Graphs of Changing Areas

Stage: 5 Challenge Level: Challenge Level:1

Matthew from QE Boys' School sent us the following clear analysis:
 
By imagining $x$ and $y$ and the length and width of a rectangle, where $y=10/x$, the obvious connection made is to relate the rectangle's dimensions to something constant. By 'reverse engineering' the expression $y=\frac{10}{x}$ into the form $xy=10$, it is immediately obvious that the area of all described rectangles is $10$.

The graph displays symmetry through $y=\pm x$, as making the implied substitutions results in: $$\pm x=\frac{10}{\pm y}$$

This is immediately and obviously the same function, as a factor of $\pm1$ is present on both sides, and can be happily and immediately cancelled, and a multiplication by $\frac{y}{x}$ on both sides restores the original form.

The symmetry in $y=-x$ has the consequence that solutions also exist in the negative $x$ and $y$ ranges, despite not being applicable to the original problem.

As $x$ gets very large, $y$ becomes very small. This can be seen intuitively from the nature of the division function, but also becomes obvious because $xy=10$. Were $x$ to increase, but $y$ remain the same, the relationship above would no longer hold true, because $(x+\delta x)y=10+\delta xy$, and so, for all non-zero $\delta$, the new $xy$ would exceed 10.

For a similar reason and demonstrated by a similar expansion, $y$ cannot increase, and so, hence, must decrease for any and all increases in $x$.

It is also of note that $y$ must remain positive, as a positive $x$ multiplied by a negative $y$ will be negative (and hence cannot equal positive 10). As 0 is technically the smallest positive number, $y$ must decrease towards 0, but, as nothing can multiply by 0 to create 10, $y$ must asymptotically approach (and hence never actually reach) 0.

 

The class of graph $y=\frac{a}{x}$, where $a$ is any number, is a scaling of the graph $y=\frac{10}{x}$. By observing that all $y$ values are $\frac{a}{10}$ times those for corresponding $x$ values on the original graph AND that all $x$ values are the same scaling of those for corresponding $y$ values on the original graph, it is impossible that any graph of the described class can intercept $y=\frac{10}{x}$ unless $a=10$, and so the scaling is 1-to-1.

 

The line $y=\frac{P}{2}-x$, equivalent to the observation that $P=2(x+y)$, and hence identical to the definition of perimeter, describes all rectangles with perimeter $P$ when plotted. By substituting the expression for $y$ into the equation $y=10/x$, a (disguised) quadratic in $x$ is formed: $$\frac{10}{x}=\frac{P}{2}-x \Rightarrow 2x^2-Px+20=0 $$

In the normal way for quadratics, it is possible to observe that the equation only has solutions when the discriminant is non-negative: $$P^2-4 \times 2 \times 20 \geq 0 \Rightarrow P^2 \geq 160$$ Hence, the graphs only intersect (and therefore solutions only exist) for $P \geq \sqrt{160}$ for the original graph, or for $P \geq 4\sqrt{a}$ when the reciprocal graph is $y=\frac{a}{x}$.

As stated above, $P \geq \sqrt{160}$ for the graph $y=\frac{10}{x}$, and, as this describes rectangles with area 10, the smallest value for $P$ must be $\sqrt{160}=4\sqrt{10}$.

Well done to Patrick from Woodbridge School, who sent us this solution, and to Lewis from Colchester Royal Grammar School, who sent us this solution.