Build the quadratic first — the vertex does the rest.
The problem
A farmer has 200 m of fencing for a rectangular field. What is the largest area he can enclose?
The perimeter gives l + w = 100, and a product with a fixed sum is largest when the parts are equal: 50 × 50 = 2,500 m². D uses 100 × 100 — that is the perimeter mistaken for a side length.
The fencing is the perimeter, and it cannot change.
2l + 2w = 200
the constraint
l + w = 100
divide by 2
w = 100 − l
one letter in terms of the other
Every problem of this kind has two relationships: one fixed (the constraint) and one to be maximised. Find the fixed one first.
A = l × w
area of a rectangle
A = l(100 − l)
substitute for w
A = 100l − l²
a quadratic in l
a = −1, so the parabola opens downward and its vertex is a maximum. Exactly what the question wants.
A = −l² + 100l
a = −1, b = 100
l = −b ÷ 2a = −100 ÷ (−2)
the axis of symmetry
l = 50
metres
w = 100 − 50 = 50
back-substitute
A = 50 × 50 = 2500
square metres
“What dimensions” wants 50 by 50. “What is the largest area” wants 2500. Stopping at w = 50 answers the first question, not the second — on an exam with a 50 among the choices, that is the trap. Squaring the 100 instead (100 × 100 = 10,000) treats the half-perimeter as a side, which is how 10,000 m² ends up in the opener's choices.
Test a few shapes with the same 200 m of fence.
| Length | Width | Area |
|---|---|---|
| 10 | 90 | 900 |
| 30 | 70 | 2100 |
| 40 | 60 | 2400 |
| 50 | 50 | 2500 |
| 60 | 40 | 2400 |
Symmetric either side of 50 — which is the parabola showing itself in the table.
For a fixed perimeter the square always wins. Worth knowing as a sanity check, though the working still has to be shown.
At ₱20 a ticket 300 people come; every ₱1 rise loses 10 of them.
price = 20 + x
x is the rise in pesos
attendance = 300 − 10x
the trade-off
R = (20 + x)(300 − 10x)
revenue
R = −10x² + 100x + 6000
expand
x = −100 ÷ (−20) = 5
price ₱25, revenue ₱6,250
Same three steps: name the variable, write the product, find the vertex. Only the story changes.
An upward parabola (a > 0) has a lowest point — that is the minimum.
numbers x and 10 − x
the constraint
S = x² + (10 − x)²
the quantity to minimise
S = 2x² − 20x + 100
a = 2 > 0, so a minimum
x = 20 ÷ 4 = 5
−b2a
S = 25 + 25 = 50
both numbers are 5
Products behave the same way. Two numbers that differ by 10 have product P = x(x + 10) = x² + 10x, which is least at x = −5: the numbers are −5 and 5 and the least product is −25.
The method never changes; these are the set-ups that make students stall.
Fence on three sides: 100 m of fence, the fourth side is a wall
2w + l = 100, so l = 100 − 2w
two widths, one length
A = w(100 − 2w) = −2w² + 100w
a = −2, a maximum
w = −100 ÷ (−4) = 25, l = 50
−b2a
A = 25 × 50 = 1,250 m²
not a square
With one side free, the best shape is twice as long as it is wide. “The square is best” only holds when all four sides use fence.
Count every piece of fence. A pen of 90 m with one fence across the middle, parallel to the width, uses three widths and two lengths: 3w + 2l = 90, so l = 45 − 1.5w and A = w(45 − 1.5w) = 45w − 1.5w². Then w = 45 ÷ 3 = 15, l = 22.5 and A = 337.5 m².
The best change is a cut: 200 buy at ₱50; each ₱1 rise loses 5 buyers
R = (50 + x)(200 − 5x)
x is the rise in pesos
R = −5x² − 50x + 10,000
200x − 250x = −50x
x = −(−50) ÷ (−10) = −5
a negative rise
price ₱45, R = 45 × 225 = ₱10,125
200 − 5(−5) = 225 buyers
A negative x is not a wrong answer here: it says lower the price by ₱5. Rejecting it and keeping ₱50 gives ₱10,000, which is less.
A vertex between whole numbers: two numbers add to 15; find their largest product
P = x(15 − x) = −x² + 15x
the constraint, substituted
x = 152 = 7.5, P = 7.5 × 7.5 = 56.25
any real numbers allowed
integers only: 7 × 8 = 56
the nearest whole numbers to 7.5
When the count must be whole (people, items, integers), test the whole numbers either side of the vertex; the parabola is symmetric, so both give the same product here.
Similar triangles link the width and the height.
h6 = 10 − x10
the small triangle beside the rectangle is similar
h = 6 − 3x5
A = x(6 − 3x5) = 6x − 3x²5
a = −35, a maximum
x = −6 ÷ (2 · (−35)) = 5
−b2a
h = 6 − 3 = 3, A = 5 × 3 = 15
half the triangle’s area of 30
The best rectangle in a right-angle corner always takes half of each leg — which is why its area comes out at exactly half the triangle’s.
Max/min items are stories: fencing, ticket prices, numbers with a fixed sum or difference, a shape inside a shape. Write the quantity in one variable, use −b2a, then substitute — the answer is usually the maximum or minimum value, not the x. Quick check: try a value either side; both must be worse.
The problem
A rectangular pen is built against a wall, so only three sides need fencing. With 60 m of fence, what is the largest area?
Only three sides are fenced, so 2w + l = 60 and A = w(60 − 2w). That peaks at w = 15, giving l = 30 and A = 450 m². B assumes a square — with one side free, the best pen is twice as wide as it is deep.
