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### Re: Extremes

Posted: June 23rd, 2018, 2:08 pm
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.

### Re: Extremes

Posted: July 25th, 2018, 10:08 pm
Warning: Too long on this thread and you guys will end in a love triangle!

### Re: Extremes

Posted: July 26th, 2018, 3:34 pm

More like viscous cycles if you ask me!

### Re: Extremes

Posted: August 12th, 2018, 1:22 pm
minimise [$]x + y[$] subject to [$]x^2 + y^2 - 2 = 0[$]

### Re: Extremes

Posted: August 12th, 2018, 1:55 pm
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 [$]y=\sqrt{2}\sin\theta[$]
[$]x+y=\sqrt{2}(\cos\theta+\sin\theta)=2\sin(\theta+\pi/4)[$]

### Re: Extremes

Posted: August 20th, 2018, 4:45 pm
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 [$]y=\sqrt{2}\sin\theta[$]
[$]x+y=\sqrt{2}(\cos\theta+\sin\theta)=2\sin(\theta+\pi/4)[$]
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 [$] y \geq 0[$]?
[$]2\sin(\theta+\pi/4)=-2[$] 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 [$]x=-\sqrt{2}[$],[$]y=0[$]