$Find the extrema of $f(x,y)=x$ subject to the constraint $x^2+2y^2=2$$

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$Find the extrema of $f(x,y)=x$ subject to the constraint $x^2+2y^2=2$$

Find the extrema of $f(x,y)=x$ subject to the constraint $x^2+2y^2=2$.

We should use the Lagrange multipliers method.

Calculus Multivariable Calculus Optimization

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At the extreme points we must have
\[\nabla f= \lambda \nabla g,\]
where $g=x^2+2y^2.$ Hence
\[(1,0)=(2\lambda x,4 \lambda y).\]
So $4 \lambda y=0$ and $2 \lambda x=1$. So $\lambda \neq 0$, and from the fist equation we get
\[y=0.\]
Substituting $y=0$ in the constraint, we get
\[x^2=2 \Rightarrow x= \pm\sqrt{2}.\]
We have
$$f(\sqrt{2},0)=\sqrt{2}, f(-\sqrt{2},0)=-\sqrt{2}.$$
So $(\sqrt{2},0)$ is the maximum and $(-\sqrt{2},0)$ is the minimum point.

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$Find the extrema of $f(x,y)=x$ subject to the constraint $x^2+2y^2=2$$

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