Answer
The maximum value of $f\left( {x,y} \right)$ on the constraint is $4\sqrt 2 $ and the minimum value is $ - 4\sqrt 2 $.
Work Step by Step
We have $f\left( {x,y} \right) = \left( {{x^2} + 1} \right)y$ and the constraint ${x^2} + {y^2} = 5$. So, $g\left( {x,y} \right) = {x^2} + {y^2} - 5 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {2xy,{x^2} + 1} \right) = \lambda \left( {2x,2y} \right)$
$2xy = 2\lambda x$, ${\ \ \ }$ ${x^2} + 1 = 2\lambda y$
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
From Step 1, we obtain
(1) ${\ \ \ \ }$ $2x\left( {y - \lambda } \right) = 0$, ${\ \ \ }$ ${x^2} + 1 = 2\lambda y$
Since the left-hand side of the second equation satisfies ${x^2} + 1 \ge 1$, therefore, $y \ne 0$ and $\lambda \ne 0$.
Step 3. Solve for $x$ and $y$ using the constraint
Case 1. $x=0$
The second equation of (1) gives $y = \frac{1}{{2\lambda }}$.
Substituting $x=0$ and $y = \frac{1}{{2\lambda }}$ in $g\left( {x,y} \right) = {x^2} + {y^2} - 5 = 0$ gives
$\frac{1}{{4{\lambda ^2}}} - 5 = 0$
$\lambda = \pm \frac{1}{{2\sqrt 5 }}$
So, $y = \pm \sqrt 5 $.
The critical points are $\left( {0,\sqrt 5 } \right)$ and $\left( {0, - \sqrt 5 } \right)$.
Case 2. $x \ne 0$
The first equation of (1) gives $y=\lambda$.
Substituting it in the second equation of (1) gives
${x^2} + 1 = 2{\lambda ^2}$
${x^2} = 2{\lambda ^2} - 1$
Substituting $y=\lambda$ and ${x^2} = 2{\lambda ^2} - 1$ in $g\left( {x,y} \right) = {x^2} + {y^2} - 5 = 0$ gives
$2{\lambda ^2} - 1 + {\lambda ^2} - 5 = 0$
$3{\lambda ^2} - 6 = 0$
The solutions are $\lambda = \pm \sqrt 2 $.
So $x = \pm \sqrt 3 $ and $y = \pm \sqrt 2 $.
The critical points are $\left( {\sqrt 3 ,\sqrt 2 } \right)$, $\left( {\sqrt 3 , - \sqrt 2 } \right)$, $\left( { - \sqrt 3 ,\sqrt 2 } \right)$, $\left( { - \sqrt 3 , - \sqrt 2 } \right)$.
Step 4. Calculate the critical values
We evaluate $f$ at the critical points and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }points}}}&{f\left( {x,y} \right)}\\
{\left( {0,\sqrt 5 } \right)}&{\sqrt 5 }\\
{\left( {0, - \sqrt 5 } \right)}&{ - \sqrt 5 }\\
{\left( {\sqrt 3 ,\sqrt 2 } \right)}&{4\sqrt 2 }\\
{\left( {\sqrt 3 , - \sqrt 2 } \right)}&{ - 4\sqrt 2 }\\
{\left( { - \sqrt 3 ,\sqrt 2 } \right)}&{4\sqrt 2 }\\
{\left( { - \sqrt 3 , - \sqrt 2 } \right)}&{ - 4\sqrt 2 }
\end{array}$
From this table we conclude that the maximum value of $f\left( {x,y} \right)$ on the constraint is $4\sqrt 2 $ and the minimum value is $ - 4\sqrt 2 $.
