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HW10 – gilbert – (55035)
1
This printout should have 16 questions.
Multiplechoice questions may continue
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choices before answering.
001
10.0 points
In the contour map below identify the
points P, Q, and R as local minima, local
maxima, or neither.
3
2
1
0
1
2
P
Q
0
R
2
1
0
1
2
3
A. local maximum at Q,
B. local minimum at R,
C. local minimum at P .
1. A and C only
2. A and B only
3. B and C only
4. C only correct
5. none of them
6. A only
7. B only
8. all of them
Explanation:
A.
FALSE: the point Q lies on the 0
contour and this contour divides the region
near Q into two regions. In one region
the contours have values increasing to
0, while in the other the contours have
values decreasing to 0. So the surface
does not have a local minimum at Q.
B.
FALSE: the contours near R are closed
curves enclosing R and the contours increase
in value as we approch R. So the surface has
a
local maximum at R, not a local minimum.
C.
TRUE: the contours near P are closed
curves enclosing P and the contours decrease
in value as we approch P . So the surface has
a
local minimum at P .
keywords: contour map, local extrema,
True/False,
002
10.0 points
Locate and classify all the local extrema of
f (x, y)
=
x
3
− y
3
+ 3xy − 2 .
1. local max at (1, −1),
local min at (0, 0)
2. local max at (1, −1),
saddle point at (0, 0)
3. local min at (0, 0),
saddle point at (1, −1)
4. local min at (1, −1),
saddle point at (0, 0) correct
5. local max at (0, 0),
saddle point at (1, −1)
Explanation:
Since f has derivatives everywhere, the
critical points occur at the solutions of
∇
f (x, y)
=
f
X
i + f
Y
j
=
0 .
But f
X
= 0 when
∂f
∂x
=
3x
2
+ 3y
=
0
,
i.e.,
y
=
−x
2
,
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View Full DocumentHW10 – 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
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 Spring '07
 Sadler

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