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

1 Answer
Aug 23, 2016

Answer:

#{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}#