Answer
The maximum value of $f\left( {x,y,z} \right)$ on the constraint is $2\sqrt 2 $ and the minimum value is $ - 2\sqrt 2 $.
Work Step by Step
We have $f\left( {x,y,z} \right) = xy + xz$ and the constraint $g\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2} - 4 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {y + z,x,x} \right) = \lambda \left( {2x,2y,2z} \right)$
(1) ${\ \ \ \ }$ $y + z = 2\lambda x$, ${\ \ }$ $x = 2\lambda y$, ${\ \ }$ $x = 2\lambda z$
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Case 1: $x=0$
The first equation of (1) gives $y+z=0$. So, $z=-y$.
The constraint $g\left( {x,y,z} \right)$ becomes
${y^2} + {y^2} - 4 = 0$
$2{y^2} = 4$
$y = \pm \sqrt 2 $
So, $z = \mp \sqrt 2 $.
The second and the third equation of (1) imply that in this case $\lambda=0$.
So, the critical points are $\left( {0,\sqrt 2 , - \sqrt 2 } \right)$, $\left( {0, - \sqrt 2 ,\sqrt 2 } \right)$.
Case 2: $y=0$
The second equation of (1) gives $x=0$. As a result, the first equation of (1) gives $z=0$.
Since $\left( {0,0,0} \right)$ does not satisfy the constraint, there is no critical point in this case.
Case 3: $z=0$
The third equation of (1) gives $x=0$. As a result, the first equation of (1) gives $y=0$.
Similar to case 2 above, since $\left( {0,0,0} \right)$ does not satisfy the constraint, there is no critical point in this case.
Case 4: $x \ne 0$, $y \ne 0$, $z \ne 0$
From equation (1) we obtain $\lambda = \frac{{y + z}}{{2x}} = \frac{x}{{2y}} = \frac{x}{{2z}}$.
We have $\frac{x}{{2y}} = \frac{x}{{2z}}$. So, $z=y$.
We have $\frac{{y + z}}{{2x}} = \frac{x}{{2y}}$. Using $z=y$, we get $\frac{{2y}}{{2x}} = \frac{x}{{2y}}$.
${y^2} = \frac{1}{2}{x^2}$
$y = \pm \frac{1}{{\sqrt 2 }}x$
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2, we obtain $y = \pm \frac{1}{{\sqrt 2 }}x$ and $z=y$. Substituting these in the constraint $g\left( {x,y,z} \right)$ gives
${x^2} + {y^2} + {z^2} - 4 = 0$
${x^2} + \frac{1}{2}{x^2} + \frac{1}{2}{x^2} - 4 = 0$
$2{x^2} = 4$
$x = \pm \sqrt 2 $
So, the critical points are $\left( {\sqrt 2 ,1,1} \right)$, $\left( {\sqrt 2 , - 1, - 1} \right)$, $\left( { - \sqrt 2 ,1,1} \right)$, $\left( { - \sqrt 2 , - 1, - 1} \right)$.
Step 4. Calculate the critical values
We evaluate $f$ at the critical points and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }points}}}&{f\left( {x,y,z} \right)}\\
{\left( {0,\sqrt 2 , - \sqrt 2 } \right)}&0\\
{\left( {0, - \sqrt 2 ,\sqrt 2 } \right)}&0\\
{\left( {\sqrt 2 ,1,1} \right)}&{2\sqrt 2 }\\
{\left( {\sqrt 2 , - 1, - 1} \right)}&{ - 2\sqrt 2 }\\
{\left( { - \sqrt 2 ,1,1} \right)}&{ - 2\sqrt 2 }\\
{\left( { - \sqrt 2 , - 1, - 1} \right)}&{2\sqrt 2 }
\end{array}$
From this table we conclude that the maximum value of $f\left( {x,y,z} \right)$ on the constraint is $2\sqrt 2 $ and the minimum value is $ - 2\sqrt 2 $.
