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How would one go about trying to calculate the highest/lowest values of a two variable function given that $(x,y)$ must lie on the upper half of the unit circle?

If it was just the unit circle, I'd set the partiel derivatives of $f$ equal to the partial derivatives of $x^2 + y^2$ times $\lambda$, and solve a system of equations with 3 equations (the two with the partial derivatives, and then the constraint itself). But given that they must lie on the upper half only, I've essentially got two constraints (that $y=0$). But they don't both need to be satisfied at the same time so.... How does one solve this?

Jackson Dan
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Use gradient to determine critical points in the interior of upper unit circle. Then find local extrema of $f(x,0)$ (i.e. potential max/min on $y=0$), and local extrema of $f(x,\sqrt{1-x^2})$ (i.e. potential max/min on the upper half border). Or use Lagrange multipliers. Finally compare all the extrema you found.

Milly
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  • Any specific reason you chose $f(x,\sqrt(1-x^2))$? Could you also have chosen to do that for y? – Jackson Dan Nov 01 '14 at 19:04
  • To find potential max/min on $y=\sqrt{1-x^2}$ (border of upper half circle). Or use Lagrange multipliers for this. Also you can find max/min for $f(\sqrt{1-y^2},y)$ Answer edited. – Milly Nov 01 '14 at 19:07