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I want to find the parameters $a$ $(1/2<a<2/3)$ and $b$ $(0<b<1)$ such that the maximum of functions $f_1$, $f_2$, $f_3$, and $f_4$ becomes minimized, i.e., I am looking for $min(max(f_1, f_2, f_3, f_4))$:

$f_1(a) = 1.5+\frac{4a^2-4a+1}{6-4a}$

$f_2(b) = 1.19+\frac{b}{1+b}$

$f_3(a, b) = 1+\frac{1}{1+b}+\frac{4+9a^2-12a}{3+9a}$

$f_4(b) = \frac{5}{12}+\frac{2}{1+b}$

What's the best way to do so?

Thanks in advance.

Pooya
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  • Hint: find and write down the maximum value each function can take. You can leave it in terms of a and b. – evaristegd Jul 12 '19 at 03:10
  • How can it help? I already know that: $max(f_1) < 1.53$ (where $a=2/3$), $max(f_2) < 1.69$ (where $b=1$), $max(f_3) < 2.03$ (where $a=1/2$ and $b=0$), and $max(f_4) < 2.42$ (where $b=0)$. – Pooya Jul 12 '19 at 03:20
  • I see. That seems correct. In that case, there's only one maximum for each function $f1, f2, f4$. The only function where you can adjust parameters is $f_3$. – evaristegd Jul 12 '19 at 03:33
  • Sorry, I think I misinterpreted the question. The first answer below seems to be right. – evaristegd Jul 12 '19 at 17:20
  • @evaristegd Though there is some though behind my answer, I would actually be interested to know if there is some more general procedure to tackle this kind of question. – Fimpellizzeri Jul 12 '19 at 21:24
  • @Fimpellizieri I didn't text a thank-you comment below your precise answer and just tagged it as the correct answer because the policy of the site is no sending a thank-you comment! Thanks, anyway. However, it was also my question that if there is any general procedure to solve such questions?! – Pooya Jul 14 '19 at 04:02

1 Answers1

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Notice that $f_2$ is increasing and and $f_4$ is decreasing. At $b_0 = 92/133 \approx 0.691729$ we have $f_2(b_0) = f_4(b_0)$. Let $v = f_2(b_0)$ and observe that $v = 1439/900 \approx 1.59889$.
It follows that $\max(f_1, f_2,f_3, f_4)\geqslant \max(f_2,f_4) \geqslant v$. The best we can hope for our minimum is hence $v$.

Notice that $f_1$ is increasing. When $a$ approaches $2/3$, $f_1(a)$ approaches $1.53 < v$. It follows from our calculations above that regardless of our choice of $a$, $f_1$ will always be less than $\max(f_2,f_4)$. Hence,

$$\max(f_1,f_2,f_3,f_4) = \max(f_2,f_3,f_4).$$

Notice that $f_3$ is decreasing in $b$ and in $a$. When $a$ approaches $2/3$, $\frac{4+9a^2-12a}{3+9a}$ approaches $0$.
We have that $f_3(2/3,b_0) \approx 1.59111 < v$. It follows that indeed $\min(\max(f_1,f_2,f_3,f_4)) =v$, with the minimum being attained at $b=b_0$ and $a$ near $2/3$.

Indeed, we have

$$f_3(a,b_0) = \frac{358}{225} + \frac{4+9a^2-12a}{3+9a}.$$

Checking for $\frac{358}{225} + \frac{4+9a^2-12a}{3+9a} = \frac{1439}{900}$ when $1/2<a<2/3$, we find that any $a\geqslant a_0$ works, where

$$a_0 = \frac{1207 - \sqrt{25249}}{1800} \approx 0.582278$$

Fimpellizzeri
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