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?

## Comments

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TopNewestDear 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. – حكيم الفيلسوف الضائع · 2 years, 11 months ago

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– Sanjeet Raria · 2 years, 11 months ago

Hey, thanks for clearing my doubt regarding those two formulae.Log in to reply

– حكيم الفيلسوف الضائع · 2 years, 11 months ago

You're welcome! ;-)Log in to reply

– Milly Choochoo · 2 years, 11 months ago

Thank you so much Hakim! I finally got a response!Log in to reply

– حكيم الفيلسوف الضائع · 2 years, 11 months ago

You're welcome! Glad I could help! \(\overset{\cdot\cdot}{\smile}\)Log in to reply

You mean AM-GM – David Lee · 2 years, 11 months ago

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A minor detail: AGM is more commonly called

AM-GM. – Daniel Liu · 2 years, 11 months agoLog in to reply

– Milly Choochoo · 2 years, 11 months 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, \(\large \frac{\sum \limits_{i=1}^n a_n}{n} \geq \sqrt[n]{\prod \limits_{i=1}^{n} } a_n \)Log in to reply

Thanks for the multitude of informative and helpful responses! – Milly Choochoo · 3 years, 2 months ago

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– Jayakumar Krishnan · 2 years, 11 months ago

WHere are the responses? I don't see any.Log in to reply

– Milly Choochoo · 2 years, 11 months ago

There aren't. I was being sarcastic.Log in to reply