#"We could use the Lagrange multiplier L :"#

#f(x,y,L) = y exp(x) - x exp(y) + L(xy - 4)#

#(df)/(dx) = y exp(x) - exp(y) + L y = 0#

#(df)/(dy) = exp(x) - x exp(y) + L x = 0#

#(df)/(dL) = xy - 4 = 0 => y = 4/x#

#=> (4/x) exp(x) - exp(4/x) + 4 L / x = 0#

#=> exp(x) - x exp(4/x) + L x = 0#

#"Multiply the last equation by (4/x) : "#

#=> (4/x) exp(x) - 4 exp(4/x) + 4 L = 0#

#"Subtract this equation from the first :"#

#=> 3 exp(4/x) + 4 L / x - 4 L = 0#

#=> 4 L (1 - 1/x) = 3 exp(4/x)#

#=> L = (3/4) exp(4/x) / (1 - 1/x)#

#"Fill in this value for L in the second equation :"#

#=> exp(x) - x exp(4/x) + (3/4) exp(4/x) x^2 / (x - 1) = 0#

#=> exp(x) + exp(4/x) [ (3/4) x^2 / (x-1) - x ] = 0#

#=> exp(x) = exp(4/x) [ x(x - 1) - (3/4) x^2 ] / (x - 1)#

#=> exp(x) = exp(4/x) x [ (1/4) x - 1 ] / (x - 1)#

#=> exp(x-4/x) = x [(1/4) x - 1]/(x - 1)#

#"This looks like a transcendental equation."#

#"I am stopping here, but if one solves this numerically, one gets"#

#"x and then L and also y."#