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\textbf{Applications of Partial Differentiation}

\textit{\textbf{Extremes}}
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\textbf{Question}

\begin{description}
\item{(a)}
$$f(x,y)=(y-x^2)(y-3x^2)$$
Show that the origin is a critical point of $f$ and that the
restriction of $f$ to every straight line through the origin has a
local minimum value at the origin.

(i.e. show that $f(x,kx)$ has a local minimum value at $x=0$ $\forall
k$, and that $f(0,y)$ has a local minimum value at $y=0$.)

\item{(b)}
Is there a local minimum value of $f(x,y)$ at the origin?

\item{(c)}
On the curve $y=2x^2$, what happens to $f$?

What does the second derivative test say about this situation?
\end{description}


\textbf{Answer}

\begin{description}
\item{(a)}

\begin{eqnarray*}
f(x,y) & = & (y-x^2)(y-3x^2) = y^2-4x^2y+3x^4\\
f_1(x,y) & = & -8xy +12x^3 = 4x(3x^2-2y)\\
f_2(x,y) & = & 2y - 4x^2
\end{eqnarray*}
Since $f_1(0,0)=f_2(0,0)=0$, therefore $(0,0)$ is a critical point of $f$.

\item{(b)}
Let $g(x)=f(x,kx)=k^2x^2-4kx^3+3x^4$. Then
\begin{eqnarray*}
g'(x) & = & 2k^2x-12kx^2+12x^3\\
g''(x) & = & 2k^2-24kx+36x^2.
\end{eqnarray*}
Since $g'(0)=0$ and $g''(0)=2k^2>0$ for $k \ne 0$, $g$ has a local
minimum value at $x=0$. Thus $f(x,kx)$ has a local minimum at $x=0$
if $k\ne 0$.

Since $f(x,0)=3x^4$ and $f(0,y)=y^2$ both have local minimum values at
$(0,0)$, $f$ has a local minimum at $(0,0)$ when restricted to any
straight line through the origin.

\item{(c)}

However, on the curve $y=2x^2$ we have
$$f(x,2x^2)=x^2(-x^2)=-x^4$$
which has a local maximum value at the origin. Therefore $f$ does
\it{not} have an (unrestricted) local minimum value at the origin.

Note that
\begin{eqnarray*}
A & = & f_{11}(0,0) = \left. (-8y+36x^2) \right |_{(0,0)} = 0\\
B & = & f_{12}(0,0) = \left. -8x \right|_{(0,0)} -0\
\end{eqnarray*}
Thus $AC=B^2$ and the second derivative test at the origin is indeterminate.

\end{description}

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