using classical curly d notation, since x and z are independent variables and y
is a function of x and z, we have
0
∂
F
(
x
,
y
,
z
)
∂
z
∂
F
∂
x
∂
x
∂
z
∂
F
∂
y
∂
y
∂
z
∂
F
∂
z
∂
z
∂
z
⇒
0
∂
F
∂
y
∂
y
∂
z
∂
F
∂
z
⇒
∂
y
∂
z
∂
F
/
∂
z
∂
F
/
∂
y
.
Second, if we use subscript notation for the partial derivatives, we may set
g(x,z) = F(x, h(x,z), z). Then since
∂
x/
∂
z = 0,
0 = g
z
(x,z) = F
y
(x, h(x,z), z)h
z
(x, z) + F
z
(x, h(x,z), z),
implying
∂
y/
∂
z = h
z
(x,z) = -F
z
(x, h(x,z), z)/F
y
(x, h(x,z), z).
(b)
Is f defined by
f(
x
,
y
)
sin(5(
x
2
y
2
))
x
2
y
2
, (
x
,
y
)
≠
(0,0)
5
, (
x
,
y
)
(0,0)
continuous at (0,0)? A complete explanation is required. Details & definitions
are at the heart of it.// Since f(0,0) = 5 and
lim
(
x
,
y
)
→
(0,0)
f(
x
,
y
)
lim
(
x
,
y
)
→
(0,0)
sin(5(
x
2
y
2
))
x
2
y
2
lim
r
→
0
sin(5
r
2
)
r
2
(
L
H
)
lim
r
→
0
10
r
cos(5
r
2
)
2
r
5
by using L’Hopital’s rule after doing a conversion to polar coordinates, f is
continuous at (0,0).
_________________________________________________________________
10.
(10 pts.)
Obtain and locate the absolute extrema of the function
f(x,y) = (x - y)
2
on the closed disk D with radius five centered at the origin.
Observe that D is given in set builder notation as follows:
D = { (x,y) : x
2
+ y
2
≤
25 }. In performing this magic, use Lagrange multipliers
to deal with the behavior of f on the boundary, B = { (x,y) : x
2
+ y
2
= 25 }.
//
Since f
x
(x,y) = 2(x - y) and f
y
(x,y) = -2(x - y), the critical points of f
are the points of the form (x,x) in the interior of B. There we have f(x,x) = 0,
the obvious minimum.
Set g(x,y) = x
2
+ y
2
- 25. Then (x,y) is on the circle defined by
x
2
+ y
2
= 25 precisely when (x,y) satisfies g(x,y) = 0. Since
∇
g(x,y) = <2x, 2y>,
∇
g(x,y)
≠
<0, 0> when (x,y) is on the circle given by g(x,y) = 0. Plainly f and
g are smooth enough to satisfy the hypotheses of the Lagrange Multiplier Theorem.
Thus, if a constrained local extremum occurs at (x,y), there is a number
λ
so
that
∇
f(x,y) =
λ∇
g(x,y). Now
∇
f(x,y) =
λ∇
g(x,y)
⇒
<2(x-y),-2(x-y)> =
λ
<2x,2y>
⇒
x
2
-y
2
= 0
⇒
x = y or x = -y by performing a little routine algebraic magic.
Solving each of the systems consisting of (a) x
2
+ y
2
= 25 and x = y , and
(b) x
2
+ y
2
= 25 and x = -y , yields (5/2
1/2
,5/2
1/2
), (-5/2
1/2
,-5/2
1/2
),
(5/2
1/2
,-5/2
1/2
), and (-5/2
1/2
,5/2
1/2
). It turns out f is zero, the minimum value, at
the first two points, and f is 50, the maximum value, at the last two points.