KBAT Problems
KBAT Problem: A Region from Constraints
A KBAT problem where real limits (budget, time, capacity) become linear inequalities and a feasible region, modelling before graphing.
Turn limits into inequalities
A workshop can make two products with limits on materials and hours. Each limit becomes one inequality; "at least" and "at most" decide the direction.
Writing all the inequalities correctly is the KBAT step.
Shade the feasible region
Graph each inequality, shade the region that satisfies all of them at once, and read the answer, such as the greatest number of a product, from the corner points. Show the shaded region clearly for full marks.
Understand the problem
A stall makes gift box A and gift box B. Each A uses 2 ribbons and each B uses 1 ribbon, with at most 10 ribbons available.
Each box takes 1 hour to assemble, with at most 7 hours available. Profit is RM 3 per A and RM 2 per B.
How many of each maximise profit? What it really asks: convert the limits into inequalities and a region before any graphing.
Plan and solve
- Let x = number of box A and y = number of box B, with x ≥ 0 and y ≥ 0.
- Ribbons: 2x + y ≤ 10. Time: x + y ≤ 7. Objective to maximise: profit P = 3x + 2y.
- The corner points of the feasible region are (0, 0), (5, 0), (0, 7) and the intersection of 2x + y = 10 with x + y = 7, which is (3, 4).
- Test P at each corner: (0,0)→0, (5,0)→15, (0,7)→14, (3,4)→9 + 8 = 17. Maximum profit is RM 17 by making 3 of A and 4 of B.
Check and a variant
Check that (3, 4) obeys both limits: ribbons 2(3) + 4 = 10 ≤ 10 and time 3 + 4 = 7 ≤ 7, both exactly met, so it sits on the boundary as an optimum should. A variant the examiner could add: change the profit to RM 2 per A and RM 3 per B.
Retesting the corners gives (0,7)→21 as the new maximum, showing the best corner shifts when the objective changes even though the region does not.
Frequently asked questions
How do I know which way to shade for each inequality?
Draw the boundary line, then test a point not on it, usually the origin. If the point satisfies the inequality, shade its side; if not, shade the other side.
Do this for each inequality and the feasible region is where all the shaded sides overlap. Testing the origin is fast and reliable.
Why do I only test the corner points and not every point in the region?
For a linear objective, the maximum or minimum always occurs at a corner of the feasible region, so checking the corners is enough. Testing every point would take forever and is unnecessary.
Identify each corner precisely, evaluate the objective at all of them, and pick the best. This is the standard, marker-approved method.
What if the best corner gives fractions but the answer must be whole items?
When the items must be whole, such as boxes or people, and a corner is fractional, you check the whole-number points inside the region nearest that corner and pick the one that satisfies every constraint with the best objective value. State clearly that the quantities must be integers, since that reasoning itself earns KBAT marks.