Applications Of Polynomial Functions
Apply polynomial functions to real problems for Queensland Year 11 Mathematical Methods (QCAA). Many practical situations, such as the volume of an open box, are modelled neatly by a cubic.
You will learn to build a volume model from a description, evaluate it, choose a sensible domain and physically reasonable zeros, and locate a maximum by reading a graph or testing values — a genuine taste of QCAA mathematical modelling.
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), applications of polynomials build a rule — usually a cubic volume model — from a real or geometric situation, then use it to answer questions. This page shows how to construct the model, evaluate it, find the physically sensible zeros and domain, and locate a maximum by reading a graph or testing whole-number values — all without calculus.
A polynomial model is an equation, such as a volume \(V(x)\), that describes a quantity in terms of one variable. For an open box made by cutting a square of side \(x\) from each corner of a sheet and folding up the sides, the volume is \(V=x(a-2x)(b-2x)\).
The zeros of the model are the inputs that make the quantity \(0\); only those that make physical sense are kept. The sensible domain is the set of inputs for which every dimension stays positive — for the box, \(0<x<\dfrac{b}{2}\) (the smaller side).
The maximum or minimum value is found in Year 11 by reading a graph over the sensible domain, or by testing values — not by differentiating.
Open box from an \(a\times b\) sheet, square of side \(x\) cut from each corner:
Sensible domain (every dimension positive, \(b\) the smaller side):
How to model and solve a polynomial problem
- Build the rule: write each dimension in terms of the variable and multiply to form the model (e.g. \(V=x(a-2x)(b-2x)\)).
- Evaluate: substitute the given value and compute, keeping units.
- Zeros and domain: find where the model is \(0\) and keep only the physically sensible interval.
- Maximum/minimum: read the peak from a graph, or test whole-number inputs across the domain and compare.
Model — base \((10-2x)\) by \((10-2x)\), height \(x\):
| \(V\) | \(=\) | \(x(10-2x)(10-2x)\) |
| \(=\) | \(x(10-2x)^2\) |
Evaluate — substitute \(x=2\):
| \(V(2)\) | \(=\) | \(2(10-4)^2\) |
| \(=\) | \(2(6)^2\) | |
| \(=\) | \(2\times36\) | |
| \(=\) | \(72\) |
Sensible domain — need \(x>0\) and \(10-2x>0\):
| \(10-2x\) | \(>\) | \(0\) |
| \(x\) | \(<\) | \(5\) |
\(V=x(10-2x)^2\); \(V(2)=72\text{ cm}^3\); sensible domain \(0<x<5\).
Zeros — set each factor to \(0\):
| \(x(12-2x)(8-2x)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(0,\ 6,\ 4\) |
Sensible domain — every dimension positive (the \(8\text{ cm}\) side runs out first):
| \(8-2x\) | \(>\) | \(0\) |
| \(x\) | \(<\) | \(4\) |
So the model only makes sense for \(0<x<4\); the zero at \(x=6\) is outside this.
Evaluate — substitute \(x=2\):
| \(V(2)\) | \(=\) | \(2(12-4)(8-4)\) |
| \(=\) | \(2(8)(4)\) | |
| \(=\) | \(64\) |
Zeros \(0,4,6\); sensible domain \(0<x<4\); \(V(2)=64\text{ cm}^3\).
Evaluate at each whole value:
| \(V(1)\) | \(=\) | \(1(10)^2=100\) |
| \(V(2)\) | \(=\) | \(2(8)^2=128\) |
| \(V(3)\) | \(=\) | \(3(6)^2=108\) |
| \(V(4)\) | \(=\) | \(4(4)^2=64\) |
| \(V(5)\) | \(=\) | \(5(2)^2=20\) |
Comparing the values, the largest volume is \(128\text{ cm}^3\), at \(x=2\).
Greatest volume \(128\text{ cm}^3\) at a cut of \(x=2\text{ cm}\).
Model — base \((18-2x)\) by \((10-2x)\), height \(x\):
| \(V\) | \(=\) | \(x(18-2x)(10-2x)\) |
Sensible domain — the \(10\text{ cm}\) side runs out first:
| \(10-2x\) | \(>\) | \(0\) |
| \(x\) | \(<\) | \(5\) |
So \(0<x<5\); test the whole values \(x=1,2,3,4\).
Evaluate at each whole value:
| \(V(1)\) | \(=\) | \(1(16)(8)=128\) |
| \(V(2)\) | \(=\) | \(2(14)(6)=168\) |
| \(V(3)\) | \(=\) | \(3(12)(4)=144\) |
| \(V(4)\) | \(=\) | \(4(10)(2)=80\) |
\(V=x(18-2x)(10-2x)\), domain \(0<x<5\); greatest volume \(168\text{ cm}^3\) at \(x=2\text{ cm}\).
Common pitfalls
Frequently asked questions
How do you build a volume model for an open box?
Cutting a square of side \(x\) from each corner of an \(a\times b\) sheet leaves a base \((a-2x)\) by \((b-2x)\) and height \(x\), so \(V=x(a-2x)(b-2x)\).
How do you find a sensible domain for a model?
Keep only the inputs that make every dimension positive. For the box that means \(0<x<\dfrac{b}{2}\), where \(b\) is the smaller side.
Which zeros of a model do you keep?
Only the physically sensible ones. A zero that makes a length negative or lies outside the sensible domain is discarded.
How do you find the maximum without calculus?
Read the peak of the graph over the sensible domain, or test whole-number inputs and compare the results.
Why does the largest zero sometimes not count?
Because the sensible domain stops earlier — once a side length would go negative, larger inputs (including a further zero) are not physically possible.
Do the answers need units?
Yes. If the sheet is measured in centimetres, the volume is in cubic centimetres; always include the units.