problem 1
. First compute:
f
x
=

y
(
x
)
(
x
2
+
y
2
+ 1)
2
= 0 and
f
y
=
(
x
2
+
y
2
+ 1)

y
(2
y
)
(
x
2
+
y
+ 2 + 1)
2
=
x
2

y
2
+ 1
(
x
2
+
y
2
+ 1)
2
Thus, at a critical point,
xy
= 0
and
x
2

y
2
+ 1 = 0
.
The ﬁrst equation says that
x
= 0 or
y
= 0.
If
x
= 0, then by solving the second equation, we ﬁnd that
y
=
±
1.
If
y
= 0, then the second equation has no solutions.
Thus the only critical points are (0
,
1) and (0
,

1). It is easy to check that
f
(0
,
1) =
1
2
and
f
(0
,

1) =

1
2
.
Also, note that
f
(
x, y
)
→
0 as
r
=
p
x
2
+
y
2
→ ∞
. This is true because

y

=
p
y
2
≤
p
x
2
+
y
2
+ 1 =
√
r
2
+ 1, so that

f
(
x, y
)
 ≤
√
r
2
+ 1
r
2
+ 1
=
1
√
r
2
+ 1
.
Then it is easy to check that the limit as
r
→ ∞
= 0, as we asserted above.
It follows that
f
(0
,
1) =
1
2
is the absolute maximum, and
f
(0
,

1) =

1
2
is
the absolute minimum.
problem 2
. Let
x, y
and
z
denote the lengths of the three sides of the
rectangle. Since they are lengths, none of them can be negative.
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 Spring '08
 B.PIGGOTT
 Critical Point, dy, dx dy, gthe ﬁrst octant

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