Solve applications involving hyperbolas

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

QUESTION 2

QUESTION 3