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$$f(x,y)=\frac{4}{x^2+y^2+1}+2 xy \\ \text{within the domain: }1/5\leq x^2+y^2\leq 4$$

I am able to find the maximum of the function at $x^2+y^2=4$ by substituting x,y for $\cos(t)$ and $\sin(t)$ and I am able to start working for a solution in the lower boundary:

$$ x=\frac{1}{\sqrt5}\cos(t), \ y=\frac{1}{\sqrt5}\sin(t) \\ \frac{4}{\frac{6}{5}}+\frac{2}{\sqrt5}(\cos(t)\sin(t))\\ \frac{df}{dt}=\frac{2}{\sqrt5}\cos(2t)=0\\ \tan(t)=1\\ x=\frac{1}{\sqrt5}\lvert\cos(\frac{\pi}{4})\rvert \ , \ \ y=\frac{1}{\sqrt5}\lvert\sin(\frac{\pi}{4})\rvert \\ x=\pm\frac{1}{\sqrt{10}},y=\pm\frac{1}{\sqrt{10}}$$

When i resubstitute this into the original function I get the wrong answer.

$$f\left(\frac{1}{\sqrt{10}},\frac{1}{\sqrt{10}}\right)=\frac{4}{\frac{12}{10}}+\frac{2}{10}=53/15 $$

Michael Hardy
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Strange Brew
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2 Answers2

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$f_x(x,y) = 0 \to \dfrac{-8x}{(x^2+y^2+1)^2} = 2y$

$f_y(x,y) = 0 \to \dfrac{-8y}{(x^2+y^2+1)^2} = 2x$.

Thus $(x,y) = (0,0)$ is a critical point, also $\dfrac{x}{y} = \dfrac{y}{x} \to x = \pm y$

Consider $g(x) = f(x,x) = \dfrac{4}{2x^2+1} + 2x^2$ with $1/5 \leq 2x^2 \leq 4$. let $u = 2x^2$, then $h(u) = g(x,x) = \dfrac{4}{u+1} + u$, $1/5 \leq u \leq 4\to h'(u) = 1-\dfrac{4}{(u+1)^2} = 0 \iff u+1 = \pm 2 \to u =1,-3 \to u = 1$, since $u > 0$. $h(1) = 3, h(1/5) = 53/15, h(4) = 24/5$.

And $r(x) = f(x,-x) = \dfrac{4}{2x^2+1} - 2x^2 = \dfrac{4}{u+1}-u = q(u)$, for $1/5 \leq u \leq 4$.

$q'(u) = -1-\dfrac{4}{(u+1)^2} < 0 \to q(u) \geq q(4) = 4/5-4$.

Thus the min value of $f(x,y)$ is $4/5-4$.

DeepSea
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  • Thanks alot for that thorough answer. I think I almost get it - however I don't understand your reasons for deducing that the critical points are $ \frac{x}{y} =\frac{y}{x}$. I guess you either subtract or add the partial derivative of y to x?

    Also how can we be sure that $f(x,x)$ will suffice to solve the equation? I mean don't you have to account for the other variable y aswell?

    Lastly I was unable to understand how you performed the last step where you introduce $p(u)\geq g(4)$ What is p(u)?.

     – Strange Brew Nov 11 '14 at 19:09
  • It should be $q(u) \geq q(4)$. Also, $x = y \to f(x,x)$, and $x = -y \to f(x,-x)$. From these cases we are led to the answer. – DeepSea Nov 12 '14 at 00:17
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You know that $1/5\leq x^2+y^2\leq 4$, so:

$$\frac{4}{x^2+y^2+1} \geq \frac{4}{4+1}=\frac{4}{5}$$

In addiction by AM-GM:

$$2|xy| \leq x^2+y^2 \leq 4$$

$$|xy| \leq 2$$

And:

$$2xy \geq -4$$

So let's try to find such a $x,y$ that:

$$xy=-2,x^2+y^2=4$$

But for $x=-\sqrt{2}$ and $y=\sqrt{2}$ both equalities are satisfy, so minimum is:

$$\frac{4}{5}-4$$

agha
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