# Help with 'max' and 'min'

Every time a problem comes up that involves these two functions

$$max(x,y)$$ or $$min(x,y)$$

I have no idea what to do (I only started seeing that recently). I am not talking about finding local minima or maxima (or saddle points) of a function using calculus. I usually see that in plain-old arithmetic problems actually.

Also, AM-GM (Arithmetic Mean - Geometric Mean) is still a bit new to me.

Could anybody link me to some problems (that aren't too difficult) on these two topics?

Note by Milly Choochoo
4 years, 3 months ago

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Dear Milly;

What is $$\operatorname{max}\{a,b\}$$? It's just a function that gives us the biggest number from $$a,b$$. The same thing for $$\operatorname{min}\{a,b\}$$, except that this time, it gives us the smallest number from $$a,b$$. Let's begin with some few examples to wrap our heads around those two functions. From the definitions we got $$\operatorname{max}\{-3,8\}$$ is equal to $$8$$. Since out of $$-3$$ and $$8$$ it's the latter that is the biggest. And for $$\operatorname{min}\{-7,1\}$$ we can see that it is equal to $$-7$$ since this last one is the smallest number out of $$-7$$ and $$1$$.

Now that we have understood those functions, we need to determine a way that will reveal to us a formula to find the min and max of any two numbers. Let's take two arbitrary numbers as an example like $$x,y$$ such that $$\displaystyle\underline{x\lt y}$$, which is a very important assumption for what will come. If we graph those into the real number line then we will see that $$x$$ is in the left in regards of $$y$$ as shown in this image:

http://i.stack.imgur.com/OpjU2.png

One of the numbers we can define based on $$x$$ and $$y$$ is the their average, $$(x+y)/2$$. What this average graphically means, is that it represent the number whose corresponding point on the number line is a midpoint of the segment $$[x,y]$$, as this image show:

http://i.stack.imgur.com/x10g3.png

We've made some neat progress till now, but we'll get to the next point only by seeing that this midpoint unlocks to us a new possibility. This midpoint makes an axis of symmetry, and so with the rightly chosen way we can go from the quantity $$(x+y)/2$$ to the smallest number, that is $$x$$, or to the biggest number $$y$$. To see what I mean look carefully at this image:

http://i.stack.imgur.com/ozkyR.png

So by adding the distance from $$(x+y)/2$$ to $$y$$ to $$(x+y)/2$$ itself we will obtain $$y$$. And if we substract the distance from $$(x+y)/2$$ to $$x$$ to $$(x+y)/2$$ itself we will obtain $$x$$. So we actually found a way to get $$x$$ and $$y$$ out of some formula that will soon discover. Note however that those two distances are the same. And how is distance represented? I mean how to find the distance from $$a$$ to $$b$$? We use absolute values! To get: $$\text{distance from }a\text{ to}b=|a-b|.$$ So if we use that property we will get that:

\eqalign{ \operatorname{max}\{x,y\}&=\dfrac{x+y}{2}+\left|\dfrac{x+y}{2}-y\right|\\ &=\dfrac{x+y}{2}+\left|\dfrac{x+y-2y}{2}\right| \\ &=\dfrac{x+y}{2}+\left|\dfrac{x-y}{2}\right| \\ &=\dfrac{x+y}{2}+\left|\dfrac{1}2(x-y)\right| \\ &=\dfrac{x+y}{2}+\dfrac12\left|(x-y)\right| \quad\text{using properties of the }|\,\cdot\,| \\ &=\dfrac{x+y}{2}+\dfrac{\left|(x-y)\right|}2 \\ &=\dfrac{x+y+\left|x-y\right|}2\quad\blacksquare \\ }

Now I'm sure you can do the same yourself to prove that:

$\operatorname{min}\{x,y\}=\dfrac{x+y-\left|x-y\right|}2.$

So now whenever you see a problem involving $$\operatorname{min}\{x,y\}$$ or $$\operatorname{max}\{x,y\}$$ just replace it with the formulas we found above, and the rest would be easy.

## Exercises:

1) Prove that $$\operatorname{min}\{x,x\}=\operatorname{max}\{x,x\}=x.$$

2) Derive a formula for $$\operatorname{min}\{x,y,z\}$$ and $$\operatorname{max}\{x,y,z\}.$$

I hope this helps. Best wishes, $$\cal H$$akim.

Hey, thanks for clearing my doubt regarding those two formulae.

- 4 years, 1 month ago

You're welcome! ;-)

Thank you so much Hakim! I finally got a response!

- 4 years, 1 month ago

You're welcome! Glad I could help! $$\overset{\cdot\cdot}{\smile}$$

You mean AM-GM

- 4 years, 1 month ago

A minor detail: AGM is more commonly called AM-GM.

- 4 years, 1 month ago

Oh yeah, woops. I don't know why I just put 'Arithmetic/Geometric Mean'. I meant to say that I wasn't too comfortable with problems involving the AM-GM inequality, $$\large \frac{\sum \limits_{i=1}^n a_n}{n} \geq \sqrt[n]{\prod \limits_{i=1}^{n} } a_n$$

- 4 years, 1 month ago

Thanks for the multitude of informative and helpful responses!

- 4 years, 3 months ago

WHere are the responses? I don't see any.

- 4 years, 1 month ago

There aren't. I was being sarcastic.

- 4 years, 1 month ago