Exercises
14
.
7
:
2)
+(x
,
y)
=
(x
-
3
+
ye
,
Be s
X
=
3
See
y
=
0
2)
minimize
xy2
on
x2
+
y
+
z
=
1
boundary
points
:
com
a
xy
=
122
(1
,
1)
using
y
=
x2
-
f(x
,
xz)
=
(x
-
3)
+
x 4x3
+
2x
-
3
=
x
=
E
= =
f
(1
,
1)
=
(1
-
3)2
+
1
=
S
cross
multiply
-
>
222
=
2x22
results
in
-
>
y
=
X
absolute
max
on
a
cross
mulyx
x
,
y2243
10)
f(x
,
y)
=
xz
-
2xy
+
y2
,
D
=
E(x
,
4)
:
02x26
,
0
=
y
=
12
-
2x3
↳
5
X
S
boundary
PS
(0
,
0
,
16
,
6)
,
10
,
12)
3)
maximize
2x
+
31
+
S2
on
xty
+
z
=
19
!
f10
,
07 0
,
f16
,
b)
=
0
,
flort
=
144
8
f (x
,
y
,
z)
=
(2
,
3
,
5)
=
x8g(x
,
y
,
2)
=
(2x
,
24
,
22)
absolute
min
of
2
at
1010)
=
absolute
minoto
where
absolute
max
of
6
at
boundary
points
.
3)
f(x
,
y)
=
y(x
-
3)
,
D
=
E(x
,
y)
:
x
+
y2293
2y
=
3x
+
y
= 2
x
&
Sx
=
2 -
2
=
Ex
Xy
(xzo
a
X
=
y
=
z
=
t
e
x2
+
x
+
Ex2
=
19
-
x
=
19
=
3
↳
x
=
2
-
>
3sin(G)(3cos(0)
-
3)
=
volume
=
x yz
=
(t)3
=
1/2
al
positive b
maintaineriz
=
192
9
SinEcosO-9
sinG
=
-AsinG
+
9 cOPO-acost
=
O
X
=
0
So
stationary
point
:
10
,
0
↳
coso
=
/2
,
o
=
13
,
5π/3
b)xz
+
y2
+
z2
=
1
,
P(z)
,
2)
1)
3
faces
on
planes
&
Vertex
on
2
=
4-x2y2
.
Maximize
Volume
:
a
=
M
x
(2
+
(y
-
12
+
(2
=
2)2
=
x
=
y
-
x
=
y
&
Of
=
((2x
-
u)
,
(zy
-
2)
,
(22
-
4))
-g
=
(2x
,
zy
,
zz)
en
a
a
E
So
S
coso
=
Sinocoso
~
absolute
max
a
a
2
-
2
=
xz
2
=
2/1
-
X
x
=
z
&
y
=
2/53
Sincecos(o)
=
0
at
0
=
0
,
12
,
i
=
first
actant
-
12
=
2
0
=
0
-
,
o
=
-
, =
=
-
t
absolute
max
of
ats
a
10-1
v
=
(F
,
3)(y)(2)
=
8/3
we
