# Finding Parameters to Minimize a Maximum of Multiple Functions

I want to find the parameters $$a$$ $$(1/2 and $$b$$ $$(0 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?

• Hint: find and write down the maximum value each function can take. You can leave it in terms of a and b. Commented Jul 12, 2019 at 3:10
• 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 $b=0)$. Commented Jul 12, 2019 at 3:20
• 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 $f_3$. Commented Jul 12, 2019 at 3:33
• Sorry, I think I misinterpreted the question. The first answer below seems to be right. Commented Jul 12, 2019 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. Commented Jul 12, 2019 at 21:24

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 $$1/2, we find that any $$a\geqslant a_0$$ works, where

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