[$]x^2 + y^2 - 2 = 0[$] is circle radius 2
[$]x=\sqrt{2}\cos\theta[$] and [$]y=\sqrt{2}\sin\theta[$]
[$]x+y=\sqrt{2}(\cos\theta+\sin\theta)=2\sin(\theta+\pi/4)[$]
answer is -2

[$]x^2 + y^2 - 2 = 0[$] is circle radius 2
[$]x=\sqrt{2}\cos\theta[$] and [$]y=\sqrt{2}\sin\theta[$]
[$]x+y=\sqrt{2}(\cos\theta+\sin\theta)=2\sin(\theta+\pi/4)[$]
answer is -2

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 [$] y \geq 0[$]?

Re: Extremes

Posted: August 21st, 2018, 5:47 am

by ppauper

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