What is other method to obtain a function min/max value without any use of derivative? For example in this function:

$f(x) = 4x + \dfrac{9\pi^2}{x} + \sin(x)$

My teacher used a method that goes something like this:

$a = 4x$, $\ \ \ $ $b = \dfrac{9\pi^2}{x}$, $\ \ \ \min = 2\sqrt{a \cdot b} - \sin(x)$.

Can anyone tell me name of such method?

  • $\begingroup$ If $a \geqslant 0, b \geqslant 0$, then $$(\sqrt{a}-\sqrt{b})^2 \geqslant 0,$$ $$a+b \geqslant 2\sqrt{a \cdot b}.$$ $\endgroup$ – Oleg567 Jun 26 '13 at 21:39
  • $\begingroup$ Hey, I asked the same question. Check out this: math.stackexchange.com/questions/426569/… $\endgroup$ – EricAm Jun 27 '13 at 16:11

There is a large collection of standard Inequalities that can be used in max/min problems.

One of the more useful ones is the Arithmetic Mean Geometric Mean Inequality (AM-GM). In the case $n=2$ it says that if $a$ and $b$ are positive then $$\frac{a+b}{2}\ge \sqrt{ab},$$ with equality only if $a=b$. That is essentially the fact that you quoted. Note that the inequality can be proved simply by starting from the fact that $(\sqrt{a}-\sqrt{b})^2\ge 0$, and then doing some manipulation.

The General AM-GM says that if the $a_i$ are positive then $$\frac{a_1+a_2+\cdots+a_n}{n}\ge (a_1a_2\cdots a_n)^{1/n},$$ with equality only when the $a_i$ are all equal.

There are many other inequalities useful for proving max/min results. For example, you might want to look up the Cauchy-Schwarz Inequality. It so happens that the inequality your teacher used can be thought of as using the case $n=2$ of AM-GM or as using the case $n=2$ of C-S, though for larger $n$ they differ.

There are many other "named" inequalities. For a long list, look [here.] (http://en.wikipedia.org/wiki/List_of_inequalities)

Remark: I strongly recommend the book Maxima and Minima Without Calculus by Ivan Niven.

  • $\begingroup$ Thanks a lot i will check it out, how do i know what is A and what is B? and why is Sin(x) thrown off? $\endgroup$ – nikola-miljkovic Jun 26 '13 at 21:56
  • $\begingroup$ It doesn't matter which of the first two terms you call $a$ and which you call $b$, the inequality is symmetric. The $\sin x$ does not fit nicely into the game. Indeed for a $3$ term thing like the one you are quoting, with a transcendental function mixed in, I would expect only estimates, not sharp results. $\endgroup$ – André Nicolas Jun 26 '13 at 22:00
  • $\begingroup$ In above task they required me to get minimum value, and i couldn't solve it using derivative, however this method got a correct answer which is 12π - 1. $\endgroup$ – nikola-miljkovic Jun 26 '13 at 22:03
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    $\begingroup$ That's purely by "accident." The combination of the first two terms reaches a minimum at $x=3\pi/2$. That happens to be where $\sin x$ reaches a minimum. But if you modify the coefficients $4$ and/or $9\pi^2$ a little, and we can minimize the sum of the first two terms using the non-calculus approach, but the $\sin x$ now spoils things, and we are not able to find an "exact" answer. $\endgroup$ – André Nicolas Jun 26 '13 at 22:44
  • $\begingroup$ @AndréNicolas Would you recommend that book for me (is it too hard)? $\endgroup$ – Ovi Jun 27 '13 at 6:47

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