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\textbf{Applications of Partial Differentiation}

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

Find and classify the critical points of the function

$$f(x,y)=x^3+y^3-3xy$$


\textbf{Answer}

\begin{eqnarray*}
f_1(x,y) & = & 3(x^2-y)\\
f_2(x,y) & = & 3(y^2-x)
\end{eqnarray*}
For critical points: $x^2=y$ and $y^2=x$.

Thus $x^4-x=0$, that is, $x(x-1)(x^2+x+1)=0$. Thus $x=0$ or $x=1$.

The critical points are $(0,0)$ and $(1,1)$. We have
\begin{eqnarray*}
A & = & f_{11}(x,y) =6x\\
B & = & f_{12}(x,y) =-3\\
C & = & f_{22}(x,y) =6y
\end{eqnarray*}
At $(0,0)$: $A=C=0$, $B=-3$. Thus $AC<B^2$, and $(0,0)$ is a saddle
point of $f$.

At $(1,1)$: $A=C=6$, $B=-3$, so $AC>B^2$. Thus $f$ has a local
minimum value at $(1,1)$.


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