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\textbf{Applications of Partial Differentiation}

\textit{\textbf{Extremes within restricted domains}}
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\textbf{Question}

Find the maximum and minimum values of

$$f(x,y)=xy-x^3y^2$$

Over the square $0 \le x \le 1$, $0 \le y \le 1$.


\textbf{Answer}

\begin{eqnarray*}
f_1 & = & y-3x^2y^2 =y(1-3x^2y)\\
f_2 & = & x-2x^3y = x(1-2x^2y)
\end{eqnarray*}
This gives $(0,0)$ as a critical point, with any other critical points
having to satisfy $3x^2y=1$ and $2x^2y=1$. i.e. $x^2y=0$.

Therefore $(0,0)$ is the only critical point, and so we need only
consider values of $f$ on the boundary as possible limits.

When $x=0$ or $y=0$, $f(x,y)=0$.

When $x=1$, $f(1,y)=y-y^2=g(y)$, for $0\le y \le 1$.

$g$ has a maximum value $1/4$ when $y=1/2$.

When $y=1$, $f(x,1)=x-x^3=h(x)$, for $0\le x \le 1$.

$h$ has a critical point given by $1-3x^2=0$; only $x=1/\sqrt{3}$ is
on the side of the square.
$$h \left ( \frac{1}{\sqrt{3}} \right ) = \frac{2}{3
\sqrt{3}}>\frac{1}{4}.$$
On the square, $f(x,y)$ has a minimum value of 0. (At point $(1,10$
and the sides $x=0$ and $y=0$.)
$f(x,y)$ has a maximum value of $2/(3\sqrt{3})$ (at point
$(1/\sqrt{3},1)$, with a smaller local maximum point at $(1,\frac{1}{2})$.



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