HOMEWORK #5 – SOLUTIONS Page 920 2 3 6

f ( x, y, z ) = x + yz, P (1, 3,1) , u =
, , 7 7 7
10) 1
1
1
iˆ +
⋅ zĵ +
⋅ yk̂ 2 x + yz
2 x + yz
2 x + yz
a) ∇f =
b) ∇f ](1,3,1) =
c)  2 + 3 + 18 23
fu = ∇f ⋅ u =
=
28
28
26) 1 ˆ
1
3
1
1
3
i+
ĵ +
k̂ = iˆ + ĵ + k̂ 4
4
4
2 4
2 4
2 4
f ( x, y, z ) = tan ( x + 2y + 3z ) , ( −5,1,1)
∇f = sec 2 ( x + 2y + 3z ) , 2 sec 2 ( x + 2y + 3z ) , 3sec 2 ( x + 2y + 3z )
∇f ]( −5,1,1) = sec 2 0, 2 sec 2 0, 3sec 2 0 = 1, 2, 3 = direction
∇f
( −5,1,1)
= 14 = max. rate of change
34) z = 1000 − 0.005x 2 − 0.01y 2 , ( 60, 40, 966 ) a) 
u = − ĵ south
∇f = −.01xiˆ − .02yĵ ⎤⎦
( 60,40 )
= −.6, −.8 m

fu = ∇f ⋅ u = +.8 ascending
m
b) −1 1


v = −iˆ + ĵ ( NW ) ⇒ u =
,
2 2
.8 −.2 m
 .6
fu = ∇f ⋅ u =
−
=
, descending
2
2
2 m
c) ∇f = −.6, −.8 is the direction of largest slope
44) −.6, −.8 = 1
m
⇒ tan θ = 1 ⇒ θ = 45
m
yz = ln ( x + z ) , ( 0, 0,1)
F ( x, y, z ) = ln ( x + z ) − yz
∇F =
1
1
, −z,
− y ⇒ ∇F ( 0, 0,1) = 1, −1,1
x+z
x+z
a) 1( x − 0 ) − 1( y − 0 ) + 1( z − 1) = 0 ⇒ x − y + z = 1 
b) n = 1, −1,1 or x = t, y = −t, z = 1 + t Page 931 6) f ( x, y ) = x 3 y + 12x 2 − 8y
fx = 3x 2 y + 24x = 0
fy = x 3 − 8 = 0 ⇒ x = 2
∴ fx = 3 ⋅ 2 2 y + 24 ⋅ 2 = 0 ⇒ y = −4
critical point is ( 2, −4 ) .
fxx = 6xy + 24 ]( 2,−4 ) = −48 + 24 = −24 < 0
fyy = 0; fxy = 3 x 2 ⎤⎦ 2,−4 = 12
(
)
D = fxx fyy − fxy2 = −24 ⋅ 0 − 12 2 = −144 ⇒
( 2, −4 ) is a saddle point.
8) f ( x, y ) = e4 y− x
2
− y2
fx = e4 y− x
2
− y2
⋅ ( −2x ) = 0 ⇒ x = 0
fy = e4 y− x
2
− y2
⋅ ( 4 − 2y ) = 0 ⇒ y = 2
fxx = −2e4 y− x
2
− y2
+ ( −2x ) e4 y− x
fyy = −2e4 y− x
2
− y2
+ ( 4 − 2y ) e4 y− x
fxy = −2x ⋅ e4 y− x
(
2
− y2
2
2
− y2
)
( 0, 2 )
⋅ ( −2x ) ⎤⎦( 0,2 ) = −2e4 < 0
2
− y2
4
⎤
⎦( 0,2 ) = −2e
( 4 − 2y )⎤⎦(0,2) = 0
D = −2e4 −2e4 − 0 2 = 4e8 > 0
∴( 0, 2 ) is a local maximum. f ( 0, 2 ) = e4 ≈ 54.6
10) f ( x, y ) = 2x 3 + xy 2 + 5x 2 + y 2
fx = 6x 2 + y 2 + 10x = 0
fy = 2xy + 2y = 0 ⇒ y ( 2x + 2 ) = 0 ⇒ y = 0 or x = −1
y = 0 : 6x 2 + 10x = 0 ⇒ x = 0, −
5
⎛ 5 ⎞
⇒ ( 0, 0 ) , ⎜ − , 0 ⎟ ⎝ 3 ⎠
3
x = −1 : 6 + y 2 − 10 = 0 ⇒ y = ±2 ⇒ ( −1, ±2 )
⎛ −5 ⎞
fxx = 12x + 10 ]( 0,0 ) = 10; ⎜ , 0 ⎟ = −10; ( −1, ±2 ) = −2
⎝ 3 ⎠
⎛ −5 ⎞ −4
fyy = 2x + 2 ]( 0,0 ) = 2; ⎜ , 0 ⎟ =
; ( −1, ±2 ) = 0
⎝ 3 ⎠
3
⎛ −5 ⎞
fxy = 2 y ]( 0,0 ) = 0; ⎜ , 0 ⎟ = 0; ( −1, ±2 ) = 4 or − 4
⎝ 3 ⎠
D ( 0, 0 ) = 10 ( 2 ) − 0 2 = 20 > 0 ⇒ local min, f = 0
40
⎛ −5 ⎞
⎛ −4 ⎞
D ⎜ , 0 ⎟ = −10 ⎜ ⎟ − 0 2 =
> 0 ⇒ local max, f = 4.6
⎝ 3 ⎠
⎝ 3⎠
3
D ( −1, 2 ) = −2 ( 0 ) − 4 2 = −16 < 0 ⇒ saddle point
2
D ( −1, −2 ) = −2 ( 0 ) − ( −4 ) = −16 < 0 ⇒ saddle point
18) f ( x, y ) = sin x sin y, − π < x < π , − π < y < π
fx = cos x sin y = 0
fy = sin x cos y = 0 ⇒ sin x = 0 or cos y = 0 ⇒ π
π
⇒x=±
2
2
⎛ π π⎞
(0, 0)
⎜⎝ ± , ± ⎟⎠
2 2
⎛π π⎞
fxx = − sin x sin y ]( 0,0 ) = 0; ⎜ , ⎟ = −1 < 0
⎝ 2 2⎠
⎛ π −π ⎞
⎛ −π π ⎞
⎛ −π −π ⎞
, ⎟ = 1 > 0; ⎜
,
= −1 < 0 ⎜⎝ ,
⎟⎠ = 1 > 0; ⎜⎝
⎝ 2 2 ⎟⎠
2 2
2 2⎠
fyy = − sin x sin y ⇒ same as above.
x=0⇒y=0
y=±
fxy = cos x cos y ]( 0,0 ) = 1;
]± ⎛⎜⎝ π ,± π ⎞⎟⎠
2
=0
2
D ( 0, 0 ) = 0 ⋅ 0 − 12 = −1 < 0 ⇒ saddle po int
⎛ π π⎞
D⎜ ± ,± ⎟ = 1− 0 = 1> 0 ⇒
⎝ 2 2⎠
⎛π π⎞
⎛ π π⎞
⎜⎝ , ⎟⎠ and ⎜⎝ − , − ⎟⎠ local maxima, f = 1
2 2
2 2
⎛ π −π ⎞
⎛ π π⎞
⎜⎝ ,
⎟⎠ and ⎜⎝ − , ⎟⎠ local minima, f = −1
2 2
2 2
46) V = 1000 cm 3 = xyz ⇒ z =
1000
xy
min S = 2xy + 2xz + 2yz
⎛ 1000 ⎞
⎛ 1000 ⎞
2000 2000
f ( x, y ) = 2xy + 2x ⎜
+ 2y ⎜
= 2xy +
+
⎟
⎟
⎝ xy ⎠
⎝ xy ⎠
y
x
2000
1000
fx = 2y − 2 = 0 ⇒ y = 2
x
x
2000
x4
fy = 2x − 2 = 0 ⇒ ( from above) x −
=0⇒
y
1000
x 3 = 1000 ⇒ x = 10
x = 10 ⇒ y = 10, z = 10cm and S = 600cm 2
4000
4000
fxx = 3 , fyy = 3 , fxy = 2
x
y
D (10,10 ) = 4 ⋅ 4 − 2 = 14 > 0 and fxx > 0 ⇒ local minimum.