Answer
The maximum value of $f\left( {x,y} \right)$ on the constraint is $8$ and the minimum value is $0$.
Work Step by Step
We have $f\left( {x,y} \right) = {x^2}{y^4}$ and the constraint $g\left( {x,y} \right) = {x^2} + 2{y^2} - 6 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {2x{y^4},4{x^2}{y^3}} \right) = \lambda \left( {2x,4y} \right)$
(1) ${\ \ \ \ }$ $2x{y^4} = 2\lambda x$, ${\ \ \ }$ $4{x^2}{y^3} = 4\lambda y$
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Case 1: $x=0$, $y \ne 0$
Since $x=0$, the second equation of (1) implies that $\lambda=0$.
Substituting $x=0$ in the constraint $g\left( {x,y} \right)$ gives $2{y^2} - 6 = 0$.
$y = \pm \sqrt 3 $
The critical points are $\left( {0,\sqrt 3 } \right)$ and $\left( {0, - \sqrt 3 } \right)$.
Case 2: $x \ne 0$, $y=0$
Since $y=0$, the first equation of (1) implies that $\lambda=0$.
Substituting $y=0$ in the constraint equation $g\left( {x,y} \right)$ gives ${x^2} - 6 = 0$.
$x = \pm \sqrt 6 $
The critical points are $\left( {\sqrt 6 ,0} \right)$ and $\left( { - \sqrt 6 ,0} \right)$.
Case 3: $x \ne 0$ and $y \ne 0$
From Step 1, we obtain $\lambda = {y^4} = {x^2}{y^2}$. So, $\lambda \ne 0$.
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2, we obtain ${y^2} = {x^2}$. So, $y = \pm x$.
Substituting it in the constraint $g\left( {x,y} \right)$ gives
${x^2} + 2{x^2} - 6 = 0$, ${\ \ \ }$ $3{x^2} - 6 = 0$
So, $x = \pm \sqrt 2 $.
Thus, the critical points are $\left( {\sqrt 2 ,\sqrt 2 } \right)$, $\left( {\sqrt 2 , - \sqrt 2 } \right)$, $\left( { - \sqrt 2 ,\sqrt 2 } \right)$, $\left( { - \sqrt 2 , - \sqrt 2 } \right)$.
Step 4. Calculate the critical values
We evaluate $f$ at the critical points and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }point}}}&{f\left( {x,y} \right)}\\
{\left( {0,\sqrt 3 } \right)}&0\\
{\left( {0, - \sqrt 3 } \right)}&0\\
{\left( {\sqrt 6 ,0} \right)}&0\\
{\left( { - \sqrt 6 ,0} \right)}&0\\
{\left( {\sqrt 2 ,\sqrt 2 } \right)}&8\\
{\left( {\sqrt 2 , - \sqrt 2 } \right)}&8\\
{\left( { - \sqrt 2 ,\sqrt 2 } \right)}&8\\
{\left( { - \sqrt 2 , - \sqrt 2 } \right)}&8
\end{array}$
From this table we conclude that the maximum value of $f\left( {x,y} \right)$ on the constraint is $8$ and the minimum value is $0$.
