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