Quadratics - the Australian Mathematical Sciences Institute

A guide for teachers – Years 11 and 12 • {17}
Example
A farmer wishes to build a large rectangular enclosure using an existing wall. She has
40 m of fencing wire and wishes to maximise the area of the enclosure. What dimensions
should the enclosure have
Solution
40 – 2x
x
x
Existing wall
With all dimensions in metres, let the width of the enclosure be x. Then the three sides
to be made from the fencing wire have lengths x, x and 40−2x, as shown in the diagram.
Thus the area A is given by
A = x(40 − 2x) = 40x − 2x 2 .
This gives a quadratic relationship between the area and the width, which can be plotted.
A
200
0 10 20
x
Clearly the area will be a maximum at the vertex of the parabola. Using x = − b , we see
2a
that the x-value of the vertex is x = 10.
Hence the dimensions of the rectangle with maximum area are 10 m by 20 m, and the
maximum area of the enclosure is 200 m 2 .
Note. Differential calculus can be used to solve such problems. While this is quite acceptable,
it is easier to use the procedure outlined above. You may use calculus, however, to
solve the following exercise.