Answer
The minimum value of $f\left( {x,y} \right)$ on the circle is $ - 2\sqrt {13} $ and the maximum value is $ 2\sqrt {13} $.
Work Step by Step
We have $f\left( {x,y} \right) = 3x - 2y$ and the constraint $g\left( {x,y} \right) = {x^2} + {y^2} - 4 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {3, - 2} \right) = \lambda \left( {2x,2y} \right)$
$3 = 2\lambda x$, ${\ \ \ \ }$ $ - 2 = 2\lambda y$
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Since $\left( {0,0} \right)$ does not satisfy the constraint, we may assume that $x \ne 0$ and $y \ne 0$.
From Step 1, we obtain $\lambda = \frac{3}{{2x}} = - \frac{1}{y}$. So $\lambda \ne 0$.
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2, we obtain $y = - \frac{2}{3}x$. Substituting it in the constraint $g\left( {x,y} \right)$ gives
${x^2} + \frac{4}{9}{x^2} - 4 = 0$
$x = \pm \frac{6}{{\sqrt {13} }}$
So, the critical points are $\left( {\frac{6}{{\sqrt {13} }}, - \frac{4}{{\sqrt {13} }}} \right)$, $\left( { - \frac{6}{{\sqrt {13} }},\frac{4}{{\sqrt {13} }}} \right)$.
Step 4. Calculate the critical values
We evaluate $f$ at the critical points and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }points}}}&{f\left( {x,y} \right)}\\
{\left( {\frac{6}{{\sqrt {13} }}, - \frac{4}{{\sqrt {13} }}} \right)}&{2\sqrt {13} }\\
{\left( { - \frac{6}{{\sqrt {13} }},\frac{4}{{\sqrt {13} }}} \right)}&{ - 2\sqrt {13} }
\end{array}$
From this table we conclude that the minimum value of $f\left( {x,y} \right)$ on the circle is $ - 2\sqrt {13} $ and the maximum value is $ 2\sqrt {13} $.
