex of the feasible region, including intersection points between two constraint lines.
• Mixing up a maximization objective with a minimization objective; always reread the word problem to confirm whether you want the largest or smallest corner value.
• Shading the wrong side of a constraint line; always test a point like \((0,0)\) in the original inequality to check which side is feasible.

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