# Finding absolute minimum and maximum of a multi-variable function

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!

• Hint: You have to study the boundary. – Ertxiem - reinstate Monica Apr 10 '19 at 20:51
• 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. – 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

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

• 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. – Martin Pham Apr 10 '19 at 21:08
• @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 $(1,-\frac 12)$. – PierreCarre Apr 10 '19 at 21:24

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 $$y=0$$, $$x_1=3/2$$; $$x_2=1/2$$;