Applications of Functions (VCE SSCE Mathematical Methods): Revision Notes

Worked Example: Taxi Fare

Problem: The cost of a taxi trip in a particular city is $1.75 up to and including 1 km. After 1 km, the passenger pays an additional 75 cents per kilometre. We need to find the function $f$ that describes this payment method and sketch the graph of $y = f(x)$.

Solution:

Let $x$ represent the length of the trip in kilometres.

For trips up to and including 1 km, the cost is simply $1.75 (a flat rate).

For trips longer than 1 km, we need to add 75 cents for each kilometre beyond the first kilometre.

The additional distance beyond 1 km is $(x - 1)$ kilometres.

The additional cost is $0.75(x - 1)$ dollars.

The total cost is the initial $1.75 plus this additional amount.

Therefore, the cost in dollars is given by the piecewise function:

The graph shows a horizontal line segment at $y = 1.75$ for $0 \leq x \leq 1$, then continues as a straight line with positive gradient for $x > 1$. Notice that when $x = 2$, the cost is $f(2) = 1.75 + 0.75(2-1) = 1.75 + 0.75 = 2.50$ dollars, as shown by the point $(2, 2.50)$.

Worked Example: Box Volume

Problem: A rectangular piece of cardboard has dimensions 18 cm by 24 cm. Four squares, each $x$ cm by $x$ cm, are cut from the corners. An open box is formed by folding up the flaps. Find a function $V$ that gives the volume of the box in terms of $x$, and state the domain of the function.

Solution:

When we cut out squares of side length $x$ from each corner and fold up the flaps, we create a box.

The length of the box becomes: $24 - 2x$ (we remove $x$ from each end)

The width of the box becomes: $18 - 2x$ (we remove $x$ from each side)

The height of the box is: $x$ (the size of the square we cut out)

The volume of a box is length × width × height, so:

Now we need to determine the domain by considering what values of $x$ make physical sense.

For the box to be formed, all dimensions must be non-negative:

• $24 - 2x \geq 0$
• $18 - 2x \geq 0$
• $x \geq 0$

From the first inequality: $x \leq 12$

From the second inequality: $x \leq 9$

From the third inequality: $x \geq 0$

The most restrictive upper bound is $x \leq 9$ (since we need all three conditions to be satisfied).

Therefore, the domain of $V$ is $[0, 9]$.

Worked Example: Rectangle Inscribed in Triangle

Problem: A rectangle is inscribed in an isosceles triangle with the dimensions shown below. Find an area function for the rectangle and state the domain.

Solution:

Let the height of the rectangle be $y$ cm and the width be $2x$ cm.

First, we need to find the height of the triangle. The base of the triangle is 18 cm, so the distance from the centre to either vertex B or C is 9 cm. The triangle is isosceles with slant heights of 15 cm.

Using Pythagoras' theorem:

Now we need to relate $x$ and $y$. Notice that triangle $AYX$ is similar to triangle $ABD$ (they share angle A and both have right angles).

For similar triangles, corresponding sides are in proportion:

Cross-multiplying:

The area of the rectangle is length × width:

Substituting our expression for $y$:

Expanding:

Factoring:

For the rectangle to be formed, we need both $x \geq 0$ and $y \geq 0$.

From $y = 12 - \frac{12x}{9} \geq 0$:

Combined with $x \geq 0$, the domain is $[0, 9]$.

Therefore, the complete function is: $A: [0, 9] \to \mathbb{R}$, $A(x) = \frac{24x}{9}(9-x)$