How linear programming solves real business problems

Every linear programming question begins the same way in real life, even though the exam version is dressed in numbers: a business (or a person, or a factory) wants to get the most out of a limited set of resources, or spend the least while still meeting a set of requirements. A bakery has a limited number of hours of oven time and a limited amount of flour; a delivery company has a limited number of vans and a limited number of drivers; a student planning revision has a limited number of hours before the exam.

Linear programming is the method Add Math offers for turning that kind of everyday limitation into a problem with a definite, checkable answer.

The two moving parts are always the same. Constraints describe the limits, they turn into inequalities.

The objective function describes what is being maximised or minimised, profit, cost, distance, time, and it turns into an expression to evaluate once the constraints have narrowed down the possibilities.

• Constraints come from the resources or requirements that limit a decision, and are written as linear inequalities.
• The objective function is the quantity being optimised, written as a linear expression in the same two variables as the constraints.
• The optimum value, the maximum or minimum, is found by checking the objective function only at the corners of the feasible region, never by testing every point inside it.

The hardest step for most students is not the algebra, it is translating a paragraph of business context into a short list of inequalities. It helps to ask the same fixed questions every time: what are the two quantities being decided (these become $x$ and $y$), and what limits are placed on them by the situation described?

A furniture workshop making chairs ($x$) and tables ($y$) each week might face constraints such as a maximum of 40 working hours available, a minimum production requirement to satisfy a standing order, and the simple fact that neither $x$ nor $y$ can be negative.

$$x \geq 0, \quad y \geq 0$$

Those two conditions are worth writing down as a habit on every linear programming question, they are easy to lose marks on simply by forgetting them, even though they follow from the situation without any extra working.

Once every constraint is graphed as a line, with the correct region shaded (or left unshaded, following the exam convention used) to show which side satisfies the inequality, the overlap of all the shaded regions is the feasible region, every point inside it satisfies every constraint at once, and represents a combination of $x$ and $y$ that is actually possible given the limits described.

The objective function is a straight line too, and as its value changes, the line slides across the graph while keeping the same gradient. The key fact that makes the whole method work is that the optimum value of a linear objective function over a region bounded by straight lines always occurs at a vertex (a corner) of that region, never somewhere in the middle of an edge, and never inside the region.

That is why the standard approach is to identify the corners of the feasible region and evaluate the objective function at each one, rather than searching the whole shaded area.

A small catering business makes packed lunches ($x$) and snack boxes ($y$) each day. Preparation time limits production to $x+y\leq 50$, ingredient stock limits it to $2x+y\leq80$, and at least 10 packed lunches must be made to fill a standing order, so $x\geq10$, with $y\geq0$.

The profit on each packed lunch is RM4 and on each snack box is RM3, so the objective function to maximise is:

$$P = 4x + 3y$$

1. Plot the boundary lines $x+y=50$, $2x+y=80$, $x=10$, and $y=0$, shading the region that satisfies all four inequalities.
2. Identify the corners of the resulting feasible region, the points where boundary lines intersect, checked against every constraint.
3. Substitute each corner's coordinates into $P=4x+3y$ and compare the results.

Working through the intersections gives corners at, among others, $(10,40)$ and $(30,20)$. At $(10,40)$, $P=4(10)+3(40)=160$; at $(30,20)$, $P=4(30)+3(20)=180$.

Checking every corner this way, not guessing from the shape of the region, is what the method requires.

The method has several steps, and Add Math linear programming questions tend to lose marks at predictable points rather than through a single hard idea.

• Shading the wrong side of a boundary line, which produces the wrong feasible region entirely.
• Forgetting the non-negativity constraints $x\geq0$ and $y\geq0$, which are implied by context even when not stated outright.
• Testing only one or two corners instead of every vertex of the feasible region.
• Mixing up whether the objective function is being maximised or minimised, and picking the wrong extreme value as a result.

Linear programming rewards a particular kind of practice, translating word problems into inequalities cleanly, drawing an accurate graph, and checking every corner without skipping one out of habit. A teacher working with you one-to-one can watch exactly where a word problem gets misread into the wrong inequality, or where a corner gets missed on the graph, and correct it in the moment rather than after a mark has already been lost.

Our teachers are experienced; lessons run online and are taught in English, while SPM papers themselves are set bilingually in Bahasa Melayu and English.

If linear programming, or the graph-reading skills it depends on, is a sticking point, a one-hour paid trial class at the teacher's own rate (from RM50 per hour, depending on experience) is a reasonable place to start.