 Cuchulainn
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### Re: Extremes

cuch, your pic shows up as a sign that says "no hotlinking"
I used the bounding box as a rough approximation to the real constraints. Hopefully the minimum does not land up in the empty quarter. http://www.datasimfinancial.com
http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself
Jean Piaget Collector
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### Re: Extremes

Warning: Too long on this thread and you guys will end in a love triangle! Cuchulainn
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### Re: Extremes More like viscous cycles if you ask me!
http://www.datasimfinancial.com
http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself
Jean Piaget Cuchulainn
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### Re: Extremes

minimise $x + y$ subject to $x^2 + y^2 - 2 = 0$
http://www.datasimfinancial.com
http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself
Jean Piaget ppauper
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### Re: Extremes

minimise $x + y$ subject to $x^2 + y^2 - 2 = 0$
$x^2 + y^2 - 2 = 0$ is circle radius 2
$x=\sqrt{2}\cos\theta$ and $y=\sqrt{2}\sin\theta$
$x+y=\sqrt{2}(\cos\theta+\sin\theta)=2\sin(\theta+\pi/4)$ Cuchulainn
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Posts: 61477
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### Re: Extremes

minimise $x + y$ subject to $x^2 + y^2 - 2 = 0$
$x^2 + y^2 - 2 = 0$ is circle radius 2
$x=\sqrt{2}\cos\theta$ and $y=\sqrt{2}\sin\theta$
$x+y=\sqrt{2}(\cos\theta+\sin\theta)=2\sin(\theta+\pi/4)$
I agree. Nice approach. At which point does it reach a minimum based on this analysis?
Another approach is to use Lagrange multipliers $L(x,y,\lambda) = x + y - \lambda(x^2 + y^2 -2)$
Taking the gradient results in 4 solutions $x^2 = 1, y^2= 1, \lambda = +1/2, \lambda = -1/2$, one of which (-1.-1) is the minimum.

Can your approach be applied to the additional inequality constraint $y \geq 0$?
http://www.datasimfinancial.com
http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself
Jean Piaget ppauper
Posts: 70239
Joined: November 15th, 2001, 1:29 pm

### Re: Extremes

$2\sin(\theta+\pi/4)=-2$ when $\theta=5\pi/4$ which is $x=y=-1$

$y\ge 0$ would be $0\le\theta\le\pi$ so the constrained minimum would be $x+y=-\sqrt{2}$ at $\theta=\pi$ or $x=-\sqrt{2}$,$y=0$  