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MATH 210
Selfquiz 6
March 01, 2009
1. Find and classify the critical points of the function
f
(
x, y
) =
x
3

3
xy
+
y
3
.
Solution:
We need to ﬁnd the ﬁrst partial derivatives and set them equal
to 0. We have:
f
x
(
x, y
) = 3
x
2

3
y
f
y
(
x, y
) = 3
y
2

3
x
If we set the above two functions equal to 0, then we get:
x
2

y
= 0
y
2

x
= 0
This means that;
x
2
=
y
y
2
=
x
By subsituting the expression for
y
from the ﬁrst equation, into the second
we get:
x
4

x
= 0. This can be factorized as:
x
(
x

1)(
x
2
+
x
+ 1) = 0
that has roots 0
,
1. If
x
= 0 then
y
= 0 (by the ﬁrst equation). Also if
x
= 1, then
y
= 1. We, thus, get two critical points (0
,
0) and (1
,
1). We
will apply the 2nd derivative test for each one of them. To this end we need
to compute the 2nd order partial derivatives. We have:
f
xx
(
x, y
) = 6
x

3
f
xy
(
x, y
) =

3
f
yy
(
x, y
) = 6
y

3
We begin our computation with (0
,
0). At that point, we get that:
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=
±
±
±
±

3

3

3

3
±
±
±
±
= 0
The 2nd derivative test is inconclusive for this point. The dominant term
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 Spring '08
 SLODOWSKI
 Critical Point, Derivative

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