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 asymptote towards (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:
$$\frac{10}{x}=\frac{P}{2}-x \Leftarrow 2x^2-Px+20=0 $$
In the normal way for quadratics, it is possible to observe that the equation only has solutions when the descriminant is non-negative:
$$P^2-4 \times 2 \times 20 \geq 0 \Leftarrow P^2 \geq 160$$
Hence, the graphs only intersect (and therefore solutions only exist) for $P \geq \sqrt{160}$ for the original graph, or for $P \geq 2 \sqrt{2a}$ 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 $\sqrt{160}=4\sqrt{10}$.
The logical extension of this (into 3D) would result predictably in the equation $z=\frac{V}{xy}$, where $V$ is the volume of choice. However, as there now exist the measure of surface area and length of edge (ie. perimeter), the number of equations can still be retained equal to the number of variables.
As the length of the edges of a cuboid, $P$, is now the sum of quadruple the size of the cube in each dimension, $P=4x+4y+4z$. In a completely new style of equation, however, the total surface area $A$ can be shown to be $A=2xy+2xz+2yz$, as double the sum of the area of each face.
By re-writing the equations above as $xy=\frac{A}{2}-z(x+y)$ and $x+y=\frac{P}{4}-z$, the depth, $z$ of a cuboid with volume $V$, surface area $A$ and edge-length $P$ is:
$$z=\frac{V}{\frac{A}{2}-z(x+y)}$$
$$z^2(x+y)+z \frac{A}{2}=V$$
$$z^2(\frac{P}{4}-z)+z \frac{A}{2}=VV$$
$$4z^3-Pz^2-2Az+V=0$$
The exact same equation, excepting the presence of $z$, results from considering $x$ and $y$ with the conditions above. Hence, it is evident that each of the 3 solutions corresponds to one of $x$, $y$, and $z$. It shouldn't matter which is which, as the ordering has no consequence upon edge length, surface area or volume.