Answer
The global minimum of $f$ is $f\left( {\frac{1}{{\sqrt 3 }},1} \right) \simeq - 0.385$ and the global maximum is $f\left( {1,0} \right) = 1$.
Work Step by Step
We have $f\left( {x,y} \right) = {x^3} - xy - {y^2} + y$ and the domain is the square $0 \le x \le 1$, $0 \le y \le 1$.
Step 1. Find the critical points on the domain and evaluate $f$ at these points
The partial derivatives are
${f_x} = 3{x^2} - y$, ${\ \ \ }$ ${f_y} = - x - 2y + 1$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
$3{x^2} - y = 0$, ${\ \ \ }$ $ - x - 2y + 1 = 0$
From the second equation, we obtain $y = \frac{{1 - x}}{2}$.
Substituting $y = \frac{{1 - x}}{2}$ in the first equation gives
$3{x^2} - \frac{{1 - x}}{2} = 0$
$6{x^2} + x - 1 = 0$
$x = \frac{{ - 1 \pm \sqrt {1 - 4\cdot6\cdot\left( { - 1} \right)} }}{{2\cdot6}} = \frac{{ - 1 \pm 5}}{{12}}$
Using $y = \frac{{1 - x}}{2}$, we obtain the critical points: $\left( {\frac{1}{3},\frac{1}{3}} \right)$, $\left( { - \frac{1}{2},\frac{3}{4}} \right)$.
The extreme values corresponding to these critical points: $f\left( {\frac{1}{3},\frac{1}{3}} \right) = 0.148$ and $f\left( { - \frac{1}{2},\frac{3}{4}} \right) = 0.438$.
Step 2. Check the boundaries
1. On the bottom edge of the square, $y=0$ and $0 \le x \le 1$. Write $g\left( x \right) = f\left( {x,0} \right) = {x^3}$.
Find the extreme values of $f$ on the bottom edge.
The critical points of $g$ can be found by solving the equation $g'\left( x \right) = 0$.
$g'\left( x \right) = 3{x^2} = 0$
$x = 0$
For $0 \le x \le 1$, the minimum value of $g$ is $0$ and the maximum value of $g$ is $1$. Thus, at the bottom edge, the minimum value of $f$ is $0$ and the maximum value of $f$ is $1$.
2. On the top edge of the square, $y=1$ and $0 \le x \le 1$. Write $h\left( x \right) = f\left( {x,1} \right) = {x^3} - x$.
Find the extreme values of $f$ on the top edge.
The critical points of $h$ can be found by solving the equation $h'\left( x \right) = 0$.
$h'\left( x \right) = 3{x^2} - 1 = 0$, ${\ \ \ }$ $h{\rm{''}}\left( x \right) = 6x$
$x = \pm \frac{1}{{\sqrt 3 }}$
For $0 \le x \le 1$, the critical point of $h$ is $x = \frac{1}{{\sqrt 3 }}$. Since $h{\rm{''}}\left( {\frac{1}{{\sqrt 3 }}} \right) > 0$, $h\left( x \right)$ has a minimum at $x = \frac{1}{{\sqrt 3 }}$. So, the minimum of $f$ along the top edge is $h\left( {\frac{1}{{\sqrt 3 }}} \right) = f\left( {\frac{1}{{\sqrt 3 }},1} \right) \simeq - 0.385$.
Since $h\left( x \right)$ has a minimum, the curve is concave up. Thus, for $0 \le x \le 1$ the maximum of $f$ occurs at $x=1$. So, the maximum of $f$ along the top edge is $h\left( 1 \right) = f\left( {1,1} \right) = 0$.
3. On the left edge of the square, $x=0$ and $0 \le y \le 1$. Write $m\left( y \right) = f\left( {0,y} \right) = - {y^2} + y$.
Find the extreme values of $f$ on the left edge.
The critical points of $m$ can be found by solving the equation $m'\left( y \right) = 0$.
$m'\left( y \right) = - 2y + 1 = 0$, ${\ \ \ }$ $m{\rm{''}}\left( x \right) = - 2$
$y = \frac{1}{2}$
Since $m{\rm{''}}\left( y \right) < 0$, $m$ has a maximum at $y = \frac{1}{2}$. So, the maximum of $f$ on the left edge is $m\left( {\frac{1}{2}} \right) = f\left( {0,\frac{1}{2}} \right) = \frac{1}{4}$. Since $m\left( y \right)$ has a maximum, the curve is concave down. Thus, for $0 \le y \le 1$, the minimum of $f$ occurs at $y=0$ and $y=1$. The minimum value of $f$ on the left edge is $m\left( 0 \right) = f\left( {0,0} \right) = 0$ and $m\left( 1 \right) = f\left( {0,1} \right) = 0$.
4. On the right edge of the square, $x=1$ and $0 \le y \le 1$. Write $n\left( y \right) = f\left( {1,y} \right) = 1 - {y^2}$.
Find the extreme values of $f$ on the right edge.
The critical points of $n$ can be found by solving the equation $n'\left( y \right) = 0$.
$n'\left( y \right) = - 2y = 0$
$y = 0$
For $0 \le y \le 1$, $n{\rm{''}}\left( y \right) = - 2 < 0$. So, $n\left( y \right)$ has a maximum at $y=0$. The maximum of $f$ on the right edge is $n\left( 0 \right) = f\left( {1,0} \right) = 1$. Since $n\left( y \right)$ has a maximum, the curve is concave down. Thus, for $0 \le y \le 1$, the minimum of $f$ occurs at $y=1$. The minimum value of $f$ on the right edge is $n\left( 1 \right) = f\left( {1,1} \right) = 0$.
The extrema on the edges are summarized in the following table:
$\begin{array}{*{20}{c}}
{}&{{\rm{Restriction{\ }of}}}\\
{{\bf{Edge}}}&{f\left( {x,y} \right){\rm{to{\ }Edge}}}\\
{Bottom:y = 0,0 \le x \le 1}&{g\left( x \right) = f\left( {x,0} \right) = {x^3}}\\
{Top:y = 1,0 \le x \le 1}&{h\left( x \right) = f\left( {x,1} \right) = {x^3} - x}\\
{Left:x = 0,0 \le y \le 1}&{m\left( y \right) = f\left( {0,y} \right) = - {y^2} + y}\\
{Right:x = 1,0 \le y \le 1}&{n\left( y \right) = f\left( {1,y} \right) = 1 - {y^2}}
\end{array}\begin{array}{*{20}{c}}
{{\rm{min{\ }of}}}&{{\rm{max{\ }of}}}\\
{{\rm{f{\ }on{\ }Edge}}}&{{\rm{f{\ }on{\ }Edge}}}\\
0&1\\
{ - 0.385}&0\\
0&{\frac{1}{4}}\\
0&1
\end{array}$
Step 3. Compare the results
Comparing our results in Step 1 and Step 2, we conclude that the global minimum of $f$ is $f\left( {\frac{1}{{\sqrt 3 }},1} \right) \simeq - 0.385$ and the global maximum is $f\left( {1,0} \right) = 1$.
