I have to find the extremas of $f(x, y)=3x+2y$ subject to $2x^2+3y^2 \leq 3$.
Since the region $2x^2+3y^2 \leq 3$ is closed, $f$ has a maximum and a minimum, which is either at the boundary or at the critical points of the function, right??
So, we have to find first the critaical points and then using Lagrange multipliers theorem we are looking if the maximum and the minimum value of $f$ is at the boundary $2x^2+3y^2=3$.
Is this correct??
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EDIT:
To find the critical points we do the following:
$$\nabla f(x, y)=\overrightarrow{0} \Leftrightarrow (3, 2)=(0, 0)$$
That is a contradiction. That means that there are no critical points.
From Lagrange multipliers theorem we get the extremas $\left (\frac{9}{2}\sqrt{\frac{2}{35}}, 2\sqrt{\frac{2}{35}}\right )$ and $\left (-\frac{9}{2}\sqrt{\frac{2}{35}}, -2\sqrt{\frac{2}{35}}\right )$.
calculating the values of $f$, we have $$f\left (\frac{9}{2}\sqrt{\frac{2}{35}}, 2\sqrt{\frac{2}{35}}\right )=\sqrt{\frac{35}{2}} \\ f\left (-\frac{9}{2}\sqrt{\frac{2}{35}}, -2\sqrt{\frac{2}{35}}\right )=-\sqrt{\frac{35}{2}}$$
So, $f$ has the absolute maximum at $\left (\frac{9}{2}\sqrt{\frac{2}{35}}, 2\sqrt{\frac{2}{35}}\right )$ and the absolute minimum at $\left (-\frac{9}{2}\sqrt{\frac{2}{35}}, -2\sqrt{\frac{2}{35}}\right )$.
Is this correct??