The following examples may look different, but the fundamental idea of each application is the same. If the product of two variables, say $x$x and $y$y remains constant, then the size of any one of the variables is inversely proportional to the size of the other.
In mathematical terms, if $xy=k$xy=k, then both $y=\frac{k}{x}$y=kx and $x=\frac{k}{y}$x=ky. If $k$k is non-zero, neither $x$x nor $y$y can disappear (become zero). However the variables can become very small or very large, but never so together.
Application 1
A rider has been allowed to agist her horses on $400m^2$400m2 of grazing land on a wealthy farmer's extensive property. The farmer (the father of the rider) is willing to build a rectangular enclosure that will contain the horses. The farmer also allows the rider to decide on the dimensions of the enclosure, but insists that the area is no more than that agreed to. What are the riders options, and if the enclosure is to be fenced at $\$35$$35 per metre, how much will it cost the farmer?
At first, it may seem that the fencing costs will not vary with the dimensions, but it turns out they do.
The basic problem here is to decide the length $l$l and width $w$w given that $l\times w=400$l×w=400. We could set up a table of values for $l$l and see how $w=\frac{400}{l}$w=400l varies. Note that $w=\frac{400}{l}$w=400l is a hyperbola centred on the origin, but because the length and width are positive numbers, only the arc in the first quadrant is relevant to our problem.
| $l$l | $5$5 | $10$10 | $20$20 | $40$40 | $80$80 |
|---|---|---|---|---|---|
| $w=\frac{400}{l}$w=400l | $80$80 | $40$40 | $20$20 | $10$10 | $5$5 |
We could also determine fencing costs as $C=\left(2l+2w\right)\times35$C=(2l+2w)×35 and see how the costs vary. The second table shows this;
| $l$l | $5$5 | $10$10 | $20$20 | $40$40 | $80$80 |
|---|---|---|---|---|---|
| $w=\frac{400}{l}$w=400l | $80$80 | $40$40 | $20$20 | $10$10 | $5$5 |
| $C$C | $5950$5950 | $3500$3500 | $2800$2800 | $3500$3500 | $5950$5950 |
It seems that a square enclosure would be the cheapest option, but there may be other considerations that the rider may need to take into account.
Application 2
A drive to Campbelltown from Sam's place is $250$250 km. How would Sam's average speed (in km/h) vary with the time taken to get to Cambelltown if she decided to take a trip tomorrow?
To answer this, we need to know that the speed s in km/h is related to the distance d kilometres and the time t hours by the equation:
$s=\frac{d}{t}$s=dt
Since the distance is given as the constant $250$250, the average speed, using function notation, can be expressed as:
$s\left(t\right)=\frac{250}{t}$s(t)=250t
Thus the time varies inversely to the speed of Sam's car, as demonstrated in this graph:
The two highlighted points have coordinates $\left(1,250\right)$(1,250) and $\left(5,50\right)$(5,50) determined as $s\left(1\right)=\frac{250}{1}=250$s(1)=2501=250 km/h and $s\left(5\right)=\frac{250}{5}=50$s(5)=2505=50 km/h.
If it were possible to travel at a constant speed of $110$110 km/h (the legal speed limit) for the entire journey, we would need to solve for t in the equation $110=\frac{250}{t}$110=250t:
| $110$110 | $=$= | $\frac{250}{t}$250t |
| $110t$110t | $=$= | $250$250 |
| $t$t | $=$= | $\frac{250}{110}$250110 |
| $t$t | $=$= | $2.273$2.273 |
This can interpreted as about $2$2 hours and $16$16 minutes. Of course in practice this would be very hard to achieve.
Worked Examples
QUESTION 1
rA group of architecture students are given the task of designing the layout of a house with a rectangular floorplan. There are no restrictions on the length and the width of the house, but the floor area must be $120$120 square metres. Each student will be allocated a rectangle with a different pair of dimensions to any other student's.
Complete the table for the various widths given:
Width in metres ($x$x) $5$5 $10$10 $15$15 $20$20 $25$25 Length in metres ($y$y) $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Form an equation for $y$y in terms of $x$x.
As the width of the house increases, what happens to the length of the house?
It increases.
AIt decreases.
BIt stays the same.
CIf the width is $24$24 metres, what will be the length of the floor area?
Graph the relationship relating the width and length of the house.
Loading Graph...Theoretically, how many students could be given unique dimensions, if the dimensions do not have to be whole number values?
$67$67
A$10$10
B$100$100
CAn infinite number.
D
QUESTION 2
A truck driver is to cover a $480$480 km journey.
If the driver travels at a constant speed of $89$89km/h, how many hours will the journey take? Round to one decimal place.
The driver's speed and therefore time taken may vary. Write down the relationship between the distance $480$480 (km), the speed $s$s (km/h) and the time $t$t (hours).
Fuel consumption also needs to be considered such that the driver must always be going slower than $98$98km/h. What is the fastest speed the driver can travel such that the speed and time taken are integer values?
fastest integer speed= $\editable{}$
QUESTION 3
A group of people are trying to decide whether to charter a yacht for a day trip to the Great Barrier Reef. The total cost of chartering a yacht is $\$1200$$1200. The cost per person if $n$n people embark on the trip is $C=\frac{1200}{n}$C=1200n
Fill in the following table of values for $C=\frac{1200}{n}$C=1200n.
Express your answer correct two decimal places where necessary.
$n$n $1$1 $2$2 $4$4 $6$6 $8$8 $10$10 $12$12 $14$14 $C$C $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Choose the correct graph of $C=\frac{1200}{n}$C=1200n
Loading Graph...ALoading Graph...BLoading Graph...CLoading Graph...DAn alternative option, the day tour, costs $\$120$$120 per person to run. Using the graph or otherwise, determine how many people will be needed to charter the yacht so that the two options cost the same for each person.