5. Let
f
(
x,y,z
) =
x
2
+
xy
+
y
2
+
z
3

4.
a
) Find a point (
x
0
,y
0
,z
0
) satisfying
f
(
x
0
,y
0
,z
0
) = 0
.
Answer:
There are many solutions: The point (
x
0
,y
0
,z
0
) = (1
,
1
,
1) seemed
to be the consensus choice, but (0
,
2
,
0), (2
,
0
,
0), (5
,
1
,

3), (1
,
5
,

3), (

2
,
1
,
1),
(1
,

2
,
1), and (2
,
2
,
2) are other integral solutions. Of course, you are not restricted
to integers, but they often make calculation easier.
b
) Can
x
be expressed as a function
g
(
y,z
) in some neighborhood of (
x
0
,y
0
,z
0
)?
Answer:
We use the point (
x
0
,y
0
,z
0
) = (1
,
1
,
1). Here
∂f/∂x
= 2
x
0
+
y
0
= 3
6
= 0.
The Implicit Function Theorem now implies that we can write
x
=
g
(
y,z
) in some
neighborhood of (1
,
1
,
1).
Note: In this case, we can use the quadratic formula to ﬁnd an expression for
g
,
g
(
y,z
) =

y
+
p
16

3
y
2

4
z
3
2
.
Of course, this is only valid for 16

3
y
2

4
z
3
≥
0. Notice that the positive square
root is chosen so that
g
(
y
0
,z
0
) =
g
(1
,
1) = 1 =
x
0
. Had we used (

2
,
1
,
1) as our
reference point, we would use the negative square root.
c
) Compute
dg
(
y
0
,z
0
).
Answer:
By the Implicit Function Theorem,
dg
(
y
0
,z
0
) =

±
∂f
∂x
²

1
³
∂f/∂y
∂f/∂z
´
=

1
2
x
0
+
y
0
³
x
0
+ 2
y
0
3
z
0
´
=

1
3
³
3
3
´
=
³

1

1
´
.
Notice that this is an easier computation than using the formula for
g
given in
part (
b
).
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 Spring '08
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 Economics, Optimization, ........., 2J, 10L

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