I am trying to solve the following problem, and I think the solution manual's conclusion is wrong:
Find the extrema of f subject to the stated constraints: $$f(x,y) = x$$, subject to $$x^2 + 2y^2 = 3$$
Obviously, the Lagrange multipliers method failed. The solution manual concluded that $f$ has no maximum or minimum value in the given set $U$, but by theorem, since we study the function on a closed and bounded set, absolute and minimum values must exist
Am I getting something wrong?
The set $\{(x,y)\in\Bbb R^2\mid x^2+2y^2=3\}$ is compact and therefore any continuous function defined on it has to have a maximum and a minimum.
And the Lagrange method does not fail. Let $g(x)=x^2+2y^2$. The the method leds to the equations$$\left\{\begin{array}{l}1=2\lambda x\\0=4\lambda y\\x^2+2y^2=3.\end{array}\right.$$It follows from the second equality that $\lambda=0$ or that $y=0$. But you cannot have $\lambda=0$, since $2\lambda x=1$. So, $y=0$, and therefore $x=\pm\sqrt3$. The maximum is then $\sqrt3$, which is attained at $\left(0,\sqrt3\right)$, and the minimum is $-\sqrt3$, which is attained at $\left(0,-\sqrt3\right)$.
All this was simply meant to show that the method works. Of course, it is easy and more natural to solve the problem without it.