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This is a somewhat vague and philosophical question.

Consider the following three problems:

Problem 1:

Minimize over all real-valued $x,$ the function $f(x) = bx-ax^2$ where $a,b>0.$

Ans: $x^* = \frac{b}{2a}$ and $f(x^*) = b\frac{b}{2a} - a\left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a}~.$

Problem 2:

Minimize over $x> 0,$ the function $f(x) = ax+\frac{b}{x}$ where $a,b>0.$

Ans: $x^* = \sqrt{\frac ba}$ and $f(x^*) = a\sqrt{\frac ba} + \frac{b}{\sqrt{\frac ba}} = 2\sqrt{ab}~.$

Problem 3:

Minimize over $x> 0,$ the function $f(x) = \sqrt{a^2+x^2}-bx$ where $a>0, 0<b<1.$

Ans: $x^* = \frac{a}{\sqrt{1-b^2}}$ and $f(x^*) = \sqrt{a^2 + \frac{a^2}{1-b^2}} - b\frac{a}{\sqrt{1-b^2}} = \frac{a(\sqrt{2-b^2}-b)}{\sqrt{1-b^2}}~.$

In both cases, we may obtain $x^*$ using differentiation, and plug it in to find that we can collect the terms in a neat way.

This is not unusual: in many (but not all) optimization problems, we find that when we compute the function at the optimizer $x^*,$ we are able to gather the different terms together in a non-clumsy way. This is especially likely to happen if the function is a sum of only two terms.

So overwhelming is my expectation, that if I don't see "nice" things happening when evaluating $f(x^*),$ I have an instinct to re-check my calculation to make sure the $x^*$ I obtained was actually right.

Has anyone else felt the same way? Why does this happen? Thanks!

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    $\begingroup$ My guess is it's because you can solve these optimization problems using the AM-GM inequality. See jwilson.coe.uga.edu/EMT725/AMGM/SSMA.Bnghm.html $\endgroup$
    – Aeryk
    Commented Sep 4, 2015 at 17:58
  • $\begingroup$ Edited to include a third example. I think that this phenomenon is more common; it happens with rather complicated functions in many cases. $\endgroup$
    – Hedonist
    Commented Sep 4, 2015 at 20:02

1 Answer 1

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As a physicist, I always thought of this as a consequence of dimensional analysis. If $f(x)=ax^k+bx^n$ has units $[f]$ and $x$ has units $[x]$, then $[a]=[f][x]^{-k}$ and $[b]=[f][x]^{-n}$. Therefore, the minimum or maximum value of $f$, having units of $[f]$ must, by dimensional analysis, be proportional to $(a^{1/k}/b^{1/n})^{1/(1/k-1/n)}$.

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