Pre-calc optimization of a rational function A student I am tutoring is in a pre-calc class and they just had a test on rational functions. One of the questions my student said they got was to minimize the area of the outer rectangle, given that the area of the inner rectangle is a fixed 48, and the distance to the outer rectangle on either side is 1.5 horizontally and 1 vertically. 

If we call the horizontal side length of the inner rectangle x, and the vertical y, then we can get the area for the outer rectangle (SB, surface area big) as
SB = 144/y + 2y + 56
Without needing much convincing, I explained (since 56 is const., and factor of 2 is constant as well) we can reduce it to optimizing 72/y+y. And then I am stuck.
After thinking about it for a bit, I tried plotting f(y)=72/y and f(y)=y together with f(y)=72/y+y, and the minimum appears at 72/y=y, which incidentally is the answer we find using calculus. But I got no idea if this is coincidental or if this is indeed a key to answering this question.
Anyways, the problem is to minimize the area of the outer rectangle without the use of calculus. Any hints / ideas?
 A: Explanation for the surface area:
\begin{align}
\text{Surface area of outer rectangle} &= (x+2)(y+3) \\
&= \left(\frac{48}{y} + 2 \right)(y+3) \\
&= 48 + 2y + \frac{144}{y} + 6 \\
&= \color{red}{54} + 2y + \frac{144}{y}
\end{align}
Though OP wrote $56$ instead, it didn't affect the minimiser (the value of $y$ which minimises the surface area of the outer rectangle)

Idea of why optimising $144/y + 2y + 56$ is equivalent to optimising $72/y + y$:
\begin{align}
\frac{144}{y} + 2y + 56 &= 2 \Bigg(\color{blue}{\frac{72}{y} + y} - \underbrace{2 \sqrt{\frac{72}{y} \cdot y}\Bigg) + 56 + 2\sqrt{72}}_{\mbox{constant}} \\&= 2\underbrace{\left( \frac{6\sqrt2}{\sqrt{y}} - \sqrt{y} \right)^2}_{\mbox{minimized when } \frac{6\sqrt2}{\sqrt{y}}
= \sqrt{y}} + 56 + 24 \sqrt2
\end{align}
The square term on the right is minimised when $\dfrac{6\sqrt2}{\sqrt{y}} = \sqrt{y}$, i.e. $y = 6\sqrt2$.

Shorter solution using a.m.-g.m.-inequality
$$54 + 2y + \frac{144}{y} \ge 54 + 2 \sqrt{2y \cdot \frac{144}{y}} = 54 + 24 \sqrt2$$
and equality holds iff
$$2y = \frac{144}{y} \iff y = \sqrt{72} = 6\sqrt2$$
or even better
\begin{align}
A &= (x+2)(y+3) = xy + (3x + 2y) + 6 = (3x + 2y) + 54 \\
&\ge 2 \sqrt{6xy} + 54 = 2\sqrt{6 \cdot 48} + 54 = 24 \sqrt{12} + 54,
\end{align}
and equality holds iff $3x = 2y$.  Solving with the constraint $xy = 48$ gives $y = \sqrt{72}$.
A: I found a relatively simple solution as well.

If you look at the graph of 72/y+y, you will find that the minimum and maximum occur when a horizontal line y=c intersects with the graph at 1 point (is tangent). If we set the graph = c, we get:
\begin{align}
 \frac{72}{y}+y=c \implies\ y^2 - cy+72 =0
\end{align}
thus:
\begin{align}
y=\frac{c\pm \sqrt{(-c)^2-4(72)}}{2}
\end{align}
Since we want the intersection to occur at one point, this implies that 
\begin{align}
c = \pm 2\sqrt{72}
\end{align}
(ie. 0 discriminant), which implies that
\begin{align}
y=\pm \sqrt{72} = \pm 6 \sqrt{2}
\end{align}
Ruling out the negative answer as not a length, we are left with 
\begin{align}
y= 6 \sqrt{2}
\end{align}
and may be use test points to demonstrate it to be a minimum (since no calculus)
