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Help me to find general solution of the following optimization problem: minimize $F(x,y)=(x^2-x_0)^2+y^2 $ subject to $y^2+x^2=Q$ where $x, y \in R$.

Thanks in advice!

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The general technique uses Lagrange multiplies. Here, you could also parametrize the circle $x^2+y^2=Q$. – Dirk Nov 7 '11 at 20:54
May be you have miss-typed the question, should it be $(x-x_0)^2+y^2$? Else (in the current form), your objective function becomes $(Q-y^2-x_0)^2+y^2$ and so... – Tapu Nov 7 '11 at 22:02
This is the right-typed question. – Max Nov 8 '11 at 3:38
up vote 1 down vote accepted

Edit: Note that the problem reduces to minimizing $(a-x_0)^2+b$ subject to $a+b=Q$ which in turn reduces to minimizing $(Q-b-x_0)^2+b$ i.e., of $x(x-p)$ subject to $x\in[0,Q]$. The point of (global) minima is given by $$x=\left\{\begin{array}{cc} 0,&\text{ if }p\le0\\\frac{p}{2},&\text{ if }0\le p\le2Q\\Q,&\text{ if }p>2Q\end{array}\right.$$

May be you have mistyped the question, should it be $(x-x_0)^2+y^2$?

Else (in the current form), the answer is quite easy (unless I missed something). Your objective function (OF) becomes $$(Q-y^2-x_0)^2+y^2$$ So writing $y^2=p$ and assuming $Q-x_0=a\ge\dfrac{1}{2}$, the OF becomes

$$(p-a)^2+p$$ $$=\left[p-(a-\dfrac{1}{2})\right]^2+a^2-(a-\dfrac{1}{2})^2$$ So the minimum value is $\quad a^2-(a-\dfrac{1}{2})^2=a-\dfrac{1}{4}$

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So... In substitution above how we would take into account that $Q-y^2>=0$? – Max Nov 8 '11 at 3:46
Good question! Please note that I have assumed $a=Q-x_0\ge\frac{1}{2}$, to confirm that $p=y^2$ can take the value $a-\frac{1}{2}$. But it looks for my answer to be a valid answer, an additional assumption that $x_0\ge-\frac{1}{2}$ is also necessary. Please let other users to try for a better solution (without these assumption). So, please don't accept this answer. – Tapu Nov 8 '11 at 4:05
@Max, does my answer somehow fits your need? Else, I will try to see using different method. I believe the answer should depend on the value of $Q$ and $x_0$ (like one I assumed). Please feel free to ask. – Tapu Nov 9 '11 at 19:48
Yep, that's right. It would be nice if you consider all cases. – Max Nov 10 '11 at 19:16
@Max, I have edited my answer to fix everything. – Tapu Nov 10 '11 at 22:45

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