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Find the absolute maximum and minimum values of the function: $$ f(x,y) = \sqrt{(x-1)^2+y^2}-x^2+2x $$

restricted to the area

$$ D = \{ (x,y): \sqrt{(x-1)^2+y^2} \leq (1/2)\} $$

I've already found the critical points of f, being ((3/2),0) and ((1/2),0) by taking the partial derivatives, setting them to zero and substituting them back in.

Would appreciate some help on how I can go onwards to find the min/max. Thanks!

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    $\begingroup$ Hint: You have to study the boundary. $\endgroup$ – Ertxiem - reinstate Monica Apr 10 '19 at 20:51
  • $\begingroup$ The boundary of the area? I've drawn the area as a circle with center in (1,0), and r=1/2. And then tried putting in the (x,y) of the extremas in the original function f. Each time I got the same values; 1.25. $\endgroup$ – Martin Pham Apr 10 '19 at 21:04
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You found some critical points, but they are not in the interior of $D$. In this case the extrema will occur at the boundary of $D$. There you can use Lagrange multipliers or just notice that on the boundary you have that $f(x,y)=\frac 12 -x^2+2x$ and $x \in [\frac 12, \frac 32]$. The minimum is attained at $x = \frac 12, \frac 32$ and the maximum at $x = 1$.

We conclude that the global minimum is $$ m = f(\frac 12,0) = f(\frac 32, 0) = \frac 54 $$

and the global maximum is

$$ M = f(1,\frac 12) = f(1,-\frac 12) = \frac 32. $$

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  • $\begingroup$ Thanks for the comment! I've already tried what you said; writing f(x,y) as you said and I then drew the area (which I saw was a circle with x from 0.5 to 1.5). I then tried putting the values I saw in the original function, but the values I got were all the same; 1.25. I'm submitting the answers on an online platform, where I can "verify" my answers, and 1.25 isn't correct for neither maxima nor minima. I'm not good with the Lagrange multipliers, although I've looked around and I see there's a way to solve my problem with that, as you suggested. $\endgroup$ – Martin Pham Apr 10 '19 at 21:08
  • $\begingroup$ @MartinPham I don't understand the problem... There are no critical points in the interior of $D$, so the max/min are attained at the boundary. Since on the boundary $f=\frac 12 -x^2 +2x$, the min is attained when $x = \frac 12, \frac 32$ (the points from before), and the maximum is attained when $x =1$ ( with $y = \pm \frac 12$). The minimum is attained at $(\frac 12 ,0)$ and $(\frac 32, 0)$ and the maximum is attained at $(1,\frac 12)$ and $(1,-\frac 12)$. $\endgroup$ – PierreCarre Apr 10 '19 at 21:24
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Attempt:

Domain is a circle center $(1,0)$, radius 1/2.

$f(x)=$

$\sqrt{(x-1)^2+y^2}-(x-1)^2+1$;

Maximum:

at $x=1$, $y=\pm 1/2$;

$f_{max}=1/2+1=3/2$.

Minimum: Set $y=0$;

$f=|x-1|-(x-1)^2+1=-[(|x-1|^2-|x-1|)+1=$

$-[|x-1|-1/2]^2+1/4+1$ ;

$f_{min}= 5/4$; at $y=0$, $|x-1|=1/2$, i.e. at $y=0 $, $x_1=3/2$; $x_2=1/2$;

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