# If xy=64 and y log(x)+x log(y)=2.5, how do you find x and y?

Aug 23, 2016

$\left\{x , y\right\} = \left\{0.976141 , 65.5643\right\}$

#### Explanation:

Calling

$f \left(x , y\right) = \left(\begin{matrix}{f}_{1} \\ {f}_{2}\end{matrix}\right) = \left(\begin{matrix}x y - a \\ y {\log}_{e} x + x {\log}_{e} y - b\end{matrix}\right)$

with $a = 64 , b = 2.5$

Expanding near the point ${p}_{0} = \left\{{x}_{0} , {y}_{0}\right\}$ in Taylor series for the linear term

$f \left(x , y\right) = f \left({x}_{0} , {y}_{0}\right) + \nabla {f}_{0} \cdot \left(p - {p}_{0}\right) + {O}^{2} \left(p - {p}_{0}\right)$ where

$\nabla f = \left(\begin{matrix}\frac{{\mathrm{df}}_{1}}{\mathrm{dx}} & \frac{{\mathrm{df}}_{1}}{\mathrm{dy}} \\ \frac{{\mathrm{df}}_{2}}{\mathrm{dx}} & \frac{{\mathrm{df}}_{2}}{\mathrm{dy}}\end{matrix}\right)$
$p - {p}_{0} = \left(\begin{matrix}x - {x}_{0} \\ y - {y}_{0}\end{matrix}\right)$

${O}^{2} \left(p - {p}_{0}\right) \to \left\{0 , 0\right\}$ as $\left\mid p - {p}_{0} \right\mid \to 0$

For $\left\mid p - {p}_{0} \right\mid$ small then

$f \left(x , y\right) \approx f \left({x}_{0} , {y}_{0}\right) + \nabla {f}_{0} \cdot \left(p - {p}_{0}\right)$

If $\left\mid p - {p}_{0} \right\mid$ is small then $f \left(x , y\right) \approx \left\{0 , 0\right\}$
and

$p = {p}_{0} - {\left(\nabla {f}_{0}\right)}^{- 1} {p}_{0}$

or

${p}_{k + 1} = {p}_{k} - {\left(\nabla {f}_{k}\right)}^{- 1} {p}_{k}$

Here

$\nabla f = \left(\begin{matrix}y & x \\ \frac{y}{x} + L o g \left(y\right) & \frac{x}{y} + L o g \left(x\right)\end{matrix}\right)$

(grad f_0)^{-1} = (((x/y + Log(x))/(x - y + y Log(x) - x Log(y)), x/(-x + y - y Log(x) + x Log(y))),((y + x Log(y))/( x (-x + y - y Log(x) + x Log(y))), y/(x - y + y Log(x) - x Log(y))))

Begining with
${p}_{0} = \left\{1 , 20\right\}$ we obtain
${p}_{1} = \left\{0.877443 , 66.4511\right\}$
${p}_{2} = \left\{0.970524 , 65.8899\right\}$
${p}_{3} = \left\{0.9761 , 65.5652\right\}$
${p}_{4} = \left\{0.976141 , 65.5643\right\}$
${p}_{5} = \left\{0.976141 , 65.5643\right\}$