MATH 215 – Winter 2005
SECOND EXAM
Problem 1.
(5+5+10=20 points) Throughout this problem
f
is the function
f
(
x, y
) =
xye

x

2
y
.
(a) Find all critical points of
f
.
(b) A calculation shows that
f
xx
=
e

x

2
y
(

2
y
+
xy
)
,
f
xy
=
e

x

2
y
(
1

x

2
y
+ 2
xy
)
,
f
yy
=
e

x

2
y
(

4
x
+ 4
xy
)
.
Using these formulas, classify the critical points of
f
.
You do not have to compute the second partials
yourself.
Solution
.
The partial derivatives are:
f
x
=
e

x

2
y
(
y

xy
)
,
f
y
=
e

x

2
y
(
x

2
y
)
.
Since the exponential does not vanish anywhere, the critical points are the solutions to the system
y

xy
= 0
,
x

2
y
= 0
.
If
y
6
= 0
then the first equation implies that
x
= 1
. If
y
= 0
then
x
must be zero, by the 2nd equation.
Therefore, the critical points are:
(0
,
0)
and
(1
,
1
/
2)
. The Hessian is
e

x

2
y

2
y
+
xy
1

2
y

x
+ 2
xy
1

2
y

x
+ 2
xy

4
x
+ 4
xy
¶
At
(0
,
0)
, the Hessian is
e
1
/
2
0
1
1
0
¶
and therefore this point is a saddle point.
At
(1
,
1
/
2)
, the Hessian is
e

2

1
/
2
0
0

2
¶
and therefore this point is a local maximum.
/
(c) Find the absolute maximum of the function
f
on the triangle with vertices at (0
,
0), (1
,
0) and
(0
,
1
/
2).
Hint:
There are not any critical points in the interior of the triangle.
Solution
.
There are no critical points in the
interior
of the region. We must check the boundary, which
consists of 3 sides. On each of the two sides that lie on the coordinate axes, the function is identically
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 Winter '08
 Fish
 Critical Point, Formulas, Multiple integral

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