HW6 Solutions
14.7 Extreme Values and Saddle Points
3.
critical point
(
2
3
,
4
3
)
⇒
local maximum of
f
(
2
3
,
4
3
)
= 0
14.
critical point (1
,
0)
⇒
local minimum of
f
(1
,
0) = 0.
28.
f
x
(
x, y
) =
−
1
x
2
+
y
= 0 and
f
y
(
x, y
) =
x
−
1
y
2
= 0
⇒
x
= 1 and
y
= 1
⇒
the critical point is (1,1);
f
xx
=
2
x
3
, f
yy
=
2
y
3
, f
xy
= 1;
f
xx
(1
,
1) = 2
, f
yy
(1
,
1) = 2
, f
xy
(1
,
1) = 1
⇒
f
xx
f
yy
−
f
2
xy
= 3
>
0 and
f
xx
= 2
>
0
⇒
local minimum of
f
(1
,
1) = 3
34.
Absolute maximum is 19 at (5
,
3) and the absolute minimum is
−
12 at (4
,
−
2)
41.
T
x
(
x, y
) = 2
x
−
1 = 0 and
T
y
(
x, y
) = 4
y
= 0
⇒
x
=
1
2
and
y
= 0 with
T
(
1
2
,
0) =
−
1
4
;
on the boundary
x
2
+
y
2
= 1;
T
(
x, y
) =
−
x
2
−
x
+ 2 for
−
1
≤
x
≤
1
⇒
T
′
(
x, y
) =
−
2
x
−
1 = 0
⇒
x
=
−
1
2
and
y
=
±
√
3
2
;
T
(
−
1
2
,
√
3
2
) =
9
4
, T
(
−
1
2
,
−
√
3
2
) =
9
4
, T
(
−
1
,
0) = 2
,
and
T
(1
,
0) = 0
⇒
the
hottest is 2
1
4
◦
at (
−
1
2
,
√
3
2
) and (
−
1
2
,
−
√
3
2
); the coldest is
−
1
4
◦
at (
1
2
,
0)
46.
f
will have a saddle point at (0
,
0) if 4
−
k
2
<
0
⇒
k >
2 or
k <
−
2;
f
will have a local minimum
at (0
,
0) if 4
−
k
2
>
0
⇒ −
2
< k <
2; the test is inconclusive if 4
−
k
2
= 0
⇒
k
=
±
2
47.
No, for example
f
(
x, y
) =
xy
has a saddle point at (
a, b
) = (0
,
0) where
f
x
=
f
y
= 0
54. (a)
t
=
π
4
for 0
≤
t
≤
π
i.
On the semi-ellipse: the absolute minimum is
f
(
−
3
,
0) =
−
6 when
t
=
π
and the absolute
maximum is
f
parenleftBig
3