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 December 24th, 2017, 02:05 AM #1 Banned Camp   Joined: Dec 2017 From: Tel Aviv Posts: 87 Thanks: 3 implicit function in inequality Can somebody, please, give me an example of problem that involve implicit function in inequality
 December 24th, 2017, 03:04 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,851 Thanks: 1075 Math Focus: Elementary mathematics and beyond ax + by < c with a, b, c real. Thanks from policer
 December 24th, 2017, 03:16 AM #3 Banned Camp   Joined: Dec 2017 From: Tel Aviv Posts: 87 Thanks: 3 Nice. thanks!!! Thanks from greg1313
 December 24th, 2017, 04:44 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,851 Thanks: 1075 Math Focus: Elementary mathematics and beyond Here's one with a twist! Minimize $\frac{2x^3+1}{4y(x-y)}$ subject to $x\ge-\frac12$ and $\frac xy>1$. I'll post my solution in a few days.
December 27th, 2017, 09:56 PM   #5
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Quote:
 Originally Posted by greg1313 Here's one with a twist! Minimize $\frac{2x^3+1}{4y(x-y)}$ subject to $x\ge-\frac12$ and $\frac xy>1$. I'll post my solution in a few days.
$$\frac{2x^3+1}{4y(x-y)}=M$$

$$\frac{\partial M}{\partial y}=(2x^3+1)\frac{8y-4x}{(4y(x-y))^2}=0\implies2y=x$$

Note that $2x^3+1=0$ does not yield a solution as $2^{-1/3}\lt-\frac12$. Continuing,

$$\frac{2x^3+1}{4y(x-y)} \Leftrightarrow x=2y \Rightarrow \left(4y+\frac{1}{4y^2}\right)'=0 \Rightarrow 4-\frac{1}{2y^3}=0\Rightarrow y=\frac12,\quad x=1,\quad \min(M)=3$$

To verify,

$$\left(4y+\frac{1}{4y^2}\right)''=\frac{3}{2y^4}$$

which is always positive so $M$ is concave up and we have a minimum. Note that $M$ is not bounded from above.

I'll admit that a more rigorous analysis may be required to assert this is indeed the minimum of $M$. Can anyone provide one?

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