HW10 – gilbert – (55035)
2
while f
Y
= 0 when
∂f
∂y
=
−3y
2
+ 3x
=
0
,
i.e.,
x
=
y
2
.
Substituting the first into the second
yields x = (−x
2
)
2
= x
4
,
which can be written as
x(1 − x
3
)
=
0 ,
i.e.,
x
=
0,1 .
Thus, the critical points are
(0, 0), (1, −1) .
and to classify these critical points we
use the Second Derivative test. Now
f
XX
=
6x ,
f
YY
=
−6y ,
f
XY
=
3 .
But then, at (0, 0),
A = f
XX
(0, 0) = 0,
B = f
XY
(0, 0) = 3 ,
while
C = f
YY
(0, 0) = 0 .
Consequently,
D = AC − B
2
= −9 < 0 . and
so there is a
saddle point at (0, 0)
.
On the other hand, at (1, −1),
A = f
XX
(1, −1) = 6,
B = f
XY
(1, −1) = 3 ,
while
C
=
f
YY
(1, −1)
=
6 .
Thus
D = AC − B
2
= 27 > 0 . and
so, since A, C > 0, there is a
local minimum at (1, −1)
.
keywords:
003 (part 1 of 3) 10.0 points
If f is defined by
f (x, y)
=
4
3
x
3
+ x
2
+ 4xy + 2y
2
+ 2y ,
(i)
locate the critical points of f .
1.
₃
−
2
1
, 0
₃
,
₃
−1,
2
3
₃
2.
₃