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Studying differential equations I came cross through this:

Let $ \displaystyle{ r = \min \left\{ a , \frac{b}{ a+ a^2 b^3} \right\} } $, where $ a,b >0$. Find $ r_{ \max} $.

Here is what I did:

Fix $ a> 0$ and define $ \displaystyle{ g(b) = \frac{b}{ a+ a^2 b^3} ,\quad b>0 }$. Then we have that $ \displaystyle{ g'(b) = \frac{ a(1-2ab^3)}{ (a+ a^2 b^3)^2} }$.

Thus, $g$ has maximum at $ \displaystyle{b_0= \frac{1}{ \sqrt[3]{2a} }}$ which gives $ \displaystyle{ g(b) \leq g\left( \frac{1}{ \sqrt[3]{2a}} \right) = \frac{2}{ 3a \sqrt[3]{2a} }}$

So, $ \displaystyle{ r = \min \{ a , \frac{2}{ 3a \sqrt[3]{2a} } \}}$. I think that to find $a$ which gives $ r_{\max}$ I have to solve the equation $ \displaystyle{ a=\frac{2}{ 3a \sqrt[3]{2a}} }$.

Is this correct and why....?

Any ideas?

Thank's in advance!

edit: I edit the title.

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up vote 4 down vote accepted

The maximum is when $$a=(4/27)^{1/7}\;\; \text{or}\;\; a=.76125127398793390138765377686933423...$$

$a$ is increasing as a increases. The other formula decreases as $a$ increases

$$a = 2^{\frac{2}7}/3^{\frac{3}7}$$ is when both sides are equal. Any change in a in either direction will cause one of them (and therefore the minimum) to decrease.

Your work up to that point was correct.

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Thank you very much for your time! – passenger Jan 11 '13 at 19:55
no problem, thank you for accepting the answer – kaine Jan 11 '13 at 19:56
Thank you Babak – kaine Jan 11 '13 at 21:05

It looks correct to me. The reason that is correct is because you're finding when the two values are equal. If they aren't equal, then since the real numbers are totally ordered, we know that one will be less than the other.

Solving for $a$, we have $$a^{7/3}=\frac{2^{2/3}}{3}.$$ Taking the $3/7$ root of each side yields $$a=\left(\frac{2^{2/3}}{3}\right)^{3/7}=\left(\frac{4}{27}\right)^{1/7}.$$

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Thank you very much for your time! – passenger Jan 11 '13 at 19:52

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