Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
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
add comment

1 Answer

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 $\quad a^2-(a-\dfrac{1}{2})^2=a-\dfrac{1}{4}$

share|improve this answer
1  
So... In substitution above how we would take into account that $Q-y^2>=0$? –  Max Nov 8 '11 at 3:46
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.