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It is very critical to find the maximum of the following function. Assume that m_{1}, m_{2} and m are constants:

$f(x,y)=\frac{m_{1}x^4+m_{2}y^4+x^2+y^2-x^2y^2}{x^2+y^2-mx^2y^2}$

I tried to to compute gradient and then find the critical points but I ended up solving a forth degree equation which was really crazy and almost impossible to solve. Is there any method that I can find maximum or minimum points of this function? Any suggestion might help.Thanks.

user160492
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  • Did you try something like http://links.uwaterloo.ca/math227docs/set4.pdf or this:http://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/min_max/min_max.html? You should get equations of 2nd degree only. – NoChance Mar 21 '16 at 21:35
  • @NoChance - they end up with 2nd degree equations because their examples start with 2nd degree equations. You might note that user160492 is starting with a 4th degree equation. – Paul Sinclair Mar 22 '16 at 00:00

2 Answers2

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Let $a = x^2$ and $b=y^2$. Then, your problem is equivalent to choosing $a,b\geq 0$ to maximize $$ g(a,b) = \frac{m_1 a^2 + m_2 b^2 + a + b -ab}{a+b-m a b}. $$ This should allow you to get rid of the fourth order terms in the derivative steps.

However, oftentimes, this problem results in a maximized value of $\infty$. For instance, if $m_1$ is positive, then let $b=0$ and $g \rightarrow \infty$ as $a\rightarrow \infty$. Same for when $m_2> 0$.

In other cases, the derivative of this function w.r.t. $a$ is $$ \frac{a m_1 (a (-b) m+a+2 b)+b^2 \left(m_2 (b m-1)+m-1\right)}{(a (-b) m+a+b)^2}. $$

At the optimum you would need this to be equal to $0$, i.e. you would only look at the numerator, $$ a m_1 (a (-b) m+a+2 b)+b^2 \left(m_2 (b m-1)+m-1\right) = 0. $$

Similarly, derivative with respect to $b$ would require $$ \frac{a^2 (m-1)+a^2 m_1 (a m-1)+b m_2 (a (-b) m+2 a+b)}{(a (-b) m+a+b)^2} = 0, $$ i.e. $$ a^2 (m-1)+a^2 m_1 (a m-1)+b m_2 (a (-b) m+2 a+b) = 0. $$

These two equalities should give you an idea about where you would find the solution.

Matt
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Consider $f(xt, yt)$ for fixed values of $x,y$ and varying $t$. This is a ratio of 2nd degree polynomials in $t$ for which you can easily find the critical points with respect to $t$, namely, either $t = 0$ or $m_1x^4 + m_2y^4 = (1 - m)x^2y^2$ (in which case, $f = 1$ for all $t$).

Now for any critical point $(x, y)$ of $f$ other than $(0,0), f(tx, ty)$ will have $t = 1$ as a critical point, which by the above implies that $m_1x^4 + m_2y^4 = (1 - m)x^2y^2$ and therefore $f(x, y) = 1$.

Paul Sinclair
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