What is optimisation?
Optimisation means finding the best answer. Best can mean the maximum (largest) or the minimum (smallest).
- Largest area for a fixed fence.
- Largest profit for a shop.
- Smallest cost or smallest time.
When the quantity is a quadratic function, its graph is a parabola. A parabola has one turning point, the vertex. The vertex is the best value.
Modelling a situation with a quadratic function
To model means to write the problem as a formula.
- Choose a letter (say x) for the thing you can change.
- Write the other lengths or amounts using x.
- Write the quantity you want to make best as a function of x.
- Write the allowed values of x (the domain). A length cannot be negative.
Example: fence 20 m, width x, so length = 10 − x and area A(x) = x(10 − x) = −x² + 10x, with 0 < x < 10.
Finding the best value from the vertex
For f(x) = ax² + bx + c:
- The vertex is at x = −b / (2a).
- Put this x back into f to get the best value.
- If a < 0 the parabola opens down, so the vertex is a maximum.
- If a > 0 it opens up, so the vertex is a minimum.
You can also complete the square: f(x) = a(x − p)² + q. The vertex is (p, q).
When the vertex is outside the domain
If the vertex x is not allowed, the best value is at an end of the domain. Always check the ends.
Optimisation in economics: maximum profit
Businesses use optimisation every day.
- Revenue R = price × number sold.
- Profit P = revenue − cost.
When price goes up, fewer items sell. So revenue often becomes a quadratic with a < 0. The vertex gives the price or amount that earns the maximum profit.
In higher classes you will find the same point with calculus: the slope (derivative) is zero at the top.
Try it: the string rectangle
Take a 40 cm piece of string and tie the ends. Make rectangles on squared paper with widths 4, 8, 10, 12 and 16 cm. Count the squares inside each. Which one holds the most? Now predict with the vertex formula, then check with the slider in the 3D.
Key formulas and definitions
- Vertex: x = −b / (2a)
- Best value: f(−b/2a)
- Vertex form: f(x) = a(x − p)² + q, vertex (p, q)
- a < 0 → maximum, a > 0 → minimum
- Profit = Revenue − Cost
- Revenue = price × quantity
Worked examples
1. Find the maximum of f(x) = −x² + 6x + 1.
a = −1, b = 6. x = −6/(2 × −1) = 3. f(3) = −9 + 18 + 1 = 10. Since a < 0, the maximum is 10 at x = 3.
2. Find the minimum of g(x) = 2x² − 8x + 5.
x = 8/(2 × 2) = 2. g(2) = 8 − 16 + 5 = −3. Since a > 0, the minimum is −3 at x = 2.
3. 20 m of fence makes a rectangle. Find the largest area.
Width x, length 10 − x. A = −x² + 10x. Vertex x = 5. A = 5 × 5 = 25 m². The best shape is a 5 m × 5 m square.
4. A wall forms one side of a rectangular pen. 40 m of fence makes the other three sides. Find the largest area.
Let the two sides touching the wall be x. The side opposite the wall is 40 − 2x. A = x(40 − 2x) = −2x² + 40x. x = −40/(−4) = 10. A = 10 × 20 = 200 m².
5. Profit P(q) = −q² + 20q − 36 (thousand rupees) where q is hundreds of items. Find the best q and the maximum profit.
q = −20/(−2) = 10, so 1000 items. P(10) = −100 + 200 − 36 = 64 thousand rupees.
6. Tickets cost €10 and 300 people come. Each €1 rise loses 20 people. Find the price for maximum revenue.
Let n = number of €1 rises. Price = 10 + n, people = 300 − 20n. R = (10 + n)(300 − 20n) = −20n² + 100n + 3000. n = −100/(−40) = 2.5. Price = €12.50, people = 250, R = €3125.
Common mistakes
- Using x = b/(2a) and forgetting the minus sign.
- Giving the x-value as the answer when the question asks for the maximum value f(x).
- Calling every vertex a maximum: check the sign of a.
- Forgetting the domain: lengths and amounts cannot be negative, and the vertex may lie outside the allowed values.