KBAT Problems
KBAT Problem: Quadratics in a Real Situation
A worked KBAT-style problem that hides a quadratic inside a real situation, the kind of multi-step question that separates an A from an A+.
Spot the quadratic
A rectangular garden has a fixed perimeter, and you are asked for the dimensions that give the largest area. The clue that it is a quadratic is the word "largest" area as a function of one side is a parabola, and its maximum is the turning point.
Solve it in three stages
- Understand: write the area as a function of one side using the perimeter.
- Plan: recognise it as a quadratic and complete the square to find the maximum.
- Execute and check: read the turning point, and confirm the dimensions are sensible.
Understand the problem
Here is a full problem. A gardener has 60 m of fencing to enclose a rectangular garden against a straight wall, so the fence covers only the other three sides.
What dimensions give the largest area? What it really asks: express area as a function of one side, then find the maximum of that quadratic, not just any rectangle.
Plan and solve
- Let the two sides at right angles to the wall be x. The side parallel to the wall is 60 − 2x, because 2x + (that side) = 60.
- Area A = x(60 − 2x) = 60x − 2x². This is a quadratic in x that opens downward, so it has a maximum.
- Complete the square: A = −2(x² − 30x) = −2((x − 15)² − 225) = −2(x − 15)² + 450.
- Maximum at x = 15, giving A = 450. The parallel side is 60 − 2(15) = 30, so the garden is 15 m by 30 m with area 450 m².
Check and a variant
Check either side of the maximum: at x = 14, area = 14 × 32 = 448; at x = 16, area = 16 × 28 = 448, both below 450, so 450 m² is genuinely the peak. A variant the examiner could add: fence all four sides with the same 60 m.
Then 2x + 2y = 60, so A = x(30 − x), peaking at x = 15, y = 15, a 15 m square of area 225 m². Same method, a square answer.
Frequently asked questions
How do I know a word problem is secretly a quadratic?
Look for a product of two quantities that must be made largest or smallest, or a squared term hiding in area, height or profit. Words like largest, maximum or least applied to a variable amount are the strongest clue that a parabola, and a turning point, are involved.
Completing the square or the formula, which is better here?
For a maximum or minimum, completing the square is better because it hands you the turning point directly. Use the quadratic formula when you only need where the graph cuts the x-axis.
For SPM you want both fluent, but a max-area question is really asking for the vertex.
My child can solve x²−5x+6=0 but freezes on worded ones. Why?
Solving a given equation and building one from a situation are different skills. The freeze is at the translation step, not the algebra.
Practise turning one short sentence into an expression first, then solve. The calculation is the part they already own; the modelling is what needs the reps.