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Answer 1 · 2016-03-25T03:30:39

Maximum is #2/(3sqrt3)# at #(+-sqrt(2/3),1/sqrt3)#

#F(x,y,lambda) = x^2y+lambda(x^2+y^2-1)#

#F_x = 2xy+2lambdax#
#F_y = x^2+2lambday#

Setting each of these to #0# and solving for #-lambda#, we get

#-lambda = y = x^2/(2y)#.

This leads to #x^2=2y^2#

Substituting into #F_lambda = x^2+y^2-1 = 0#

We get #3y^2 = 1#, so #y=+-1/sqrt3#

Using #x^2=2y^2#, we get #x= +- sqrt(2/3)#.

Simple arithmetic gets the maximum value is #2/(3sqrt3)#