Answer
The minimum value of $f\left( {x,y} \right)$ on the constraint is $48$.
There is no maximum of $f\left( {x,y} \right)$ on the constraint.
Work Step by Step
We have $f\left( {x,y} \right) = 4{x^2} + 9{y^2}$ and the constraint $g\left( {x,y} \right) = xy - 4 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {8x,18y} \right) = \lambda \left( {y,x} \right)$
$8x = \lambda y$, ${\ \ \ \ }$ $18y = \lambda x$
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 $\lambda = 8\frac{x}{y} = 18\frac{y}{x}$. 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, $y = \pm \frac{2}{3}x$. Since $xy = 4$, therefore, $x$ and $y$ must have the same sign. Thus, $y = \frac{2}{3}x$.
Substituting it in the constraint $g\left( {x,y} \right)$ gives
$\frac{2}{3}{x^2} - 4 = 0$
So, $x = \pm \sqrt 6 $.
Thus, the critical points are $\left( { - \sqrt 6 , - \frac{2}{3}\sqrt 6 } \right)$, $\left( {\sqrt 6 ,\frac{2}{3}\sqrt 6 } \right)$.
Step 4. Calculate the critical values
We evaluate $f$ at the critical points: $\left( { - \sqrt 6 , - \frac{2}{3}\sqrt 6 } \right)$, $\left( {\sqrt 6 ,\frac{2}{3}\sqrt 6 } \right)$.
$f\left( { - \sqrt 6 , - \frac{2}{3}\sqrt 6 } \right) = f\left( {\sqrt 6 ,\frac{2}{3}\sqrt 6 } \right) = 48$
Since $f\left( {x,y} \right) = 4{x^2} + 9{y^2}$ is increasing on the constraint (please see the figure attached), we conclude that the extreme value $48$ is the minimum value of $f\left( {x,y} \right)$ on the constraint.
There is no maximum value of $f\left( {x,y} \right)$ on the constraint.
