Answer
The global minimum is $f\left( {0,0} \right) = 0$ and the global maximum is $f\left( {1,1} \right) = 3$.
Work Step by Step
We have $f\left( {x,y} \right) = {x^2} + 2{y^2}$ 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
${f_x} = 2x$, ${\ \ \ \ }$ ${f_y} = 4y$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$.
So, there is only one critical point at $\left( {0,0} \right)$. The extreme value of $f$ is $f\left( {0,0} \right) = 0$.
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:
$\begin{array}{*{20}{c}}
{}\\
{{\rm{Edge}}}\\
{Bottom:y = 0,0 \le x \le 1}\\
{Top:y = 1,0 \le x \le 1}\\
{Left:x = 0,0 \le y \le 1}\\
{Right:x = 1,0 \le y \le 1}
\end{array}\begin{array}{*{20}{c}}
{{\rm{Restriction{\ }of}}}&{{\rm{Minimum{\ }of}}}&{{\rm{Maximum{\ }of}}}\\
{f\left( {x,y} \right){\rm{to{\ }Edge}}}&{f\left( {x,y} \right){\rm{on{\ }Edge}}}&{f\left( {x,y} \right){\rm{on{\ }Edge}}}\\
{g\left( x \right) = f\left( {x,0} \right) = {x^2}}&0&1\\
{h\left( x \right) = f\left( {x,1} \right) = {x^2} + 2}&2&3\\
{m\left( y \right) = f\left( {0,y} \right) = 2{y^2}}&0&2\\
{n\left( y \right) = f\left( {1,y} \right) = 1 + 2{y^2}}&1&3
\end{array}$
Step 3. Compare the results
Comparing the value of $f$ in Step 1 and the values in this table we obtain the global minimum $f\left( {0,0} \right) = 0$ and the global maximum $f\left( {1,1} \right) = 3$.
