Answer
The maximum value of $f\left( {x,y} \right)$ on the constraint is $4\sqrt 5 $ and the minimum value is $ - 4\sqrt 5 $.
Work Step by Step
We have $f\left( {x,y} \right) = {x^2}y + x + y$ and the constraint $g\left( {x,y} \right) = xy - 4 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {2xy + 1,{x^2} + 1} \right) = \lambda \left( {y,x} \right)$
(1) ${\ \ \ }$ $2xy + 1 = \lambda y$, ${\ \ }$ ${x^2} + 1 = \lambda x$
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Since ${x^2} + 1 \ge 1$, so the second equation of (1) implies that $\lambda \ne 0$ and $x \ne 0$. As a result, the first equation of (1) implies that $y \ne 0$.
From Step 1, for $x \ne 0$ and $y \ne 0$ we obtain $\lambda = \frac{{2xy + 1}}{y} = \frac{{{x^2} + 1}}{x}$.
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2, we obtain $x\left( {2xy + 1} \right) = y\left( {{x^2} + 1} \right)$. So,
$2{x^2}y + x = {x^2}y + y$
${x^2}y - y + x = 0$
$y\left( {{x^2} - 1} \right) + x = 0$
$y = \frac{x}{{1 - {x^2}}}$
Substituting it in the constraint $g\left( {x,y} \right)$ gives
$\frac{{{x^2}}}{{1 - {x^2}}} - 4 = 0$
${x^2} - 4\left( {1 - {x^2}} \right) = 0$
$5{x^2} = 4$
So, $x = \pm \frac{2}{{\sqrt 5 }}$.
Using $y = \frac{x}{{1 - {x^2}}}$ we obtain the critical points: $\left( {\frac{2}{{\sqrt 5 }},2\sqrt 5 } \right)$ and $\left( { - \frac{2}{{\sqrt 5 }}, - 2\sqrt 5 } \right)$.
Step 4. Calculate the critical values
We evaluate $f$ at the critical points and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }point}}}&{f\left( {x,y} \right)}\\
{\left( {\frac{2}{{\sqrt 5 }},2\sqrt 5 } \right)}&{4\sqrt 5 }\\
{\left( { - \frac{2}{{\sqrt 5 }}, - 2\sqrt 5 } \right)}&{ - 4\sqrt 5 }
\end{array}$
From this table we conclude that the maximum value of $f\left( {x,y} \right)$ on the constraint is $4\sqrt 5 $ and the minimum value is $ - 4\sqrt 5 $.
