Answer
The global minimum is $f\left( { - 0.43,0.9} \right) = f\left( { - 0.43, - 0.9} \right) = - 0.52$ and the global maximum is $f\left( {0.77,0.64} \right) = f\left( {0.77, - 0.64} \right) = 1.22$.
Work Step by Step
We have $f\left( {x,y} \right) = {x^2} + 2x{y^2}$ on the domain ${x^2} + {y^2} \le 1$.
Step 1. Find the critical points on the domain and evaluate $f$ at these points
The partial derivatives are
${f_x} = 2x + 2{y^2}$ ${\ \ \ }$ ${f_y} = 4xy$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 2x + 2{y^2} = 0$ ${\ \ }$ ${f_y} = 4xy = 0$
From the second equation we obtain $x=0$ or $y=0$.
Substituting $x=0$ in the first equation gives $y=0$. Substituting $y=0$ in the first equation gives $x=0$. So, there is only one critical point at $\left( {0,0} \right)$. The extreme value is $f\left( {0,0} \right) = 0$.
Step 2. Check the boundaries
The boundary is the circle: ${x^2} + {y^2} = 1$.
We restrict the function $f$ along the boundary and obtain
$g\left( x \right) = {x^2} + 2x\left( {1 - {x^2}} \right) = - 2{x^3} + {x^2} + 2x$
We find the extreme value of $g$ by solving the equation $g'\left( x \right) = 0$:
$g'\left( x \right) = - 6{x^2} + 2x + 2 = 0$
$3{x^2} - x - 1 = 0$
The solutions are $x = \frac{{1 \pm \sqrt {1 + 12} }}{6} = \frac{{1 \pm \sqrt {13} }}{6}$.
So, the critical points are $x \simeq 0.77$ and $x \simeq - 0.43$. The corresponding extreme values are given below
$\begin{array}{*{20}{c}}
x&y&{f\left( {x,y} \right)}\\
{0.77}&{0.64}&{1.22}\\
{0.77}&{ - 0.64}&{1.22}\\
{ - 0.43}&{0.9}&{ - 0.52}\\
{ - 0.43}&{ - 0.9}&{ - 0.52}
\end{array}$
Step 3. Compare the results
Comparing the results from Step 1 and Step 2 we obtain the global minimum $f\left( { - 0.43,0.9} \right) = f\left( { - 0.43, - 0.9} \right) = - 0.52$ and the global maximum $f\left( {0.77,0.64} \right) = f\left( {0.77, - 0.64} \right) = 1.22$.
