Help with 'max' and 'min'

Every time a problem comes up that involves these two functions

max(x,y)max(x,y) or min(x,y) 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
7 years, 2 months ago

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Dear Milly;

What is max{a,b}\operatorname{max}\{a,b\}? It's just a function that gives us the biggest number from a,ba,b. The same thing for min{a,b}\operatorname{min}\{a,b\}, except that this time, it gives us the smallest number from a,ba,b. Let's begin with some few examples to wrap our heads around those two functions. From the definitions we got max{3,8}\operatorname{max}\{-3,8\} is equal to 88. Since out of 3-3 and 88 it's the latter that is the biggest. And for min{7,1}\operatorname{min}\{-7,1\} we can see that it is equal to 7-7 since this last one is the smallest number out of 7-7 and 11.

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,yx,y such that x<y\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 xx is in the left in regards of yy as shown in this image:

One of the numbers we can define based on xx and yy is the their average, (x+y)/2(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][x,y], as this image show:

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(x+y)/2 to the smallest number, that is xx, or to the biggest number yy. To see what I mean look carefully at this image:

So by adding the distance from (x+y)/2(x+y)/2 to yy to (x+y)/2(x+y)/2 itself we will obtain yy. And if we substract the distance from (x+y)/2(x+y)/2 to xx to (x+y)/2(x+y)/2 itself we will obtain xx. So we actually found a way to get xx and yy 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 aa to bb? We use absolute values! To get: distance from a tob=ab.\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:


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


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

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

I hope this helps. Best wishes, \calH\cal Hakim.

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

Milly Choochoo - 7 years ago

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

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

Sanjeet Raria - 7 years ago

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You're welcome! ;-)

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Thanks for the multitude of informative and helpful responses!

Milly Choochoo - 7 years, 2 months ago

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

Jayakumar Krishnan - 7 years ago

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There aren't. I was being sarcastic.

Milly Choochoo - 7 years ago

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

Daniel Liu - 7 years ago

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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, i=1nanni=1nnan\large \frac{\sum \limits_{i=1}^n a_n}{n} \geq \sqrt[n]{\prod \limits_{i=1}^{n} } a_n

Milly Choochoo - 7 years ago

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You mean AM-GM

David Lee - 7 years ago

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