Answer
The minimum value of $f\left( {x,y} \right) = {x^2}y$ on the ellipse $4{x^2} + 9{y^2} = 36$ is $ - 4\sqrt 3 $ and the maximum value is $4\sqrt 3 $.
Work Step by Step
We have $f\left( {x,y} \right) = {x^2}y$ and the constraint $g\left( {x,y} \right) = 4{x^2} + 9{y^2} - 36 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {2xy,{x^2}} \right) = \lambda \left( {8x,18y} \right)$
(1) ${\ \ \ \ }$ $2xy = 8\lambda x$, ${\ \ \ }$ ${x^2} = 18\lambda y$
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Since $\left( {0,0} \right)$ does not satisfy the constraint, we have the following cases:
Case 1. $x=0$ and $y \ne 0$
In this case $\lambda = 0$. Substituting $x=0$ in the constraint $g\left( {x,y} \right) = 4{x^2} + 9{y^2} - 36 = 0$ gives $y = \pm 2$.
So, the critical points are $\left( {0,2} \right)$ and $\left( {0, - 2} \right)$.
Notice that the second equation of (1) does not allow the case $x \ne 0$ and $y=0$.
Case 2. $x \ne 0$ and $y \ne 0$
From Step 1, we obtain $\lambda = \frac{y}{4} = \frac{{{x^2}}}{{18y}}$. So $\lambda \ne 0$.
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2, we obtain ${y^2} = \frac{2}{9}{x^2}$. So, $y = \pm \frac{{\sqrt 2 }}{3}x$.
Substituting it in the constraint $g\left( {x,y} \right) = 4{x^2} + 9{y^2} - 36 = 0$ gives
$4{x^2} + 9\left( {\frac{2}{9}{x^2}} \right) - 36 = 0$
$6{x^2} = 36$
So, $x = \pm \sqrt 6 $.
Using $y = \pm \frac{{\sqrt 2 }}{3}x$, we obtain the critical points: $\left( {\sqrt 6 ,\frac{2}{{\sqrt 3 }}} \right)$, $\left( {\sqrt 6 , - \frac{2}{{\sqrt 3 }}} \right)$, $\left( { - \sqrt 6 ,\frac{2}{{\sqrt 3 }}} \right)$, $\left( { - \sqrt 6 , - \frac{2}{{\sqrt 3 }}} \right)$.
Step 4. Calculate the critical values
We evaluate $f$ at the critical points and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }point}}}&{f\left( {x,y} \right)}\\
{\left( {0,2} \right)}&0\\
{\left( {0, - 2} \right)}&0\\
{\left( {\sqrt 6 ,\frac{2}{{\sqrt 3 }}} \right)}&{4\sqrt 3 }\\
{\left( {\sqrt 6 , - \frac{2}{{\sqrt 3 }}} \right)}&{ - 4\sqrt 3 }\\
{\left( { - \sqrt 6 ,\frac{2}{{\sqrt 3 }}} \right)}&{4\sqrt 3 }\\
{\left( { - \sqrt 6 , - \frac{2}{{\sqrt 3 }}} \right)}&{ - 4\sqrt 3 }
\end{array}$
From this table we conclude that the minimum value of $f\left( {x,y} \right) = {x^2}y$ on the ellipse $4{x^2} + 9{y^2} = 36$ is $ - 4\sqrt 3 $ and the maximum value is $4\sqrt 3 $.
