Find the maximum and minimum value of X^3Y^2(1-x-y)
Answers
Step-by-step explanation:
We can expand
f
to
f
(
x
,
y
)
=
x
y
−
x
2
y
−
x
y
2
. Next, find the partial derivatives and set them equal to zero.
∂
f
∂
x
=
y
−
2
x
y
−
y
2
=
y
(
1
−
2
x
−
y
)
=
0
∂
f
∂
y
=
x
−
x
2
−
2
x
y
=
x
(
1
−
x
−
2
y
)
=
0
Clearly,
(
x
,
y
)
=
(
0
,
0
)
,
(
1
,
0
)
,
and
(
0
,
1
)
are solutions to this system, and so are critical points of
f
. The other solution can be found from the system
1
−
2
x
−
y
=
0
,
1
−
x
−
2
y
=
0
. Solving the first equation for
y
in terms of
x
gives
y
=
1
−
2
x
, which can be plugged into the second equation to get
1
−
x
−
2
(
1
−
2
x
)
=
0
⇒
−
1
+
3
x
=
0
⇒
x
=
1
3
. From this,
y
=
1
−
2
(
1
3
)
=
1
−
2
3
=
1
3
as well.
To test the nature of these critical points, we find second derivatives:
∂
2
f
∂
x
2
=
−
2
y
,
∂
2
f
∂
y
2
=
−
2
x
, and
∂
2
f
∂
x
∂
y
=
∂
2
f
∂
y
∂
x
=
1
−
2
x
−
2
y
.
The discriminant is therefore:
D
=
4
x
y
−
(
1
−
2
x
−
2
y
)
2
=
4
x
y
−
(
1
−
2
x
−
2
y
−
2
x
+
4
x
2
+
4
x
y
−
2
y
+
4
x
y
+
4
y
2
)
=
4
x
+
4
y
−
4
x
2
−
4
y
2
−
4
x
y
−
1
Plugging the first three critical points in gives:
D
(
0
,
0
)
=
−
1
<
0
,
D
(
1
,
0
)
=
4
−
4
−
1
=
−
1
<
0
, and
D
(
0
,
1
)
=
4
−
4
−
1
=
−
1
<
0
, making these points saddle points.
Plugging in the last critical point gives
D
(
1
3
,
1
3
)
=
4
3
+
4
3
−
4
9
−
4
9
−
4
9
−
1
=
1
3
>
0
. Also note that
∂
2
f
∂
x
2
(
1
3
,
1
3
)
=
−
2
3
<
0
. Therefore,
(
1
3
,
1
3
)
is a location of a local maximum value of
f
. You can check that the local maximum value itself is
f
(
1
3
,
1
3
)
=
1
27
.
Below is a picture of the contour map (of level curves) of
f
(the curves where the output of
f
is constant), along with the 4 critical points of
f
.
Answer:
Discuss the maximum or minimum value of :
=
3
2
(1 − − )