Answer
The maximum value of $f\left( {x,y} \right)$ on the constraint is $\frac{8}{3}$ and the minimum value is $ - \frac{8}{3}$.
Work Step by Step
We have $f\left( {x,y} \right) = xy$ and the constraint $g\left( {x,y} \right) = 4{x^2} + 9{y^2} - 32 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {y,x} \right) = \lambda \left( {8x,18y} \right)$
$y = 8\lambda x$, ${\ \ \ \ }$ $x = 18\lambda y$
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 $\lambda = \frac{y}{{8x}} = \frac{x}{{18y}}$. So $\lambda \ne 0$.
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2, we obtain ${y^2} = \frac{4}{9}{x^2}$. So, $y = \pm \frac{2}{3}x$.
Substituting it in the constraint $g\left( {x,y} \right)$ gives
$4{x^2} + 9\left( {\frac{4}{9}{x^2}} \right) - 32 = 0$
So, $x = \pm 2$.
Thus, the critical points are $\left( {2,\frac{4}{3}} \right)$, $\left( {2, - \frac{4}{3}} \right)$, $\left( { - 2,\frac{4}{3}} \right)$, $\left( { - 2, - \frac{4}{3}} \right)$.
Step 4. Calculate the critical values
We evaluate $f$ at the critical points and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }point}}}&{f\left( {x,y} \right)}\\
{\left( {2,\frac{4}{3}} \right)}&{\frac{8}{3}}\\
{\left( {2, - \frac{4}{3}} \right)}&{ - \frac{8}{3}}\\
{\left( { - 2,\frac{4}{3}} \right)}&{ - \frac{8}{3}}\\
{\left( { - 2, - \frac{4}{3}} \right)}&{\frac{8}{3}}
\end{array}$
From this table we conclude that the maximum value of $f\left( {x,y} \right)$ on the constraint is $\frac{8}{3}$ and the minimum value is $ - \frac{8}{3}$.
