In the optimization problem max: $$6x+2xy-2x^2-2y^2$$ subject to $x+2y\le2$ and $-x+y^2\le1$ I need to draw the graph of the feasible region in order to determine if the problem has global solutions, but I don't know what the graph of $-x+y^2=1$ looks like.
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I have added LaTeX to your post, please make sure the edits are correct. – Jeel Shah Jul 10 '14 at 03:09
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1The function $-x+y^2=1$ is a parabola. – Raven Jul 10 '14 at 03:13
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@Rana I think y^2=ax is a parabola; does that mean that a=(x+1)/x ? Isn't it wrong for a to be expressed in terms of x? How do I recognize a parabola? I know it may sound incredible, but yes, I already know the Kuhn-Tucker method for optimizing and yet I haven't been taught about parabolas. – Werther Jul 10 '14 at 05:09
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The region is the set of all $(x,y)$ such that $y^2-1 < x < 2-2y$. It's fairly easy to visualize if you plot $x=y^2-1$ and $x=2-2y$. It looks like so:
If you plot the function $f(x,y) = 6 x + 2 xy - 2 x^2 - 2 y^2$ over this region, you get something like so:
Now, if the critical point happens to be a saddle point, then the extreme must lie on the boundary.
Mark McClure
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I've been researching, and it seems that a function whose graph is a parabola is like this: (y-k)^2=4p(x-h) , not like y^2=x+1. How do I recognize a parabola? – Werther Jul 10 '14 at 04:42
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I should have been more specific, my question is not how to solve the optimization problem, my question is how to recognize a parabola; I didn´t tag my question as optimization, someone edited the label – Werther Jul 10 '14 at 04:51
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1@Werther, if you substitute $k=0,p=\frac{1}{4},h=-1$ in the equation of the parabola $(y-k)^2=4p(x-h)$, then it becomes $y^2=x+1$. The easisest way to recognize parabola is to have aa equation involving quadratic equation of one variable among x,y and linear function of the other. Hope it helps. – Raven Jul 10 '14 at 05:22