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

1 Answer
Aug 23, 2016

{x,y} = {0.976141,65.5643}

Explanation:

Calling

f(x,y) = ((f_1),(f_2))=((x y -a),(y log_e x+x log_e y - b))

with a = 64, b = 2.5

Expanding near the point p_0 = {x_0,y_0} in Taylor series for the linear term

f(x,y) = f(x_0,y_0) + grad f_0 cdot(p - p_0) +O^2(p-p_0) where

grad f = (((df_1)/dx,(df_1)/dy),((df_2)/dx,(df_2)/dy))
p-p_0 = ((x-x_0),(y-y_0))

O^2(p-p_0)->{0,0} as abs(p-p_0)->0

For abs(p-p_0) small then

f(x,y) approx f(x_0,y_0) + grad f_0 cdot(p - p_0)

If abs(p-p_0) is small then f(x,y) approx {0,0}
and

p = p_0 - (grad f_0)^{-1}p_0

or

p_{k+1} = p_k - (grad f_k)^{-1}p_k

Here

grad f = ((y, x),(y/x + Log(y), x/y + Log(x)))

(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 = {1,20} we obtain
p_1 = {0.877443,66.4511}
p_2 = {0.970524,65.8899}
p_3 = {0.9761,65.5652}
p_4 = {0.976141,65.5643}
p_5 = {0.976141,65.5643}