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December 24th, 2017, 02:05 AM   #1
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implicit function in inequality

Can somebody, please, give me an example of problem that involve implicit function in inequality
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December 24th, 2017, 03:04 AM   #2
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ax + by < c with a, b, c real.
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December 24th, 2017, 03:16 AM   #3
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Nice. thanks!!!
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December 24th, 2017, 04:44 PM   #4
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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.
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December 27th, 2017, 09:56 PM   #5
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Quote:
Originally Posted by greg1313 View Post
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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