Graphical and transformation skills allow us to model a range of practical situations. In many cases, an accurate graph enables us to find the exact solutions to problems. At other times a sketch will allow us to visualise the situation presented. We can then use algebraic skills to find the solutions that make sense. For example, if the $x$x-axis represents a measurement such as the length of an object or the time passing in a given scenario, a negative solution cannot exist, even if the unrestricted graph associated with the event shows us these solutions.
We may also need to interpret the trends shown by a graph and relate this to the situation being modelled. The following hints may help you in making these connections:
- "initially" means that time ($t$t) is zero, which would be found at the origin
- "eventually" refers to the behaviour of the graph as time passes - this means we are being asked what is happening to the output $y$y as the input values increase
- the gradient of a curve represents the rate of change of whatever is being displayed. The steeper the gradient, the faster the rate of change, which may be either increasing or decreasing depending on its sign
- maximum and minimum points on graphs are often significant in relation to the problem being solved
Worked example
Example 1
An open chip box is made from a sheet of cardboard measuring $30$30 $cm$cm by $20$20 $cm$cm with four squares cut away from each corner as depicted in the diagram. What is the greatest volume that can be held by the box, and what are its dimensions?
Here is the constructed box when folded:
To find the volume, we must work out the dimensions in terms of $x$x. From the first diagram we can see that the length of the box will be $l=30-2x$l=30−2x and the width will be $w=20-2x$w=20−2x, while the height of the box is $x$x since whatever is cut from each corner becomes the height. This means the volume is given by $V=\left(30-2x\right)\left(20-2x\right)x$V=(30−2x)(20−2x)x or when expanded $V=4x^3-100x^2+600x$V=4x3−100x2+600x.
This is a cubic function that we can sketch: the factorised form allows us to see that there are $3$3 $x$x-intercepts at $0$0, $10$10 and $15$15 found by letting $V=0$V=0.
Of course we can't have $x$x greater than half the shortest side, so the region beyond $x=10$x=10 makes no sense in this practical situation. We are interested in finding the maximum volume in the feasible region, which the graph shows to occur when $x=4$x=4.
We can see from the scale on this graph that the maximum volume is approximately $1100$1100$\text{cm}^3$cm3. We can return to the equation to confirm the exact answer, by substituting in $x=4$x=4: $V=\left(30-2\times4\right)\left(20-2\times4\right)\times4=1056$V=(30−2×4)(20−2×4)×4=1056 $\text{cm}^3$cm3.
Finally, the original question asked us to also find the dimensions of the box. The height, $x$x, is clearly $4$4cm. The length is $l=30-2x=30-2(4)=22$l=30−2x=30−2(4)=22cm, and the width $w=20-2x=20-2(4)=12$w=20−2x=20−2(4)=12cm.
Practice questions
Question 1
The signal ratio $D$D measured in decibels of an electronic system is given by the formula $D=10\log\left(\frac{F}{I}\right)$D=10log(FI) where $F$F and $I$I are the output and input powers, measured in megawatts, MW, of the system respectively.
Find the input power $I$I in MW if the output power is equal to $10$10 MW and the signal ratio is $20$20 decibels.
The graph below shows the equation $D=10\log\left(\frac{F}{5}\right)$D=10log(F5) when the input power is $5$5 MW.
Loading Graph...Which interval contains the signal ratio when $F=9$F=9?
$\left(2,3\right)$(2,3)
A$\left(1,2\right)$(1,2)
B$\left(3,4\right)$(3,4)
C$\left(4,5\right)$(4,5)
DWhich interval contains the output power when the signal ratio is $D=1$D=1?
Loading Graph...$\left(9,12\right)$(9,12)
A$\left(0,3\right)$(0,3)
B$\left(3,6\right)$(3,6)
C$\left(6,9\right)$(6,9)
D
Question 2
Suppose you attend a fundraiser where each person in attendance is given a ball, each with a different number. The balls are numbered $1$1 through $x$x. Each person in attendance places his or her ball in an urn. After dinner, a ball is chosen at random from the urn. The probability that your ball is selected is $\frac{1}{x}$1x. Therefore, the probability that your ball is not chosen is $1-\frac{1}{x}$1−1x.
Plot the graph of $P\left(x\right)=1-\frac{1}{x}$P(x)=1−1x.
Loading Graph...As the number of people in attendance, $x$x, increases, the likelihood of your ball not being chosen:
Increases exponentially.
ADecreases toward certainty.
BStays the same.
CIncreases toward certainty.
D
Question 3
If the electricity bill is not paid by the due date, the company charges a fee for each day that it is overdue. The table shows the fees.
| Number of days after bill due ($x$x) | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 | $6$6 |
|---|---|---|---|---|---|---|
| Overdue fee in dollars ($y$y) | $4$4 | $8$8 | $16$16 | $32$32 | $64$64 | $128$128 |
At what rate is the fee increasing each day? Select all that apply.
increasing by $2$2 each day
Adoubling each day
Btripling each day
Cincreasing by a factor of $2$2 each day
DIf the bill is paid $3$3 days overdue, what overdue fee will it incur in dollars?
Which function below models the overdue fee $y$y as a function of the number of days overdue $x$x?
$y=2\left(4^x\right)$y=2(4x)
A$y=4^x$y=4x
B$y=4\left(2^{x-1}\right)$y=4(2x−1)
C$y=4\left(2^x\right)$y=4(2x)
DThe function $y=4\left(2^{x-1}\right)$y=4(2x−1) has been drawn. Why can we only use the graph for $x\ge1$x≥1?
Loading Graph...In reality, when the bill is zero days overdue, there is an overdue penalty of $\$\quad$$ $\editable{}$. According to the model, if the bill is zero days overdue, the penalty is $\$\quad$$ $\editable{}$. Therefore we only consider the graph for $x\ge1$x≥1.
A $\$128$$128 bill is overdue. How many days must it remain overdue for the overdue penalty to equal the cost of the bill itself?