minimum value of a two variable problem

Question

Suppose that $x$ and $y$ are in $(−2, 2)$ and $xy = −1$. The minimum value of $\frac4{4-x^2}+\frac9{9-y^2}$ is ?

So I have tried to manipulate this equation but what I got is a complex equation with square roots in it...

Answer 1

Consider the function$$f(x,y)=\frac4{(4-x^2)}+\frac9{(9-y^2)}$$ You are also given the condition $xy=-1$. So, replace $y$ by $-\frac 1x$ and consider now (after simplication) that you need to minimize $$g(x)=\frac{9 x^4-72 x^2+4}{9 x^4-37 x^2+4}$$ Compute the derivative (and simplify) to get

$$g'=\frac{70 x \left(9 x^4-4\right)}{\left(9 x^4-37 x^2+4\right)^2}$$ Since, it is a minimum or maximum, then $????$.

The second derivative test would tell you which points are minimum of maximum.

Answer 2

Let $x^2=\frac{2}{3}a$ and $y^2=\frac{3}{2}b$.

Hence, $ab=1$ and by C-S and AM-GM we obtain: $$\frac{4}{4-x^2}+\frac{9}{9-y^2}=\frac{4}{4-\frac{2}{3}a}+\frac{9}{9-\frac{3}{2}b}=$$ $$=6\left(\frac{1}{6-a}+\frac{1}{6-b}\right)\geq\frac{6(1+1)^2}{6-a+6-b}=$$ $$=\frac{24}{12-a-b}\geq\frac{24}{12-2}=\frac{12}{5}.$$ The equality occurs for $a=b=1$, which says that $\frac{12}{5}$ is a minimal value.

Done!

Answer 3

$x, y \in (−2, 2)$ and $xy = −1$.

Let $3x+2y=k$. Then $9x^2-12+4y^2=k^2$. That is, $9x^2+4y^2=k^2+12$.

\begin{align} f(x,y) &= \frac4{(4-x^2)}+\frac9{(9-y^2)} \\ &= \dfrac{72 - (9x^2+4y^2)}{36-(9x^2+4y^2)+1} \\ &= \dfrac{72 - (k^2+12)}{36-(k^2+12)+1} \\ &= \dfrac{60 - k^2}{25-k^2} \\ &= 1 + \dfrac{35}{25-k^2} \end{align}

This attains its smallest value, $\frac{12}{5}$, when $k=0$, for example, when $x = \frac 23$ and $y=-1$.