Answer
We show that the Lagrange equations for $f$ subject to the constraint have solutions, but that $f$ has no min or max on the constraint curve.
However, this does not contradict Theorem 1.
Work Step by Step
We have $f\left( {x,y} \right) = 2x + y$ subject to the constraint $g\left( {x,y} \right) = {x^2} - {y^2} - 1 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {2,1} \right) = \lambda \left( {2x, - 2y} \right)$
So, the Lagrange equations are
$1 = \lambda x$, ${\ \ \ }$ $1 = - 2\lambda y$
These equations imply that $x \ne 0$, $y \ne 0$ and $\lambda \ne 0$.
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
From the results in Step 1, we obtain
$\lambda = \frac{1}{x} = - \frac{1}{{2y}}$
$y = - \frac{x}{2}$
Step 3. Solve for $x$ and $y$ using the constraint
Substituting $y = - \frac{x}{2}$ in the constraint gives
${x^2} - {\left( { - \frac{x}{2}} \right)^2} - 1 = 0$
$\frac{3}{4}{x^2} = 1$
So, $x = \pm \frac{2}{{\sqrt 3 }}$.
Using $y = - \frac{x}{2}$, we obtain the critical points: $\left( {\frac{2}{{\sqrt 3 }}, - \frac{1}{{\sqrt 3 }}} \right)$, $\left( { - \frac{2}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)$.
Step 4. Calculate the critical values
We evaluate the extreme values at the critical points and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }point}}}&{f\left( {x,y} \right)}\\
{\left( {\frac{2}{{\sqrt 3 }}, - \frac{1}{{\sqrt 3 }}} \right)}&{\sqrt 3 }\\
{\left( { - \frac{2}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)}&{ - \sqrt 3 }
\end{array}$
From the table, it seems that the minimum value and the maximum value of $f$ are $ - \sqrt 3 $ and $\sqrt 3 $, respectively. However, referring to the figure attached, the point $\left( { - \frac{2}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)$ does not corresponds to the minimum value of $f$ since there are points on the left level curves subject to the constraint, which have smaller values of $f$. Similarly, the point $\left( {\frac{2}{{\sqrt 3 }}, - \frac{1}{{\sqrt 3 }}} \right)$ does not corresponds to the maximum value of $f$ since there are points on the right level curves subject to the constraint, which have larger values of $f$.
Alternatively, we can show that $f$ has no min or max on the constraint curve by writing $f$ as a function of single variable:
We have the constraint $g\left( {x,y} \right) = {x^2} - {y^2} - 1 = 0$. So, $y = \pm \sqrt {{x^2} - 1} $.
Substituting the constraint $y = \pm \sqrt {{x^2} - 1} $ in $f$ gives
$f\left( {x,y} \right) = 2x \pm \sqrt {{x^2} - 1} $
As $x \to \infty $, so $f \to \infty $. Thus, there is no maximum value of $f$.
Similarly, as $x \to - \infty $, so $f \to - \infty $. Thus, there is no minimum value of $f$. Hence, we conclude that $f$ has no min or max on the constraint curve.
Hence, the Lagrange equations have a solution but that $f$ has no min or max on the constraint curve. This result does not contradict Theorem 1. Recall from Theorem 1 which states that if $f\left( {x,y} \right)$ has a local minimum or a local maximum on the constraint curve, then the Lagrange condition holds, hence the Lagrange equations have solutions. But, the theorem does not guarantee the reverse.
