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Steve's Mapping

Steve has created two mappings. Can you figure out what they do? What questions do they prompt you to ask?

Graphs of Changing Areas

Stage: 5 Challenge Level: Challenge Level:1
We received some excellent solutions to this problem that considered the relationship between the rectangles and the graph. Amrit and Guruvignesh, both from Hymers College, and Sergio from Kings College Alicante correctly identified that the rectangles all share an area of 10 units. Here's what Guruvignesh said:

The graph shows the relationship between $x$ and $y$ when the area is $10$ square units. Because $y=\frac{10}x, xy=10$. Therefore, the rectangles all have the same area. 

The graph has reflectional symmetry in the line $y=x$. 
As $x$ becomes very large, the graph will elongate and the $y$ coordinate will near the $x$ axis but will not touch the axis, because the length of a side can never be equal to $0$. This is because as one side increases the other side decreases to maintain a constant area, so in this case an area of $10$ square units.  

The graph will never intersect with the graphs $y=\frac{20}x$ and $y=\frac5x$ because the areas of the rectangles have halved or doubled.

Pablo from Kings College Alicante said a little more about the asymptotic behaviour of the graph:

As $x$ gets larger, $y$ tends to $0$
if $x = 10, y = 1$
if $x = 100, y = 0.1$
if $x = 1000, y = 0.01$
etc.

Agathiyan, also from Hymers College, explained why the line $y=\frac{1}{2} P-x$ does not always intersect with the curve $y=\frac{10}x$:

The graph would not intersect for all values of $P$ since the line has a gradient of -1, thus it slopes downwards, and since there is space underneath the curve in the positive $x$ and $y$ quadrant, there are values of the $y$ intercept ($\frac{P}{2}$) which are positive and also allow the line to pass underneath the curve, thus not all values of $P$ intersect. One example of a value that doesn’t intersect is if $P=5$.

Pablo showed that a square has the minimum perimeter:

$P = 2x + 2y$
$P = 2x + \frac{20}x$
$P = \frac{2x^2 + 20}{x}$

Differentiating using the quotient rule,
$P' = \frac{(x)(4x) - (2x^2+20)(1)}{x^2}$
$P' = \frac{2x^2-20}{x^2}$

Stationary point when P' = 0
$0 =\frac{2x^2-20}{x^2}$ 
$0 = 2x^2 - 20$
$20 = 2x^2$
$x^2 = 10$
$x = \sqrt{10}$

$y = \frac{10}{\sqrt{10}} = \sqrt{10}$

In other words, a square has the minimum perimeter of $4\sqrt{10}$. There is no maximum
perimeter because you can make rectangles with sides ${1 000 000, 0.000001}$
which would give a perimeter of over $2 000 000$.


Tim from Gosforth Academy sent us a very clear solution, which you can read as a pdf: Tim's solution.