Answer
The global minimum is $f\left( {0,0} \right) = 0$ and the global maximum is $f\left( {0,1} \right) = 2$.
Work Step by Step
We have $f\left( {x,y} \right) = {x^3} + {x^2}y + 2{y^2}$ and the domain: $x,y \ge 0$, $x + y \le 1$.
Step 1. Find the critical points on the domain and evaluate $f$ at these points
The partial derivatives are
${f_x} = 3{x^2} + 2xy$, ${\ \ \ }$ ${f_y} = {x^2} + 4y$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 3{x^2} + 2xy = 0$, ${\ \ }$ ${f_y} = {x^2} + 4y = 0$
From the second equation we get $y = - \frac{1}{4}{x^2}$.
Substituting it in the first equation gives
$3{x^2} - \frac{1}{2}{x^3} = 0$
${x^2}\left( {3 - \frac{1}{2}x} \right) = 0$
The solutions are $x=0$, $x=6$.
So, the critical points are $\left( {0,0} \right)$ and $\left( {6, - 9} \right)$. However, the critical point $\left( {6, - 9} \right)$ is not in our domain. Thus, for the critical point $\left( {0,0} \right)$ in our domain, we get $f\left( {0,0} \right) = 0$.
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:
$\begin{array}{*{20}{c}}
{}&{{\rm{Restriction{\ }of}}}&{}&{f\left( {x,y} \right)}&{f\left( {x,y} \right)}\\
{{\rm{Edge}}}&{f\left( {x,y} \right){\rm{to{\ }Edge}}}&{}&{{\rm{min}}{\rm{.}}}&{{\rm{max}}{\rm{.}}}\\
{Bottom:y = 0,0 \le x \le 1}&{g\left( x \right) = f\left( {x,0} \right) = {x^3}}&{g' = 3{x^2}}&0&1\\
{Left:x = 0,0 \le y \le 1}&{m\left( y \right) = f\left( {0,y} \right) = 2{y^2}}&{m' = 4y}&0&2\\
{Right:0 \le x \le 1,x + y = 1}&{n\left( x \right) = f\left( {x, - x + 1} \right)}&{n' = 6x - 4}&{\frac{2}{3}}&2\\
{}&{\ \ \ \ \ \ \ }{ = 3{x^2} - 4x + 2}
\end{array}$
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 $f\left( {0,1} \right) = 2$.
