(x2 + y2)ey2-x2 Solution. fx

§14.7 #15. Find the local maximum and minimum values and saddle
point(s) of the function.
Solution.
f(x, y) = (x 2 + y 2 )e y2 −x 2
fx(x, y) = 2xe y2−x2 + (x 2 + y 2 )(−2x)e y2−x2 = (−2x 3 + 2x − 2xy 2 )e y2 −x 2
fy(x, y) = 2ye y2−x2 + (x 2 + y 2 )(2y)e y2−x2 = (2y 3 + 2y + 2x 2 y)e y2 −x 2
(−2x 3 + 2x − 2xy 2 )e y2 −x 2
(2y 3 + 2y + 2x 2 y)e y2 −x 2
−2x(x 2 − 1 + y 2 ) = 0
2y(y 2 + 1 + x 2 ) = 0
= 0
= 0
If x = 0 (first equation), then the second equation becomes
2y(y 2 + 1) = 0,
so y = 0. We get a critical point (0, 0).
If y = 0 (second equation), then the first equation becomes
−2x(x 2 − 1) = 0,
so x = 0 or x = ±1. We get a critical points (0, 0), (±1, 0).
This leaves the case
x 2 − 1 + y 2 = 0
y 2 + 1 + x 2 = 0
Adding the two equations gives 2x 2 + 2y 2 = 0, which has solution (0, 0).
So the critical points are (0, 0), (1, 0), and (−1, 0).
fxx(x, y) = (−6x 2 + 2 − 2y 2 )e y2−x2 + (−2x 3 + 2x − 2xy 2 )(−2x)e y2−x2 fyy(x, y) = (6y 2 + 2 + 2x 2 )e y2−x2 + (2y 3 + 2y + 2x 2 y)(2y)e y2−x2 fxy(x, y) = (4xy)e y2−x2 + (−2x 3 + 2x − 2xy 2 )(2y)e y2−x2 1

• Critical point (0, 0).
fxx(0, 0) = 2, fyy(0, 0) = 2, fxy(0, 0) = 0
D(0, 0) =
fxx
fxy
=
2
0
0
2 = 4 > 0
fyx fyy
For the critical point (0, 0), D(0, 0) = 4 > 0, fxx(0, 0) = 4 > 0, therefore
(0, 0) is a local minimum. We have f(0, 0) = 0.
• Critical points (±1, 0).
fxx(±1, 0) = −4, fyy(±1, 0) = 2, fxy(±, 0) = 0
D(±1, 0) =
fxx
fxy
=
−4
0
0
2 = −8 < 0
fyx fyy
For the critical points (±1, 0), D(±1, 0) = −8 < 0, therefore (±1, 0)
are saddle points. We have f(±1, 0) = 1/e.
2