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!

  • 1
    $\begingroup$ 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

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

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

  • $\begingroup$ 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


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




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


Minimum: Set $y=0$;


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.