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We need to determine the global maximum and minimum of:

$f(x,y)=y^2-16x^2$ on the interval of: $\{(x,y) : y ≤ 1−x^2,y ≥ 0\}$

My initial thought was that I could use extreme value theorem, later realising that was wrong. Then I started to use Lagrange multiplier because these were constraints. I did it and go the answers to be $(-1,0)$ and $(1,0)$ as well as $(-3,-8)$ and $(3,-8)$. Although when you evaluate it comes up as all these values are equal to zero, when plugged into $f(x,y$). Some help is really appreciated!!!

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  • $\begingroup$ $f(\pm1,0)=1$ and $(\pm3,8)$ do not belong to the domain. $\endgroup$ – AugSB Apr 4 at 18:27
  • $\begingroup$ $f(\pm1,0)=-16\ne1$ $\endgroup$ – Peter Foreman Apr 4 at 18:28
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The set $\{(x,y) : y \leq 1-x^2, y \geq 0\}$ is not an interval, it is a region in the $xy$-plane.

First you need to find the critical points of $f(x,y)$, i.e., points where the gradient is zero. If any of these lie within the region above, they are candidates for the locations of the max/min.

You also need to find critical points of $f(x,y)$ along the boundaries of the region above, i.e., along the curve $y=1-x^2$ and the line $y=0$. For the first curve you should use Lagrange multipliers, and for the second you can just plug in $y=0$ and treat it as a function of $x$.

Finally, you also need to test the points where the boundary curves of the region intersect, i.e., where $y=1-x^2$ and $y=0$ intersect.

Once you have the list of all these critical points, plug them all into $f$ to see where the min and max occur.

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$y≤1−x^2,y≥0$ or,

$x^2+y ≤ 1$, or, $0≤y≤1$ and $0≤x^2≤1$ or $0≤|x|≤1$.

To maximize $|x|$ has to be minimum which is zero(x=0), in that case $y=1$ (maximum) , to minimize similarly $|x|=1$ or $x=±1$ and $y=0$.

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