Examine for extreme value
F(x,y)=(y²-x²)e^y
Answers
Answer:
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Step-by-step explanation:
Answer
Steve M
Feb 11, 2018
We have:
f
(
x
,
y
)
=
x
y
+
e
−
x
2
−
y
2
Step 1 - Find the Partial Derivatives
We compute the partial derivative of a function of two or more variables by differentiating wrt one variable, whilst the other variables are treated as constant. Thus:
The First Derivatives are:
f
x
=
y
+
e
−
x
2
−
y
2
(
−
2
x
)
=
y
−
2
x
e
−
x
2
−
y
2
f
y
=
x
+
e
−
x
2
−
y
2
(
−
2
y
)
=
x
−
2
y
e
−
x
2
−
y
2
The Second Derivatives (quoted) are:
f
x
x
=
−
2
e
−
x
2
−
y
2
+
4
x
2
e
−
x
2
−
y
2
f
y
y
=
−
2
e
−
x
2
−
y
2
+
4
y
2
e
−
x
2
−
y
2
The Second Partial Cross-Derivatives are:
f
x
y
=
1
+
4
x
y
e
−
x
2
−
y
2
f
y
x
=
1
+
4
x
y
e
−
x
2
−
y
2
Note that the second partial cross derivatives are identical due to the continuity of
f
(
x
,
y
)
.
Step 2 - Identify Critical Points
A critical point occurs at a simultaneous solution of
f
x
=
f
y
=
0
⇔
∂
f
∂
x
=
∂
f
∂
y
=
0
i.e, when:
f
x
=
y
−
2
x
e
−
x
2
−
y
2
=
0
...
[
A
]
f
y
=
x
−
2
y
e
−
x
2
−
y
2
=
0
...
[
B
]
}
simultaneously
From which we can establish:
[
A
]
⇒
y
−
2
x
e
−
x
2
−
y
2
=
0
⇒
e
−
x
2
−
y
2
=
y
2
x
[
B
]
⇒
x
−
2
y
e
−
x
2
−
y
2
=
0
⇒
e
−
x
2
−
y
2
=
x
2
y
Thus we require that:
y
2
x
=
x
2
y
∴
x
2
=
y
2
Then we have two (infinite plane) solutions:
∴
x
=
±
y
And so we conclude there are infinitely many critical points along the entire lengths of the intersection of the curve and the two planes
x
=
±
y
Step 3 - Classify the critical points
In order to classify the critical points we perform a test similar to that of one variable calculus using the second partial derivatives and the Hessian Matrix.
Δ
=
H
f
(
x
,
y
)
=
∣
∣
∣
f
x
x
f
x
y
f
y
x
f
y
y
∣
∣
∣
=
∣
∣
∣
∣
∣
∂
2
f
∂
x
2
∂
2
f
∂
x
∂
y
∂
2
f
∂
y
∂
x
∂
2
f
∂
y
2
∣
∣
∣
∣
∣
=
f
x
x
f
y
y
−
(
f
x
y
)
2
Then depending upon the value of
Δ
:
Δ
>
0
There is maximum if
f
x
x
<
0
and a minimum if
f
x
x
>
0
Δ
<
0
there is a saddle point
Δ
=
0
Further analysis is necessary
Δ
=
{
−
2
e
−
x
2
−
y
2
+
4
x
2
e
−
x
2
−
y
2
}
{
−
2
e
−
x
2
−
y
2
+
4
y
2
e
−
x
2
−
y
2
}
−
{
1
+
4
x
y
e
−
x
2
−
y
2
}
2
=
e
−
2
(
x
2
+
y
2
)
(
−
8
x
y
e
x
2
+
y
2
−
e
2
(
x
2
+
y
2
)
−
8
x
2
−
8
y
2
+
4
)
We need to consider the sign of
Δ
, and we note that
e
z
>
0
∀
z
∈
R
, so only need to consider the sign of:
Δ
'
=
−
8
x
y
e
x
2
+
y
2
−
e
2
(
x
2
+
y
2
)
−
8
x
2
−
8
y
2
+
4
So, depending upon the sign
Δ
'
we have an infinite number maxima and saddle points along the planes
x
=
±
y
