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I want to minimize the function $$f(x) = \max \left\{ \frac{a}{x^2}, \frac{b}{1 - x} \right\},$$ where $a, b > 0$ are constants, and $0 < x < 1$.

Is there a way to find an $x$ which will make $f(x)$ as small as possible?

When I plot $\frac{1}{x^2}$ and $\frac{1}{1 - x}$ together, we can see that $x \approx 0.62$ minimizes the maximum of these two functions. Will the answer always be that the minimizing $x$ occurs where $\frac{a}{x^2} = \frac{b}{1 - x}$?

plotting both functions

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Should we also assume $a>0$ $b>0$? –  leonbloy Apr 17 '12 at 15:39
    
To your last question: No the minimum will not necessarily be where the two functions equal. Take for example two decreasing functions. Or take two functions that never intersect. –  Thomas Apr 17 '12 at 15:40
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@leonbloy Yes, we should assume $a, b > 0$. Sorry for not clarifying; I'll edit the question. –  jamaicanworm Apr 17 '12 at 15:40
    
@Thomas Yes, I know this will not always be the case for arbitrary functions. But for this function $f(x) = \max\{\frac{a}{x^2}, \frac{b}{1-x}\}$, is it true? –  jamaicanworm Apr 17 '12 at 15:41
    
@jamaicanworm Ah... sorry. –  Thomas Apr 17 '12 at 15:42

3 Answers 3

up vote 11 down vote accepted

For simplicity, let's write $g(x) = \frac{a}{x^2}$ and $h(x)=\frac{b}{1-x}$. Then $f=\max\lbrace g,h\rbrace$. Note that $g$ is strictly decreasing and $h$ is strictly increasing. The function $g-f$ is a continuous function such that $\lim\limits_{x\downarrow 0}=+\infty$ and $\lim\limits_{x\uparrow1}=-\infty$. Since $g-f$ is continuous, it therefore must have a zero somewhere on $(0,1)$. (Added: this zero is unique, since $g-f$ is strictly decreasing, as it is the sum of stricly decreasing functions $g$ and $-f$.)

So the two graphs will necessarily intersect in a unique point $x_0\in(0,1)$. Now, $f(x)=g(x)$ for $x\in(0,x_0]$ and $f(x)=h(x)$ for $x\in[x_0,1)$, so $f$ is strictly decreasing on $(0,x_0]$ and strictly increasing on $[x_0,1)$. This implies the only possible point where it might achieve a minimum is $x_0$ and also that it indeed achieves a minimum there. To calculate the point $x_0$, just solve the quadratic equation you get.

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In general, if you have two functions $f_1(x),f_2(x)$ continuous in $(0,1)$ respectively decreasing/increasing, AND if there exists some $x_0 \in (0,1)$ such that $f_1(x_0)=f_2(x_0)$ (it can exist at most one), then it's obviously true that the minimum of $\max(f_1(x),f_2(x))$ occurs at $x_0$. And this is the case here.

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If a and b are both positive, consider the derivative of $\frac{a}{x^2}$ and $\frac{b}{1-x}$ being $\frac{-2a}{x^3}$ and $\frac{b}{(1-x)^2}$, respectively. In the range 0 < x < 1, $\frac{-2a}{x^3} < 0$ and $\frac{b}{(1-x)^2} > 0$ throughout. So minimum can occur only at 0, 1, or at the point where these meet. Since $a/x^2$ will tend to infinity as x tends to 0 and $\frac{b}{1-x}$ will tend to infinity as x tends to 1, the minimum will come only at the point where these meet.

If a and b are both negative, a similar argument will work because the max function will tend to 0 at both ends but will be negative elsewhere.

If only one is negative, then the function will basically become the function with the positive coefficient, so the minimum will occur at x tending to 0 (for a positive) or 1 (for b positive).

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