## Thomas' Calculus 13th Edition

$(\pm \dfrac{1}{\sqrt 2},\dfrac{1}{2})$ and $(\pm \dfrac{1}{\sqrt 2},-\dfrac{1}{2})$
Given: $f(x,y) =xy$ and $g(x,y)=x^2+2y^2-1=0$ ...(1) Consider the gradient equation as follows: $\nabla =\lambda \nabla g$ Here $y=2 \lambda x$ or, $y=\pm \dfrac{1}{2 \sqrt 2}$ and $x=4 \lambda y$ or, $x=\pm \dfrac{2}{2 \sqrt 2}y$ Now, equation (1) becomes: $g(x,y)=x^2+2y^2-1=0$ This implies that $x=\pm \dfrac{1}{\sqrt 2}$ and $y= \pm \dfrac{1}{2}$ Answer: our points are: $(\pm \dfrac{1}{\sqrt 2},\dfrac{1}{2})$ and $(\pm \dfrac{1}{\sqrt 2},-\dfrac{1}{2})$