[tex]L(x,y,\lambda)=xy^2+\lambda(x+y^2-4)[/tex]
[tex]\begin{cases}L_x=y^2+\lambda=0&(1)\\L_y=2xy+2\lambda y=0&(2)\\L_\lambda=x+y^2-4=0&(3)\end{cases}[/tex]
Multiply (1) by [tex]2y[/tex], then subtract (2) from that result:
[tex]2y^3+2\lambda y=0[/tex]
[tex]\implies(2y^3+2\lambda y)-(2xy+2\lambda y)=0[/tex]
[tex]y^3-xy=0[/tex]
[tex]y(y-\sqrt x)(y+\sqrt x)=0[/tex]
which means [tex]y=0[/tex], [tex]y=-\sqrt x[/tex], or [tex]y=\sqrt x[/tex]. But [tex]x>0[/tex], so we only take [tex]y=\sqrt x[/tex].
Substitute into (3) and solve for [tex]x[/tex].
[tex]x+(\sqrt x)^2=4\implies2x=4\implies x=2\implies y=\sqrt2[/tex]
So a [tex]Q[/tex] attains a maximum at [tex](2,\sqrt2)[/tex], giving a maximum value of [tex]2(\sqrt2)^2=4[/tex].
