# 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?

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

$$54 + 2y + \frac{144}{y} \ge 54 + 2 \sqrt{2y \cdot \frac{144}{y}} = 54 + 24 \sqrt2$$

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 $y = \sqrt{72}$.

• I think the OP's equation is wrong - it should be $A=\frac{144}{y} + 2y +6$. It doesn't affect your equation much but perhaps you should change it. – Landuros Apr 7 '18 at 8:30
• @Landuros Thanks for comment. I'm editing my answer. I interpret part of OP's question as "why is optimising SB = ... the same as optimising $\frac{72}{y} + y$", so my hint addresses this. – GNUSupporter 8964民主女神 地下教會 Apr 7 '18 at 8:35
• @Landuros No, it's definitely 54. Try it yourself. (There is going to be an xy + 6 when you expand the area) – Yaroslav Apr 7 '18 at 8:40
• @Yaroslav $A=(x+2)(y+3)-xy=xy+3x+2y+6-xy=3x+2y+6$. Then one can substitute $x=\frac{48}{y}$ and get $A=\frac{144}{y}+2y+6$. But GNU is right, it doesn't affect the final answer. – Landuros Apr 7 '18 at 8:44
• @GNUSupporter, yeah, I am still not seeing why it has to be minimized at sqrt(72/y)=sqrt(y). It seems to just go back to that observation I saw. – Yaroslav Apr 7 '18 at 8:45

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)

• +1 for your effort to answer your own question, but compare it with the three-line solution using AM-GM inequality, you'll find that it's the simplest solution. To find the minimal area, you need to substitute $y =6\sqrt2$ back into \begin{align} y=\frac{c\pm\sqrt{(-c)^2-4(72)}}{2} \end{align}. This is prone to error. \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 $3x=2y$. That gives $y=6\sqrt2$ with the minimal area $54+24\sqrt2$ at the same time. – GNUSupporter 8964民主女神 地下教會 Apr 7 '18 at 13:26
• That's nice too. I'm just worried that it may confuse my student when I show her this solution - unlike me, who has a physics degree under my belt, if I start going on about AM-GM inequality, I'll have to explain to her what it is by probably deriving it. I find presenting proofs to high school students confuses them more, so I just show it in small doses, and only at the end of a unit we cover. Anyways, I'll have plenty of solutions to show during my next lesson! I love when a question has a lot of different approaches to solve it! – Yaroslav Apr 7 '18 at 18:18