I tried this question on a site and found that none of the answers were the ones I got. I think my calculations were right but my understanding of applying the correct theorems may have been off.
I found the derivatives with respect to x and y and got:
$f_x = 2x \quad$ $f_{xx} = 2 \quad $ $ f_{y} = 2y \quad $ $ f_{yy}=2 \quad $ $ f_{xy} = 0 $
The critical values are when $ 2x = 2y $ which is the line x = y. the fancy formula:
$ D = f_{xx}*f_{yy} – (f_{xy})^2 $ gives us $D = 4$ always. So long as the point is a critical point, its a relative min. Since it is subject to the constraint, we need the values which satisfy $x^2+ 2y^2 = 1 $. Those are ones where $ x=y$ so we look at $x^2+ 2x^2 = 1 \implies 3x^2= 1 \implies x = \pm1/\sqrt{3}$. Either works so when I plug it in to the original function, since $ x=y$, I get $ 2/3 $. Thus my absolute minimum.
Is my reasoning correct? or am I missing tools or using the tools incorrectly from multivariable Calculus?
Best Answer
From our constraint $x^2+2y^2=1$ we have $x^2=1-2y^2$ hence we can rewrite our function in terms of $y$:
$$f(y)=1-2y^2+y^2=1-y^2$$
With the constraint $y \in [-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}]$ because $x^2 \geq 0$ if $x \in \mathbb{R}$.
It should be clear that $f$ attains an absolute maximum at $1$ when $y=0$, and has absolute minimums of $\frac{1}{2}$ when $y=\pm \frac{\sqrt{2}}{2}$.