Answer
The maximum value of $f\left( {x,y,z} \right)$ on the constraint is $20$ and the minimum value is $-20$.
Work Step by Step
We have $f\left( {x,y,z} \right) = xy + 2z$ and the constraint $g\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2} - 36 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {y,x,2} \right) = \lambda \left( {2x,2y,2z} \right)$
(1) ${\ \ \ \ }$ $y = 2\lambda x$, ${\ \ }$ $x = 2\lambda y$, ${\ \ }$ $2 = 2\lambda z$
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
From the third equation of (1) we notice that $\lambda \ne 0$ and $z \ne 0$. So, $z = \frac{1}{\lambda }$.
Since $\left( {0,0,0} \right)$ does not satisfy the constraint, we may have the following cases:
Case 1: $x=0$, $y=0$, $z = \frac{1}{\lambda }$
Substituting $x=0$, $y=0$, $z = \frac{1}{\lambda }$ in $g\left( {x,y} \right)$ gives $\frac{1}{{{\lambda ^2}}} - 36 = 0$. So, $\lambda = \pm \frac{1}{6}$.
So, the critical points are $\left( {0,0,6} \right)$, $\left( {0,0, - 6} \right)$.
Since $\lambda \ne 0$, the first and the second equation of (1) implies that if $x=0$ then $y=0$. Likewise, if $y=0$, then $x=0$.
Case 2: $x \ne 0$, $y \ne 0$, $z = \frac{1}{\lambda }$
Equation (1) becomes $\lambda = \frac{y}{{2x}} = \frac{x}{{2y}} = \frac{1}{z}$. So,
${y^2} = {x^2}$, ${\ \ \ }$ $y = \pm x$
$z = \frac{{2y}}{x} = \pm 2$
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2, we obtain $y = \pm x$ and $z = \pm 2$. Substituting these in the constraint $g\left( {x,y,z} \right)$ gives
${x^2} + {y^2} + {z^2} - 36 = 0$
${x^2} + {x^2} + 4 - 36 = 0$
$x = \pm 4$
Using $\frac{y}{{2x}} = \frac{x}{{2y}} = \frac{1}{z}$ as a guide for the signs, we obtain the critical points: $\left( {4,4,2} \right)$, $\left( {4, - 4, - 2} \right)$, $\left( { - 4,4, - 2} \right)$, $\left( { - 4, - 4,2} \right)$.
Step 4. Calculate the critical values
We evaluate $f$ at the critical points and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }points}}}&{f\left( {x,y,z} \right)}\\
{\left( {0,0,6} \right)}&{12}\\
{\left( {0,0, - 6} \right)}&{ - 12}\\
{\left( {4,4,2} \right)}&{20}\\
{\left( {4, - 4, - 2} \right)}&{ - 20}\\
{\left( { - 4,4, - 2} \right)}&{ - 20}\\
{\left( { - 4, - 4,2} \right)}&{20}
\end{array}$
From this table we conclude that the maximum value of $f\left( {x,y,z} \right)$ on the constraint is $20$ and the minimum value is $-20$.
