The problem is this:
The standard-form of Pringles chips is achieved by creating the hyperbolic paraboloïd $z=\frac{y^2}{4}-\frac{x^2}{2}$ on a domain given by the elliptic plate $x^2+\frac{y^2}{2}\le q^2$, $q\in\mathbb R_+$. For a special edition they change the domain to $x^2+y^4-ay^2=1$.
For which values of $a$ does the maximum of $z$ on this curve lie on the $y$-axis, but the minimum of $z$ on this curve not on the $x$-axis.
I have tried and tried but I can't seem to get past this:
$$f(x,y)=\frac{y^2}{4}-\frac{x^2}{2}=\frac{y^2}{4}-\frac{1}{2}(1-y^4+ay^2)=\frac{y^4}{2}+\frac{y^2}{4}-\frac{a}{2}y^2-\frac{1}{2}$$
Then $$\frac{df}{dy}=2y^3+\frac{y}{2}-ay=y(2y^2+\frac{1}{2}-a)=0$$ if $y=0$ or $2y^2=a-\frac{1}{2} \Leftrightarrow y^2=\frac{2a-1}{4}$ $(a\ne1/2)$.
If $y=0$ then $f(y)=-1/2$ and so $x^2=1$. (probably minima, so then $a\ne1/2$)
Is this enough? Do I have to use the second derivative test here to show that this is a minima (but then I wouldn't have showed that there aren't other minima, right?)?
If $y^2=\frac{2a-1}{4}$ where $a\ne1/2$ then $f(y)=\ldots=\frac{1}{4}(-a^2+a-\frac{17}{32})=\frac{2a-1}{8}-x^2$ $\Leftrightarrow$ $\frac{a^2}{4}+\frac{1}{128}=x^2$. Do I have to do anything with this?