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December 24th, 2017, 03:05 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, 04:04 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond 
ax + by < c with a, b, c real.

December 24th, 2017, 04:16 AM  #3 
Banned Camp Joined: Dec 2017 From: Tel Aviv Posts: 87 Thanks: 3 
Nice. thanks!!!

December 24th, 2017, 05:44 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond 
Here's one with a twist! Minimize $\frac{2x^3+1}{4y(xy)}$ subject to $x\ge\frac12$ and $\frac xy>1$. I'll post my solution in a few days. 
December 27th, 2017, 10:56 PM  #5  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond  Quote:
$$\frac{\partial M}{\partial y}=(2x^3+1)\frac{8y4x}{(4y(xy))^2}=0\implies2y=x$$ Note that $2x^3+1=0$ does not yield a solution as $2^{1/3}\lt\frac12$. Continuing, $$\frac{2x^3+1}{4y(xy)} \Leftrightarrow x=2y \Rightarrow \left(4y+\frac{1}{4y^2}\right)'=0 \Rightarrow 4\frac{1}{2y^3}=0\Rightarrow y=\frac12,\quad x=1,\quad \min(M)=3$$ 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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