Answer
The global minimum is $f\left( {1,1} \right) = - 1$ and the global maximum is $f\left( {1,0} \right) = f\left( {0,1} \right) = 1$.
Work Step by Step
We have $f\left( {x,y} \right) = {x^3} + {y^3} - 3xy$ 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
${f_x} = 3{x^2} - 3y$, ${\ \ \ }$ ${f_y} = 3{y^2} - 3x$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 3{x^2} - 3y = 0$, ${\ \ }$ ${f_y} = 3{y^2} - 3x = 0$
From the first equation we get $y = {x^2}$.
Substituting it in the second equation gives
$3{x^4} - 3x = 0$
$x\left( {{x^3} - 1} \right) = 0$
The solutions are $x=0$, $x=1$.
So, the critical points of $f$ are $\left( {0,0} \right)$ and $\left( {1,1} \right)$. The values of $f$ corresponding to these critical points are $f\left( {0,0} \right) = 0$ and $f\left( {1,1} \right) = - 1$, respectively.
Step 2. Check the boundaries
We restrict the function $f$ along the edges and find the minimum and maximum values. The results are given in the following table:
$\begin{array}{*{20}{c}}
{}&{{\rm{Restriction{\ }of}}}&{f\left( {x,y} \right)}&{f\left( {x,y} \right)}\\
{{\rm{Edge}}}&{f\left( {x,y} \right){\rm{to{\ }Edge}}}&{{\rm{min}}{\rm{.}}}&{{\rm{max}}{\rm{.}}}\\
{Bottom:y = 0,0 \le x \le 1}&{g\left( x \right) = f\left( {x,0} \right) = {x^3}}&0&1\\
{Top:y = 1,0 \le x \le 1}&{h\left( x \right) = f\left( {x,1} \right) = {x^3} - 3x + 1}&{ - 1}&1\\
{Left:x = 0,0 \le y \le 1}&{m\left( y \right) = f\left( {0,y} \right) = {y^3}}&0&1\\
{Right:x = 1,0 \le y \le 1}&{n\left( y \right) = f\left( {1,y} \right) = {y^3} - 3y + 1}&{ - 1}&1
\end{array}$
Step 3. Compare the results
Comparing the values of $f$ in Step 1 and the values in this table we obtain the global minimum $f\left( {1,1} \right) = - 1$ and the global maximum $f\left( {1,0} \right) = f\left( {0,1} \right) = 1$.
