Whenever a question asks for the largest or smallest, it is asking for the vertex.
Find the maximum height
A ball's height is h(t) = −5t² + 20t + 1 metres after t seconds. What is its greatest height?
The peak is at the vertex: t = −b2a = −202 × −5 = 2 seconds. The height then is h(2) = −5(4) + 40 + 1 = 21 m. A is the time, not the height — the commonest slip in projectile questions. D is the starting height, h(0), which is where the ball was thrown from.
t = −b2a is WHEN. Substituting it gives WHAT.
The vertex has two coordinates and questions ask for one or the other. Compute both, label them, then answer what was asked.
−5t² + 20t + 1 = 0
ground level
D = 400 + 20 = 420
use the formula
t = −20 ± √420−10
√420 ≈ 20.49
t ≈ 4.05 s (rejecting −0.05)
time cannot be negative
The negative root corresponds to a moment before the ball was thrown. It is mathematically real and physically meaningless, so it is rejected.
72 m of fence, a rectangle
largest possible area?
2w + 2l = 72, so l = 36 − w
one variable
A = w(36 − w) = −w² + 36w
a quadratic in w
w = −36−2 = 18
the vertex
A = 324 m², an 18 by 18 square
the answer
For a plain rectangle the best shape is always a square. That changes the moment a wall replaces one side — then only three sides are fenced, and the maths gives a different shape.
72 m of fence, a wall on one side
only three sides fenced
2w + l = 72, so l = 72 − 2w
different constraint
A = w(72 − 2w) = −2w² + 72w
the area function
w = −72−4 = 18, l = 36
A = 648 m²
Twice the area from the same fence, because the wall does a third of the work. The method did not change — only the constraint equation did.
At ₱200, 300 tickets sell
each ₱10 rise loses 10 buyers
price = 200 + 10x, sold = 300 − 10x
x rises of ₱10
R = (200 + 10x)(300 − 10x)
= −100x² + 1000x + 60,000
x = −1000−200 = 5
price ₱250, 250 sold
R = ₱62,500
up from ₱60,000
Name one quantity x, write the other with x, and only then decide: vertex (largest, smallest) or solve (a value is given).
Two numbers add to 18. What is the largest their product can be?
numbers x and 18 − x
one variable
P = x(18 − x) = −x² + 18x
a downward parabola
x = −18−2 = 9
the vertex
P = 9 × 9 = 81
check: 8 × 10 = 80, smaller ✓
A ball's height is h = −5t² + 30t metres. When is it 40 m up?
−5t² + 30t = 40
the height is given: solve, no vertex
t² − 6t + 8 = 0
move 40 across, divide by −5
(t − 2)(t − 4) = 0, t = 2 s or t = 4 s
check: −20 + 60 = 40 ✓
The ball passes 40 m twice: at 2 s on the way up and at 4 s on the way down. Both are real and both are physically possible — only negative times get rejected. Halfway between them, at t = 3, is the peak.
The same vertex routine with a different constraint each time.
Two numbers differ by 10. What is the smallest their product can be?
P = x(x + 10) = x² + 10x
a > 0: this one has a minimum
x = −102 = −5
the numbers are −5 and 5
P = −25
a negative minimum
Upward parabola, so the vertex is a smallest value — and it can be negative.
At ₱50 a canteen sells 100 snacks; each ₱5 price cut sells 25 more.
R = (50 − 5x)(100 + 25x)
x price cuts of ₱5
R = −125x² + 750x + 5,000
expand
x = −750−250 = 3
three cuts: price ₱35, 175 sold
R = 35 × 175 = ₱6,125
up from ₱5,000
The price moves down this time, so the price bracket has a minus. The method does not care.
60 m of fence makes a rectangle split into two pens by one fence across the middle.
3w + 2l = 60, so l = 30 − 1.5w
three widths, two lengths
A = w(30 − 1.5w) = −1.5w² + 30w
the area function
w = −30−3 = 10, l = 15
A = 150 m²
Check: 3(10) + 2(15) = 60 ✓. The extra fence means the best shape is no longer a square.
Expect a projectile height, a fixed-fence area (with or without a wall), a number pair, or a price-and-quantity revenue — asking for the greatest value, when it happens, or when a given height is reached. The fastest route is one variable, a quadratic, then −b2a for "when" and substitution for "how much"; set the function equal to the given value only when a value is given. The 10-second check: answer exactly what was asked (time or height, width or area) and test a neighbouring value to confirm it is smaller.
Find the largest area
A farmer has 40 m of fencing for a rectangular pen. What is the largest area she can enclose?
With 2w + 2l = 40, l = 20 − w, so A = w(20 − w) = −w² + 20w. The vertex is at w = 10, giving a 10 by 10 square with area 100 m². D treats 40 as one side and squares it. C uses 20 × 10 without checking the perimeter — that shape needs 60 m of fence.
Hint: write the length in terms of the width first, so the area becomes a quadratic in one variable. Then find its vertex.
