A quantity that depends quadratically on one variable is largest or smallest at the vertex, and the vertex is x = -b/(2a). The algebra is three lines; the work is building the quadratic from the situation and respecting the values the variable is actually allowed to take.
Core ideas
One turning point, so there is nothing to compare — provided the variable is free to take any value.
There is no second candidate to check and no interior comparison to make. That is why this is the one optimisation problem school algebra can settle exactly, and why calculus is not needed for it.
x = -b/(2a) says where; f(-b/(2a)) says how much. "How many should be sold" and "what is the greatest profit" have different answers, and each is wrong in the other's place.
A length is positive, a number of tickets is a whole number, a time after launch is not negative. If the vertex is unavailable, a closed interval requires comparison of its endpoints and a whole-number model requires nearby permitted integers. An excluded endpoint may leave a bound that is approached but never attained, so an optimum need not exist.
The material
The theorem, the modelling, and the cases where the vertex is not the answer.
Every problem in this lesson has the same shape: a quantity is expressed as a quadratic in one variable, and the largest or smallest value of that quadratic is wanted. “Vertex Form and the Axis of Symmetry” supplies the answer; what remains is getting to the quadratic and reading the question correctly.
Let f(x) = ax² + bx + c with a ≠ 0 and h = -b/(2a). If a > 0 then f(x) > f(h) for every x ≠ h, so f(h) is the smallest value f takes and there is no largest. If a < 0 then f(x) < f(h) for every x ≠ h, so f(h) is the largest and there is no smallest.
“Graphs of Quadratic Functions” proved f(h + t) = f(h) + at² for every real t. Every real number x is h + t for exactly one t, namely t = x - h, so this covers all of them.
t² > 0 whenever t ≠ 0. If a > 0 then at² > 0 and f(x) > f(h); if a < 0 then at² < 0 and f(x) < f(h). Either way the comparison is strict away from h, so the optimum is attained at x = h and nowhere else.
For the other direction: with a > 0, f(h) + at² grows without bound as |t| grows, so no number is an upper bound for f and there is no maximum. With a < 0 the same term decreases without bound, so there is no minimum.
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Let x be one side in metres. The perimeter fixes the adjacent side: 2x + 2y = 80 gives y = 40 - x. The area is A(x) = x(40 - x) = -x² + 40x, so a = -1 and b = 40.
a < 0, so the vertex is a maximum, at x = -40/(2·(-1)) = 20. Then A(20) = 20 · 20 = 400. The permitted values are 0 < x < 40, and 20 is among them, so the greatest area is 400 square metres, from a 20 by 20 square.
The optimal rectangle came out square. That is not proved by this example — one perimeter was tested, not all of them. Running the same three lines with a perimeter P gives A(x) = x(P/2 - x), vertex at x = P/4, and the other side P/2 - P/4 = P/4 as well; doing it once with a letter is what settles the general claim.
A ball thrown upward has height h(t) = -5t² + 20t + 1 metres after t seconds. a = -5, so the vertex is a maximum, at t = -20/(2·(-5)) = 2 seconds, and the greatest height is h(2) = -20 + 40 + 1 = 21 metres.
"When does it land" is a different question and needs the formula, not the vertex: -5t² + 20t + 1 = 0 gives t = 2 ± √105/5, and only the positive root 2 + √105/5 — about 4.05 seconds — is in the domain t ≥ 0.
Suppose profit is P(x) = -3x² + 30x - 12 in pounds from x machines, and only 4 machines exist. The vertex is at x = 5, outside the permitted 0 ≤ x ≤ 4. By the identity P(h + t) = P(h) + at² the function rises steadily as x climbs towards 5, so on 0 ≤ x ≤ 4 the best allowed value is the endpoint: P(4) = -48 + 120 - 12 = 60. The unreachable P(5) = 63 is not an answer to the question that was asked.
Whole-number constraints work the same way. If x must be an integer and the vertex sits at x = 5/2, compare the two integers either side. For f(x) = -x² + 5x those are f(2) = 6 and f(3) = 6, equal — which the symmetry predicts, since 2 and 3 are the same distance from the axis x = 5/2. A vertex at x = 5/2 is not itself an answer when only whole numbers are allowed.
Finish by answering in a sentence with units. "x = 20" is not an answer to "what are the dimensions of the largest rectangle"; "20 m by 20 m, with area 400 m²" is.
Interactive
The vertex is the optimum, with no calculus
Method
The last step is where marks are lost, and it is not the algebra.
Give it a letter and a unit. Use the constraint in the problem to write everything else in terms of it — that is what the constraint is for.
Expand to ax² + bx + c. The sign of a tells you at once whether the vertex is a maximum or a minimum, so check it before computing anything.
Lengths are positive, counts are whole numbers, times are not negative. Write the interval down; it is part of the model, not an afterthought.
If the vertex is permitted, it is the answer. Otherwise compare the permitted endpoints of a closed interval or the permitted integers nearest the vertex; if the relevant endpoint is excluded, say when only a bound and no attained optimum exists. Then report the requested location or value with units.
The constraint is used first, to get from two unknowns down to one. Nothing can be optimised while there are still two.
Both numbers were asked for, so x = 6 on its own does not answer the question. Half of the work in these problems is noticing which of h and f(h) is wanted, and here it is neither: it is x and 12 - x together.
P(5) and P(7) agreeing is not luck. 5 and 7 are equidistant from the axis x = 6, so the symmetry of “Graphs of Quadratic Functions” forces equal values — which makes it a cheap check on h. If those two came out different, h would be wrong.
For a faded rectangle model, a rectangle has perimeter 54 m. Let x be one side; the supplied constraint makes the other 27 - x. Write the area function and its physical domain, locate the vertex, then answer with both dimensions and area. Compare with a 13.5 m by 13.5 m square and area 729/4 m². Reporting only x = 13.5 leaves both the requested value and its units unanswered.
Given a described situation you should produce the variable, the constraint, the quadratic and the permitted values before computing -b/(2a). The vertex takes one line; if the setup took no longer than the vertex did, something in the problem was assumed rather than read.
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