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Given that $\frac{1}{3}<x \le \frac{1}{2}$ and $y\ge1$ Find the minimum value of $P=x^2+y^2+\frac{x^2y^2}{(4xy-x-y)^2}$

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Is this homework? What have you done, so far? – Siminore Aug 6 '12 at 10:18
Sorry, it's $y\ge1$ not $y>1$ – tangkhaihanh Aug 6 '12 at 10:59
up vote 0 down vote accepted

As it seems to be a homework question and I don't have Latex-MathType in my job's computer (I'll edit my answer late to make it formal), I will give you just an outline of how to solve it.

You have 3 inequality constraints:

$x > \frac13 \Rightarrow f_1(x,y) = x - \frac13 > 0$
$x \leq \frac12 \Rightarrow f_2(x,y) = x - \frac12 \leq 0$
$f_3(x,y) = y-1 > 0$

Form $F(x,y) = P(x,y) - f_1(x,y) - f_2(x,y) - f_3(x,y)$, and use Karush-Kuhn-Tucker conditions.

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If I'm going to have problems for doing mathematics in my job's computer at least I'll have them with LaTeX...or even better: I'll try to get a job where I can do mathematics. – DonAntonio Aug 6 '12 at 10:36



The zero of the derivative is obtained when $$\frac{1}{x}=\left(4-\frac{1}{x}-\frac{1}{y}\right)$$ So when $\frac{1}{x}=2-\frac{1}{2y}$, and by symmetry on the other derivative, you obtain $\frac{1}{y}=2-\frac{1}{2x}$.

Hence $\frac{1}{x}=\frac{4}{3}$

The global minimum is in $(\frac{3}{4},\frac{3}{4})$ Too bad, it's not in the scope.

Rewrite the zero of derivative you obtain $2y=4xy-x=x(4y-1)$ So $x=\frac{2y}{4y-1}$ (for the x derivative), and $y=\frac{2x}{4x-1}$ for the other.

The minimum is on the boundary, so :

  1. for $y=1$, we have $x=\frac{1}{2}$ that is a minimum along one coordinate.
  2. for $x=\frac{1}{2}$, we have $y=1$
  3. for $x=\frac{1}{3}$, we have $y=-\frac{1}{2}$ (outside)

So the minimum is in $x=\frac{1}{2}$ and $y=1$ and $P=\frac{9}{4}$

Verification : $P(\frac{1}{3},1)=\infty$

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