Find $f$ which will maximize $y_1$ and minimize $y_2$:
$y_1 = \dfrac{ I f}{ 1+a f} $
$y_2 = (bf^3 + cf)\dfrac{I}{y_1}$
Note : $I , a, b , c > 0 ; f > 0 $
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Generally when we have two functions and we want to maximize one and simultaneously minimize the other if we do it independently, generally we would get different $f$'s.Therefore we should assign a weight to each of them, according to their priorities. One way is to minimize (maximize) their difference, another one is minimizing one plus reciprocal of the other, etc. So generally there is no a unique answer to this question. Even saying considering trade-off is a qualitative statement and it could have lots of quantitative interpretations.
When we do a trade-off, the $f^*$ would not be global extremum of both of them anymore.
Consider we are maximizing $y_1$ and minimizing $y_2$ simultaneously where:
$$ y_1 = e^{-(x-1)^2}, \qquad y_2 = 2 - e^{-(x-2)^2} $$ which are plotted here :
Now consider these two functions:
$$ y_3 = y_1 - y_2, \qquad y_4 = y_1 + \frac1{y_2} $$ Which are plotted here :
It is easy to see that maximums of each of them occur in different places.
Also note that generally it is not possible to find an $f^*$ satisfying $y_1(f^*)<y_1(f)$ and $y_2(f^*)>y_2(f)$ simultaneously.
Mahdi Khosravi
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Thanks for your reply. Yes definitely there is a trade-off between y1 and y2 in this case. The question is what is the BEST f which maximize y1 and minimize y2 (considering trade-off) and I believe the best f is unique. – marcella Jul 26 '13 at 07:32
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Oh, You've changed the question! let me change the answer! – Mahdi Khosravi Jul 26 '13 at 07:43
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1f* is the BEST f if y1(f)>y1(f) and y2(f)<y2(f) – marcella Jul 26 '13 at 07:45
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yes y=y2+1/y1 is easier to minimize but is it possible to prove that if f* can minimize y then for any positive f y1(f)>y1(f) and y2(f)<y2(f) ? – marcella Jul 26 '13 at 07:58
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1No, That's not possible generally! That's possible only if minimum of one and maximum of the other take places for same $f$ – Mahdi Khosravi Jul 26 '13 at 08:00
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do you think dy1/df = dy2/df can give us better trade-off than minimization of y=(y2+1/y1) ? – marcella Jul 26 '13 at 08:06
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No, the solution of that is where the slope of two functions is equal!!! – Mahdi Khosravi Jul 26 '13 at 08:09
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consider y1(f1) is maximum value of y1 and y2(f2) is the minimum value of y2. Now is it possible to find f* such that [y1(f1)-y1(f)] and [y2(f)-y2(f2)] both are minimum. I call this f* best and I'm looking for this f* – marcella Jul 26 '13 at 08:20
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let us continue this discussion in chat – Mahdi Khosravi Jul 26 '13 at 08:45
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@MahdiKhosravi: I saw this well designed answer yesterday and of course yek be-ezafe. Mano tahte tasir gharar dade in jawab. Khili kaar dare ta kaamel befahmamesh Mahdi. Good job. :) – Mikasa Aug 03 '13 at 17:09
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@BabakS. Thanks. I am improving the answer hope to get better. – Mahdi Khosravi Aug 03 '13 at 17:20