Answer
(a) the function $f$ takes on its minimum value $0$ at $\left( {x,y} \right) = \left( {0,0} \right)$
(b) solving the equations ${f_x} = 0$ and ${f_y} = 0$ we obtain the critical points of $f$ consist of the origin $\left( {0,0} \right)$ and the unit circle ${x^2} + {y^2} = 1$.
(c) we prove that $f$ takes on its maximum value on the unit circle by analyzing the function $g\left( t \right) = t{{\rm{e}}^{ - t}}$ for $t > 0$, where $t \equiv {x^2} + {y^2}$.
Work Step by Step
(a) Since ${{\rm{e}}^{ - {x^2} - {y^2}}} > 0$ and ${x^2} + {y^2} \ge 0$, the function $f\left( {x,y} \right) = \left( {{x^2} + {y^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$ takes on its minimum value $0$ at $\left( {x,y} \right) = \left( {0,0} \right)$. Thus, by Theorem 1, $\left( {0,0} \right)$ is a critical point of $f$.
(b) The partial derivatives are
${f_x} = 2x{{\rm{e}}^{ - {x^2} - {y^2}}} - 2x\left( {{x^2} + {y^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
$ = 2x\left( {1 - \left( {{x^2} + {y^2}} \right)} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
${f_y} = 2y{{\rm{e}}^{ - {x^2} - {y^2}}} - 2y\left( {{x^2} + {y^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
$ = 2y\left( {1 - \left( {{x^2} + {y^2}} \right)} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 2x\left( {1 - \left( {{x^2} + {y^2}} \right)} \right){{\rm{e}}^{ - {x^2} - {y^2}}} = 0$
${f_y} = 2y\left( {1 - \left( {{x^2} + {y^2}} \right)} \right){{\rm{e}}^{ - {x^2} - {y^2}}} = 0$
Since ${{\rm{e}}^{ - {x^2} - {y^2}}} \ne 0$, we obtain two simultaneous equations
$2x\left( {1 - \left( {{x^2} + {y^2}} \right)} \right) = 0$
$2y\left( {1 - \left( {{x^2} + {y^2}} \right)} \right) = 0$
From the first equation, we obtain the solutions $x=0$, and ${x^2} + {y^2} = 1$.
Substituting $x=0$ in the second equation gives
$2y\left( {1 - {y^2}} \right) = 0$
The solutions are $y=0$, $y = \pm 1$.
Thus, the critical points are $\left( {0,0} \right)$, $\left( {0,1} \right)$, $\left( {0, - 1} \right)$ and the equation ${x^2} + {y^2} = 1$.
Notice that $\left( {0,1} \right)$ and $\left( {0, - 1} \right)$ are points on ${x^2} + {y^2} = 1$.
Together with the result from part (a), therefore, the set of critical points of $f$ consist of the origin $\left( {0,0} \right)$ and the unit circle ${x^2} + {y^2} = 1$.
(c) We have $f\left( {x,y} \right) = \left( {{x^2} + {y^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$.
We are given the fact that the Second Derivative Test fails. However, to determine the extreme values, let us analyze the function $g\left( t \right) = t{{\rm{e}}^{ - t}}$ for $t > 0$, where $t \equiv {x^2} + {y^2}$.
The derivatives are
$g'\left( t \right) = {{\rm{e}}^{ - t}} - t{{\rm{e}}^{ - t}} = \left( {1 - t} \right){{\rm{e}}^{ - t}}$
$g{\rm{''}}\left( t \right) = - {{\rm{e}}^{ - t}} - \left( {1 - t} \right){{\rm{e}}^{ - t}} = \left( { - 2 + t} \right){{\rm{e}}^{ - t}}$
The critical points of $g$ is obtained by solving $g'\left( t \right) = \left( {1 - t} \right){{\rm{e}}^{ - t}} = 0$.
Since ${{\rm{e}}^{ - t}} \ne 0$, we get $t=1$. So, the critical point of $g$ is at $t=1$.
Substituting $t=1$ in $g{\rm{''}}\left( t \right)$ gives $g{\rm{''}}\left( 1 \right) = - \frac{1}{{\rm{e}}} < 0$.
Since $g{\rm{''}}\left( 1 \right) < 0$, by Theorem 3 of Section 4.4, $g$ has local maximum at $t=1$.
Since $t = {x^2} + {y^2}$, Thus, $f\left( {x,y} \right) = \left( {{x^2} + {y^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$ has local maximum at ${x^2} + {y^2} = 1$. Hence, we conclude that $f$ takes on its maximum value on the unit circle ${x^2} + {y^2} = 1$.
