Applications of Parabolas
Parabolas model thrown balls, arches, cables and profits, and the vertex gives the greatest or least value. This NSW Year 9 Advanced maths subtopic solves real-life problems with quadratic models (NESA outcome MA5‑NLI‑C‑02).
You will find the maximum height of a ball or rocket, when it lands, the times it is above a given height, and a minimum overnight temperature. You will maximise profit and revenue, check whether a truck fits under an arch, find the lowest point of a bridge cable, design the largest yard from a set length of fencing, and prove that a square gives the greatest area for a fixed perimeter. It is part of Year 9 Advanced Coordinate Geometry and Graphs.
Every question with a fully worked solution.
- Applications of Parabolas - Video - Applications of Parabolas Watch
Theory
Many real quantities follow a parabola: the height of a thrown ball, a business's profit, the area of a fenced yard. The vertex gives the greatest or least value, and solving the quadratic tells you when a value is reached. This Year 9 Advanced maths page shows how to set up and solve these problems.
A quantity is quadratic when its rule has the form \(y = ax^2 + bx + c\). Its graph is a parabola, so it rises to a single highest value or falls to a single lowest value at the vertex.
If \(a < 0\), the vertex is a maximum: the greatest height, profit, revenue or area. If \(a > 0\), the vertex is a minimum: the lowest temperature, the lowest point of a cable.
The vertex is on the axis of symmetry, \(x = -\dfrac{b}{2a}\). The \(x\)-value tells you when or at what size the best value happens; substituting it gives the best value itself.
The value at \(x = 0\) is the starting value: the launch height, or the loss when nothing is made. The solutions of \(y = 0\) are where the object is on the ground or the profit breaks even.
Every answer must make sense in context. A negative time, length or number of items is rejected.
For \(y = ax^2 + bx + c\), the best value is at
If the rule factorises as \(y = k(x - p)(x - q)\), the vertex is halfway between the zeros:
Business models:
How to solve a parabola problem
- Set up the rule if it is not given: revenue \(=\) price \(\times\) number; area \(=\) length \(\times\) width with the fixed perimeter used to remove one variable.
- Read \(a\) and \(b\), and decide from the sign of \(a\) whether the vertex is a maximum or a minimum.
- Locate the vertex: \(x = -\dfrac{b}{2a}\), or the midpoint of the zeros.
- Substitute to get the greatest or least value, with units.
- Solve \(y = k\) for 'when' questions: rearrange to \(= 0\), take out a common factor, factorise. Reject answers that are impossible in context.
- Answer the question asked: a time of day, a price, the other side length, a duration (later time minus earlier time).
Shapes and clearances
- Arches and bridges: put the origin at the centre or the base, so the rule is \(y = a(x - p)(x - q)\). Use the highest point to find \(a\).
- Clearance: a centred load of width \(w\) has its corners at \(x = \pm\dfrac{w}{2}\). Compare the arch height there with the load height.
- Symmetry: points the same distance either side of the axis have the same height, which halves the work when adding several heights.
(i) the greatest height
(ii) when the stone hits the beach
(iii) for how long the stone is higher than \(75\) m.
(i) \(a = -5\), \(b = 30\). The greatest height is on the axis of symmetry.
| \(t\) | \(=\) | \(-\dfrac{b}{2a} = -\dfrac{30}{2 \times (-5)} = 3\) |
| \(h\) | \(=\) | \(35 + 90 - 45 = \mathbf{80}\text{ m}\) |
(ii) It hits the beach when \(h = 0\). Divide by \(-5\) to make it monic.
| \(35 + 30t - 5t^2\) | \(=\) | \(0\) |
| \(t^2 - 6t - 7\) | \(=\) | \(0\) |
| \((t - 7)(t + 1)\) | \(=\) | \(0\) |
| \(t\) | \(=\) | \(7 \quad (t = -1 \text{ rejected})\) |
(iii) Solve \(h = 75\):
| \(35 + 30t - 5t^2\) | \(=\) | \(75\) |
| \(t^2 - 6t + 8\) | \(=\) | \(0\) |
| \((t - 2)(t - 4)\) | \(=\) | \(0\) |
| \(4 - 2\) | \(=\) | \(\mathbf{2}\text{ seconds}\) |
The stone lands after \(7\) s. A negative time is before the throw, so it is rejected.
(i) Write the monthly revenue \(R\) in terms of \(n\).
(ii) Find the fee that gives the greatest revenue, the number of members and the revenue.
(i) Revenue \(=\) fee \(\times\) number of members.
| \(\text{fee}\) | \(=\) | \(40 + 2n\) |
| \(\text{members}\) | \(=\) | \(300 - 10n\) |
| \(R\) | \(=\) | \((40 + 2n)(300 - 10n)\) |
| \(\) | \(=\) | \(12\,000 + 200n - 20n^2\) |
(ii) \(a = -20 < 0\), so the vertex is a maximum.
| \(n\) | \(=\) | \(-\dfrac{200}{2 \times (-20)} = 5\) |
| \(\text{fee}\) | \(=\) | \(40 + 10 = \mathbf{\$50}\) |
| \(\text{members}\) | \(=\) | \(300 - 50 = \mathbf{250}\) |
| \(R\) | \(=\) | \(50 \times 250 = \mathbf{\$12\,500}\) |
Check one step either side: \(n = 4\) gives $12 480 and \(n = 6\) gives $12 480, both less.
(i) With the origin at the centre of the road, find its equation in the form \(y = a(x - 5)(x + 5)\).
(ii) Can a truck \(3\) m wide and \(4.5\) m high drive through the middle?
(iii) How wide can a \(4.5\) m high load be?
(i) The arch meets the road at \(x = \pm 5\). The top is \((0, 6)\); substitute it to find \(a\).
| \(6\) | \(=\) | \(a(0 - 5)(0 + 5)\) |
| \(6\) | \(=\) | \(-25a\) |
| \(a\) | \(=\) | \(\boldsymbol{-\dfrac{6}{25}}\) |
(ii) The truck's top corners are \(1.5\) m either side of the centre. Find the arch height there.
| \(y\) | \(=\) | \(-\dfrac{6}{25}(1.5 - 5)(1.5 + 5)\) |
| \(\) | \(=\) | \(5.46\text{ m}\) |
\(5.46 > 4.5\), so yes, it fits.
(iii) Find where the arch is exactly \(4.5\) m high.
| \(-\dfrac{6}{25}(x^2 - 25)\) | \(=\) | \(4.5\) |
| \(x^2 - 25\) | \(=\) | \(-18.75\) |
| \(x^2\) | \(=\) | \(6.25\) |
| \(x\) | \(=\) | \(\pm 2.5\) |
| \(\text{width}\) | \(=\) | \(\mathbf{5}\text{ m}\) |
(i) Show that the area is \(A = 60w - 2w^2\), where \(w\) m is the width.
(ii) Show that \(A = 450 - 2(w - 15)^2\), and hence find the greatest area and the dimensions that give it.
(i) Two widths and one length use all the fencing.
| \(2w + L\) | \(=\) | \(60\) |
| \(L\) | \(=\) | \(60 - 2w\) |
| \(A\) | \(=\) | \(w(60 - 2w) = 60w - 2w^2\) |
(ii) Expand the right-hand side:
| \(450 - 2(w - 15)^2\) | \(=\) | \(450 - 2(w^2 - 30w + 225)\) |
| \(\) | \(=\) | \(60w - 2w^2 = A\) |
A square is never negative, so \(2(w - 15)^2 \ge 0\) and \(A \le 450\). The area equals \(450\) only when \(w = 15\).
| \(w\) | \(=\) | \(15\text{ m}\) |
| \(L\) | \(=\) | \(60 - 30 = 30\text{ m}\) |
| \(A_{\max}\) | \(=\) | \(\mathbf{450}\text{ m}^2\) |
Common pitfalls
Frequently asked questions
How do you find the maximum height of a ball using a quadratic?
Find the time of the maximum with \(t = -\dfrac{b}{2a}\), or halfway between the two times when the height is zero. Then substitute that time into the height formula.
How do you find when a ball hits the ground?
Set the height equal to zero and solve the quadratic, usually by factorising. Reject any negative time; the positive answer is when it lands.
How do you know if a parabola has a maximum or a minimum?
Look at the coefficient of \(x^2\). If it is negative the parabola is concave down and the vertex is a maximum. If it is positive the vertex is a minimum.
How do you maximise revenue with a quadratic?
Write revenue as price \(\times\) number sold, with both in terms of the same variable. Expand to get a quadratic with a negative \(x^2\) term, then use the axis of symmetry to find the best price.
Which rectangle has the largest area for a fixed perimeter?
The square. Writing the area in terms of one side gives a quadratic whose vertex is where the two sides are equal. If one side is against a wall, the best yard is twice as long as it is wide.
Why do you reject a negative answer in a quadratic word problem?
Both solutions satisfy the equation, but a negative time, length or number of items cannot happen in the real situation, so only the answer that makes sense in context is kept.