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If $q>p>0$ show that the point $(x,y)=(0,0)$ minimizes the function $$f(x,y) = (y-px^2)(y-qx^2)$$ locally on the lines $$y=ht, x=kt$$ through the point (x,y).

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Why is this tagged 'convex-optimization'? It doesn't look convex to me... – Michael Grant Feb 4 '14 at 14:05

Just insert the definition of $y$ and $x$ so define the function $$g(t)=f(kt,ht)=\ldots$$ Now you have a function $g:R\rightarrow R$, where you can use standard one-dimensional analysis to show that $t=0$ is a minimal point (under certain conditions on $q$ and $p$ maybe). But this means that $f$ has a minimum in $(0,0)$ if you only consider sequences in $R^2$ that are linear (you walk on lines into $(0,0)$.

You did not ask this, but the natural question would be now: Is $(0,0)$ a minimal point of $f$.

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