Answer
The global minimum is $f\left( {0,1} \right) = - 2$ and the global maximum is $f\left( {1,0} \right) = 1$.
Work Step by Step
We have $f\left( {x,y} \right) = {x^3} - 2y$ 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
${f_x} = 3{x^2}$, ${\ \ \ }$ ${f_y} = - 2$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$.
Since ${f_y} = - 2$, it will never becomes zero. Hence, there is no critical point of $f$.
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:
$\begin{array}{*{20}{c}}
{}&{{\rm{Restriction{\ } of}}}\\
{{\rm{Edge}}}&{f\left( {x,y} \right){\rm{ to{\ } Edge}}}\\
{Bottom:y = 0,0 \le x \le 1}&{g\left( x \right) = f\left( {x,0} \right) = {x^3}}\\
{Top:y = 1,0 \le x \le 1}&{h\left( x \right) = f\left( {x,1} \right) = {x^3} - 2}\\
{Left:x = 0,0 \le y \le 1}&{m\left( y \right) = f\left( {0,y} \right) = - 2y}\\
{Right:x = 1,0 \le y \le 1}&{n\left( y \right) = f\left( {1,y} \right) = 1 - 2y}
\end{array}\begin{array}{*{20}{c}}
{{\rm{Minimum{\ } of}}}&{{\rm{Maximum{\ } of}}}\\
{f\left( {x,y} \right){\rm{ on{\ } Edge}}}&{f\left( {x,y} \right){\rm{on{\ } Edge}}}\\
0&1\\
{ - 2}&{ - 1}\\
{ - 2}&0\\
{ - 1}&1
\end{array}$
Step 3. Compare the results
From this table we obtain the global minimum $f\left( {0,1} \right) = - 2$ and the global maximum $f\left( {1,0} \right) = 1$.
