I have to find the maximum and minimum of the function

$$f(x)=\frac {1}{x^2+y^2}$$

when $$x^2+(y-2)^2 \leq 1$$

What I did was:

1) Find the critical points of the function when $x^2+(y-2)^2 < 1$. It is easy to derivate, find the gradient, and prove that there isn't any.

2) If $x^2+(y-2)^2 = 1$, then:




$$F(t)=f(\sqrt{1-(t-2)^2},t) = \frac {1} {\sqrt{1-(t-2)^2}^2+t^2}=\frac {1}{1-t^2-4+4t+t^2} =\frac{1}{4t-3}$$



And that is the weird part, because it seems that the function as no critical points for any possible t. But the thing is, it HAS to. I mean, the function HAS to get to a maximum value at some point of $x^2+(y-2)^2 \leq 1$. It is reasonable to believe that this point lies in the 'frontier' (I don't know how is the English word to say 'end of an interval'), that is, when $x^2+(y-2)^2 = 1$, but I can't find it. Am I doing something wrong? I don't what to use Lagrange, if possible. I could, I know, but I what to know what am I doing wrong here.

Thank you all in advance for your patience with my English and for helping me!

  • $\begingroup$ Take for example $y=x^2$ in $[2,4]$ don't have a critical point on the interval but the function is crecient on the interval so the maximum is at $x=4$. $\endgroup$ – rlartiga May 3 '14 at 20:31
  • $\begingroup$ At the boundary you have to also consider the case $y=t$, $x = -\sqrt{1 - (t-2)^2}$. Moreover you'll have to check the endpoints of the interval, i.e. when $t=1$ or $t=3$, separately. $\endgroup$ – J. J. May 3 '14 at 20:54
  • $\begingroup$ Another option would be to parametrize the boundary using trigonometric functions, that way you get rid of the piecewise mess: i.e. set $x = \cos \theta$, $y = 2 + \sin \theta$, $\theta \in \mathbb{R}$. $\endgroup$ – J. J. May 3 '14 at 20:55
  • $\begingroup$ I think that the first case changes nothing, but I hav'nt consider the endpoints, you are right... $\endgroup$ – Lessa121 May 3 '14 at 20:55

Since you have ruled out any interior critical points, then you look for the extrema on the boundary of your constraint, i.e. the extrema of $f(x,y)={1\over x^2+y^2}$ occur along $x^2+(y-2)^2=1\implies x^2=1-(y-2)^2$ at which ${1\over x^2+y^2}={1\over 1-(y-2)^2+y^2}$. Calling this last function $g(y)$ and noting from the original region of optimization, $x^2+(y-2)^2\le 1$ that $1\le y\le 3$, we want to maximize $g(y)$ on $1\le y\le 3$; the max value of $g$ occurs at $y=1\implies x=0\implies f(0,1)=1$ is the maximum we seek.

Similarly, the minimum of $g(y)$ on $1\le y\le 3$ occurs at $y=3\implies x=0\implies f(0,3)={1\over 9}$ is the minimum we seek.

Here's a picture to help you visualize the geometry of the situation. This is the surface $z=f(x,y)$ plotted subject to the constraint $x^2+(y-2)^2\le 1$. The extrema are denoted by the black dots.

enter image description here

  • $\begingroup$ Thanks. To get the maximum value of $g$, derivatives are useless, so you just use that the function is not crecient, and take the minimum possible value for y? $\endgroup$ – Lessa121 May 3 '14 at 21:55
  • $\begingroup$ Also, where did you plot it? Thanks again $\endgroup$ – Lessa121 May 3 '14 at 21:58
  • $\begingroup$ Derivatives weren't useless---you had to use them to determine if there were any interior critical points. In this case, there just happened to be none. I plotted this in Mathematica. $\endgroup$ – JohnD May 3 '14 at 22:17
  • $\begingroup$ Yes, of course, I meant just in this case. Thank you! $\endgroup$ – Lessa121 May 5 '14 at 4:12

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