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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 $2/(3sqrt3)$