Answer
The maximum value of $f\left( {x,y} \right)$ on the constraint is $\sqrt {\frac{{41}}{3}} $ and the minimum value is $ - \sqrt {\frac{{41}}{3}} $.
Work Step by Step
We have $f\left( {x,y,z} \right) = 3x + 2y + 4z$ and the constraint $g\left( {x,y,z} \right) = {x^2} + 2{y^2} + 6{z^2} - 1 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {3,2,4} \right) = \lambda \left( {2x,4y,12z} \right)$
$3 = 2\lambda x$, ${\ \ }$ $2 = 4\lambda y$, ${\ \ }$ $4 = 12\lambda z$
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Since $\left( {0,0,0} \right)$ does not satisfy the constraint, we may assume that $x \ne 0$, $y \ne 0$ and $z \ne 0$.
From Step 1, we obtain $\lambda = \frac{3}{{2x}} = \frac{1}{{2y}} = \frac{1}{{3z}}$. So $\lambda \ne 0$.
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2, we obtain $y = \frac{1}{3}x$ and $z = \frac{2}{9}x$. Substituting these in the constraint $g\left( {x,y,z} \right)$ gives
${x^2} + 2{\left( {\frac{1}{3}x} \right)^2} + 6{\left( {\frac{2}{9}x} \right)^2} - 1 = 0$
$x = \pm 3\sqrt {\frac{3}{{41}}} $
Using $y = \frac{1}{3}x$ and $z = \frac{2}{9}x$, we obtain the critical points:
$\left( {3\sqrt {\frac{3}{{41}}} ,\sqrt {\frac{3}{{41}}} ,\frac{2}{{\sqrt {123} }}} \right)$ and $\left( { - 3\sqrt {\frac{3}{{41}}} , - \sqrt {\frac{3}{{41}}} , - \frac{2}{{\sqrt {123} }}} \right)$.
Step 4. Calculate the critical values
We evaluate $f$ at the critical points and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }points}}}&{f\left( {x,y,z} \right)}\\
{\left( {3\sqrt {\frac{3}{{41}}} ,\sqrt {\frac{3}{{41}}} ,\frac{2}{{\sqrt {123} }}} \right)}&{\sqrt {\frac{{41}}{3}} }\\
{\left( { - 3\sqrt {\frac{3}{{41}}} , - \sqrt {\frac{3}{{41}}} , - \frac{2}{{\sqrt {123} }}} \right)}&{ - \sqrt {\frac{{41}}{3}} }
\end{array}$
From this table we conclude that the maximum value of $f\left( {x,y} \right)$ on the constraint is $\sqrt {\frac{{41}}{3}} $ and the minimum value is $ - \sqrt {\frac{{41}}{3}} $.
