Given that $\frac{1}{3}
Find the minimum value of $P=x^2+y^2+\frac{x^2y^2}{(4xy-x-y)^2}$
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0Sorry, it's $y\ge1$ not y>1 – 2012-08-06
2 Answers
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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2If 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. – 2012-08-06
$P=x^2+y^2+\frac{(xy)^2}{(4xy-x-y)^2}=x^2+y^2+{\left(4-\frac{1}{x}-\frac{1}{y}\right)^{-2}}$
$\frac{dP}{dx}=2x-\frac{2}{x^2}\left(4-\frac{1}{x}-\frac{1}{y}\right)^{-3}$
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 :
- for $y=1$, we have $x=\frac{1}{2}$ that is a minimum along one coordinate.
- for $x=\frac{1}{2}$, we have $y=1$
- 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 : $P(\frac{1}{3},1)=\infty$