Learn | Grasp Maths

Lesson overview

Algebra — modelling and optimisation

Many real-world situations can be modelled using quadratic functions. For example, if a rectangular garden has a fixed perimeter, its area is maximised when it is closest to a square. A quadratic function $f(x) = ax^2 + bx + c$ has a vertex at $x = -\frac{b}{2a}$ , which is the maximum (if $a < 0$ ) or minimum (if $a > 0$ ). The maximum or minimum value is found by substituting this x-value back into the function. Graphs and completing the square both reveal the vertex and help us solve optimisation problems in context.

Setting up a quadratic model from a real situation

A box with base dimensions x by x and height h has a total surface area of 120 ${cm}^{2}$ . Express the volume V in terms of x only.

Use the constraint (surface area fixed) to express one variable in terms of the other, then write the objective (volume) in one variable.

Finding the vertex of a quadratic from context

A rectangular pen is made using 40 m of fencing. The area A depends on the length x. Find the length that maximises area.

Write the objective as a quadratic. The x-coordinate of the vertex is $-\frac{b}{2a}$ . Substitute to find the maximum/minimum value.

Using completing the square to find the vertex

Find the maximum value of $f(x) = -2x^2 + 8x + 5$ .

Complete the square. The vertex form $a(x - h)^2 + k$ shows vertex at $(h, k)$ .

Worked example — optimising profit

A company sells x units of a product. Profit P (in pounds) is modelled by $P(x) = -x^2 + 100x - 1500$ . Find the number of units that maximises profit and the maximum profit.

Identify the quadratic function: $P(x) = -x^2 + 100x - 1500$ .
Since $a = -1 < 0$ , the parabola opens downward, so the vertex is a maximum.
Find the x-coordinate of the vertex using $x = -\frac{b}{2a} = -\frac{100}{2(-1)} = 50$ .
Substitute $x = 50$ into $P(x)$ : $P(50) = -(50)^2 + 100(50) - 1500$ .
Calculate: $P(50) = -2500 + 5000 - 1500 = 1000$ .
The maximum profit is £1000 when 50 units are sold.

Try it

Start by setting up the quadratic model from the context. Identify whether you are maximising or minimising, then find the vertex.

Question 1

A rectangle has a perimeter of 20 cm. Express its area A in terms of length x.

💡 If perimeter is 20 and one side is x, then the other side is $(20 - 2x)/2 = 10 - x$ .

Question 2

The function $f(x) = x^2 - 6x + 5$ has a vertex at x =

💡 Use $x = -\frac{b}{2a} = -\frac{-6}{2(1)} = 3$ .

Question 3

A rectangular pen uses 60 m of fencing. What length and width maximise area?

💡 Area $A = x(30 - x)$ . Vertex at $x = 15$ , so both dimensions are 15 m.

Question 4

Find the maximum value of $f(x) = -x^2 + 4x + 3$ .

💡 Vertex at $x = 2$ . Then $f(2) = -(2)^2 + 4(2) + 3 = -4 + 8 + 3 = 7$ .

Question 5

Profit $P = -2x^2 + 40x + 100$ . How many units maximise profit?

💡 Vertex at $x = -\frac{40}{2(-2)} = 10$ .

Common mistakes

Watch for these when working through the lesson.

Using the wrong formula for the vertex. Remember: $x = -\frac{b}{2a}$ , not $b/2a$ .
Finding the x-coordinate of the vertex but forgetting to substitute it back to find the maximum/minimum value.
Not setting up the constraint correctly. Read the problem carefully to see what is fixed (perimeter, fencing, surface area, etc.).
Confusing whether a parabola opens upward (minimum) or downward (maximum). Check the sign of a.

Optimisation and quadratic modelling

Lesson overview

Worked example — optimising profit

Try it

Common mistakes

Related topics

Independent practice will plug in here