Answer
The global minimum is $f\left( {0,0} \right) = 0$ and the global maximum is $f\left( {1,0} \right) = f\left( {0,1} \right) = 1$.
Work Step by Step
We have $f\left( {x,y} \right) = {x^2} + x{y^3} + {y^2}$ 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
${f_x} = 2x + {y^3}$, ${\ \ \ }$ ${f_y} = 3x{y^2} + 2y$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 2x + {y^3} = 0$, ${\ \ }$ ${f_y} = 3x{y^2} + 2y = 0$
From the first equation we get $x = - \frac{1}{2}{y^3}$.
Substituting it in the second equation gives
$ - \frac{3}{2}{y^5} + 2y = 0$
$y\left( {2 - \frac{3}{2}{y^4}} \right) = 0$
$y=0$, ${\ \ \ }$ ${y^4} - \frac{4}{3} = 0$
We can write the second equation as
$\left( {{y^2} - \frac{2}{{\sqrt 3 }}} \right)\left( {{y^2} + \frac{2}{{\sqrt 3 }}} \right) = 0$
For real solutions, we obtain $y=0$, $y = \pm {\left( {\frac{2}{{\sqrt 3 }}} \right)^{1/2}} = \pm \left( {\frac{{\sqrt 2 }}{{{3^{1/4}}}}} \right)$.
So, the critical points are $\left( {0,0} \right)$, $\left( { - \frac{{\sqrt 2 }}{{{3^{3/4}}}},\frac{{\sqrt 2 }}{{{3^{1/4}}}}} \right) \simeq \left( { - 0.62,1.07} \right)$ and $\left( {\frac{{\sqrt 2 }}{{{3^{3/4}}}}, - \frac{{\sqrt 2 }}{{{3^{1/4}}}}} \right) \simeq \left( {0.62, - 1.07} \right)$. However, only the critical point $\left( {0,0} \right)$ is in the interior of our domain. The extreme value of $f$ corresponding to $\left( {0,0} \right)$ 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:
$\begin{array}{*{20}{c}}
{}&{{\rm{Restriction{\ }of}}}&{f\left( {x,y} \right)}&{f\left( {x,y} \right)}\\
{{\rm{Edge}}}&{f\left( {x,y} \right){\rm{to{\ }Edge}}}&{{\rm{min}}{\rm{.}}}&{{\rm{max}}{\rm{.}}}\\
{Bottom:y = 0,0 \le x \le 1}&{g\left( x \right) = f\left( {x,0} \right) = {x^2}}&0&1\\
{Left:x = 0,0 \le y \le 1}&{m\left( y \right) = f\left( {0,y} \right) = {y^2}}&0&1\\
{Right:0 \le x \le 1,x + y = 1}&{n\left( x \right) = f\left( {x, - x + 1} \right)}&{0.55}&1\\
{}&{ = - {x^4} + 3{x^3} - {x^2} - x + 1}&{}&{}
\end{array}$
Notes:
- The critical point of $n\left( x \right)$ is $x \simeq 0.56$, so the extreme value $n\left( {0.56} \right) \simeq 0.55$.
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 $f\left( {1,0} \right) = f\left( {0,1} \right) = 1$.
