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

- Cuchulainn
**Posts:**61185**Joined:****Location:**Amsterdam-
**Contact:**

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

http://www.datasimfinancial.com

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

Warning: Too long on this thread and you guys will end in a love triangle!

- Cuchulainn
**Posts:**61185**Joined:****Location:**Amsterdam-
**Contact:**

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

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

- Cuchulainn
**Posts:**61185**Joined:****Location:**Amsterdam-
**Contact:**

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

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

[$]x^2 + y^2 - 2 = 0[$] is circle radius 2minimise [$]x + y[$] subject to [$]x^2 + y^2 - 2 = 0[$]

[$]x=\sqrt{2}\cos\theta[$] and [$]y=\sqrt{2}\sin\theta[$]

[$]x+y=\sqrt{2}(\cos\theta+\sin\theta)=2\sin(\theta+\pi/4)[$]

answer is -2

- Cuchulainn
**Posts:**61185**Joined:****Location:**Amsterdam-
**Contact:**

I agree. Nice approach. At which point does it reach a minimum based on this analysis?[$]x^2 + y^2 - 2 = 0[$] is circle radius 2minimise [$]x + y[$] subject to [$]x^2 + y^2 - 2 = 0[$]

[$]x=\sqrt{2}\cos\theta[$] and [$]y=\sqrt{2}\sin\theta[$]

[$]x+y=\sqrt{2}(\cos\theta+\sin\theta)=2\sin(\theta+\pi/4)[$]

answer is -2

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.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

[$]2\sin(\theta+\pi/4)=-2[$] 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 [$]x=-\sqrt{2}[$],[$]y=0[$]

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

GZIP: On