Additional Exercises: Solutions
4
1
Multi-Variable Functions
Problem 1
1. f (x, y) = ln(x2 + y 4 )
∂f
2x
= 2
∂x
x + y4
4y 3
∂f
= 2
∂y
x + y4
2. f (x, y) = y x
∂f
= y x · ln y
∂x
∂f
= xy x−1
∂y
3. f (x, y) = ln(y +
p
x2 + y 2
∂f
1
p
=
∂x
y + x2 + y 2
1
∂f
p
=
∂y
y + x2 + y 2
4. f (x, y, z) = xyz
∂f
= yzxyz−1
∂x
∂f
= xyz z ln x
∂y
∂f
= xyz y ln x
∂z
−1/2
x
1 2
· 2x = p
x + y2
2
2
y x + y 2 + x2 + y 2
1 2
1
2 −1/2
· 1+
x +y
· 2y = p
2
2
x + y2
·
Additional Exercises: Solutions
Problem 2
1. f (x, y) = (xy)3 + xy 2
∂f
= 3x2 y 3 + y 2
∂x
∂f
= 3x3 y 2 + 2xy
∂y
∂ 2f
= 6xy 3
∂x2
∂ 2f
= 6x3 y + 2x
2
∂y
∂ 2f
∂ 2f
=
= 9x2 y 2 + 2y
∂x∂y
∂y∂x
2. f (x, y) = 3x2 − 4y 2 + 5xy + 4y
∂f
= 6x + 5y
∂x
∂f
= −8y + 5x + 4
∂y
∂ 2f
=6
∂x2
∂ 2f
= −8
∂y 2
∂ 2f
∂ 2f
=
=5
∂x∂y
∂y∂x
3. K(x1 , x2 ) =
5x1
x2
∂K
5
=
∂x1
x2
∂K
5x1
=− 2
∂x2
x2
∂ 2K
=0
∂x21
∂ 2K
10x1
= 3
2
∂x2
x2
2
∂ K
∂ 2f
5
=
=− 2
∂x1 ∂x2
∂x2 ∂x1
x2
2
Additional Exercises: Solutions
3
2
4. g(x, y, z) = 5x2 yz 4 + 8 xy 5
∂g
y2
= 10xyz 4 − 40 6
∂x
x
16y
∂g
= 5x2 z 4 + 5
∂y
x
∂g
= 20x2 yz 3
∂z
∂ 2g
∂x2
∂ 2g
∂y 2
∂ 2g
∂z 2
∂ 2g
∂x∂y
∂ 2g
∂x∂z
∂ 2g
∂y∂z
5. f (x, y) =
= 10yz 4 + 240
=
y2
x7
16
x5
= 60x2 yz 2
∂ 2g
80y
= 10xz 4 − 6
∂y∂x
x
2
∂ g
=
= 40xyz 3
∂z∂x
∂ 2g
=
= 20x2 z 3
∂z∂y
=
x4 −3x2 y
3x+2y 2
9x4 + 8x3 y 2 − 9x2 y − 12xy 3
∂f
=
∂x
(3x + 2y 2 )2
−9x3 − 4x4 y + 6x2 y 2
∂f
=
∂y
(3x + 2y 2 )2
∂ 2f
54x4 + 96x3 y 2 + 48x2 y 4 − 24y 5
=
∂x2
(3x + 2y 2 )3
∂ 2f
−12x5 + 108x3 y + 24x4 y 2 − 24x2 y 3
=
∂y 2
(3x + 2y 2 )3
∂ 2f
∂ 2f
−24x4 y − 27x3 − 32x3 y 3 − 54x2 y 2 + 24xy 4
=
=
∂x∂y
∂y∂x
(3x + 2y 2 )3
Additional Exercises: Solutions
6. K(x1 , x2 , x3 ) = x2 · e4x1 +5x3
∂K
= 4x2 · e4x1 +5x3
∂x1
∂K
= e4x1 +5x3
∂x2
∂K
= 5x2 · e4x1 +5x3
∂x3
∂ 2K
∂x21
∂ 2K
∂x22
∂ 2K
∂x23
∂ 2K
∂x1 ∂x2
∂ 2K
∂x1 ∂x3
∂ 2K
∂x2 ∂x3
= 16x2 · e4x1 +5x3
=0
= 25x2 · e4x1 +5x3
∂ 2f
= 4 · e4x1 +5x3
∂x2 ∂x1
∂ 2f
=
= 20x2 · e4x1 +5x3
∂x3 ∂x1
∂ 2f
=
= 5 · e4x1 +5x3
∂x3 ∂x2
=
7. x(A, K) = 120 · A0,85 · K 0,3
∂x
= 102 · A−0,15 K 0,3
∂A
∂x
= 36 · A0,85 K −0,7
∂K
∂ 2x
= −15, 3 · A−1,15 K 0,3
2
∂A
∂ 2x
= −25, 2 · A0,85 K −1,7
∂K 2
∂ 2x
∂ 2x
=
= 30, 6 · A−0,15 K −0,7
∂A∂K
∂K∂A
4
Additional Exercises: Solutions
8. L(x, y, λ) = 8x0,3 y 0,7 + λ(200 − 6x − 5y)
∂L
= 2, 4 · x−0,7 y 0,7 − 6λ
∂x
0,3
∂L
x
= 5, 6
− 5λ
∂y
y
∂L
= 200 − 6x − 5y
∂λ
∂ 2L
∂x2
∂ 2L
∂y 2
∂ 2L
∂λ2
∂ 2L
∂x∂y
∂ 2L
∂x∂λ
∂ 2L
∂y∂λ
= −1, 68 · y 0,7 x−1,7
= −1, 68 · y 0,3 x−1,3
=0
∂ 2L
= −1, 68 · y −0,3 x−0,7
∂y∂x
∂ 2L
=
= −6
∂λ∂x
∂ 2L
=
= −5
∂λ∂y
=
9. f (x, y) = xy · exy
∂f
= exy y + xy 2
∂x
∂f
= exy x + x2 y
∂y
∂ 2f
xy
2
3
=
e
2y
+
xy
∂x2
∂ 2f
= exy 2x2 + x3 y
2
∂y
∂ 2f
∂ 2f
=
= exy (x2 y 2 + 3xy + 1)
∂x∂y
∂y∂x
5
Additional Exercises: Solutions
6
10.
q
L(r1 , r2 , r3 , λ1 , λ2 ) = 2 r12 + 3r22 − 5r32
+ λ1 (10 − r1 − 2r2 + r3 ) + λ2 (20 − r1 r2 r3 )
∂L
∂r1
∂L
∂r2
∂L
∂r3
∂L
∂λ1
∂L
∂λ2
=p
=p
=p
2r1
(r12 + 3r22 − 5r32 )
6r2
(r12 + 3r22 − 5r32 )
−10r3
(r12 + 3r22 − 5r32 )
− λ1 − λ2 r2 r3
− 2λ1 − λ2 r1 r3
+ λ1 − λ2 r1 r2
= 10 − r1 − 2r2 + r3
= 20 − r1 r2 r3
6r22 − 10r32
∂ 2L
p
=
∂r12
(r12 + 3r22 − 5r32 )3
∂ 2L
6r12 − 30r32
p
=
∂r22
(r12 + 3r22 − 5r32 )3
∂ 2L
−10r12 − 30r22
p
=
∂r32
(r12 + 3r22 − 5r32 )3
∂ 2L
∂ 2L
=
=0
∂λ21
∂λ22
∂ 2L
−6r1 r2
∂ 2L
=
=p 2
− λ2 r3
∂r1 ∂r2
∂r2 ∂r1
(r1 + 3r22 − 5r32 )3
∂ 2L
10r1 r3
∂ 2L
=
=p 2
− λ2 r2
∂r1 ∂r3
∂r3 ∂r1
(r1 + 3r22 − 5r32 )3
∂ 2L
∂ 2L
=
= −1
∂r1 ∂λ1
∂λ1 ∂r1
∂ 2L
∂ 2L
=
= −r2 r3
∂r1 ∂λ2
∂λ2 ∂r1 y
∂ 2L
∂ 2L
30r2 r3
=
=p 2
− λ2 r1
∂r2 ∂r3
∂r3 ∂r2
(r1 + 3r22 − 5r32 )3
∂ 2L
∂r2 ∂λ1
∂ 2L
∂r2 ∂λ2
∂ 2L
∂r3 ∂λ1
∂ 2L
∂r3 ∂λ2
∂ 2L
∂λ1 ∂λ2
=
=
=
=
=
∂ 2L
= −2
∂λ1 ∂r2
∂ 2L
= −r1 r3
∂λ2 ∂r2
∂ 2L
=1
∂λ1 ∂r3
∂ 2L
= −r1 r2
∂λ2 ∂r3
∂ 2L
=0
∂λ2 ∂λ1
Additional Exercises: Solutions
Problem 3
1
1
f (x1 , x2 , x3 , x4 ) = x21 x2 + x1 x22 + x32 − 4x2 − x23 + x3 − x24 + 2x4
3
2
∂f
∂x1
∂f
∂x2
∂f
∂x3
∂f
∂x4
= 2x1 x2 + x22 = 0
= x21 + 2x1 x2 + x22 − 4 = 0
= −x3 + 1 = 0
= −2x4 + 2 = 0
from III: x3 = 1
from IV: x4 = 1
from I:x2 (2x1 + x2 ) = 0
x2,1 = 0
x2,2 = −2x1
x2,1 in III: x21 = 4
x1,1 = −2
x1,2 = 2
x1 in x2,2 = −2x1
x2,3 = 4
x2,4 = −4
→ possible extreme points:
1. (2, 0, 1, 1)
2. (−2, 0, 1, 1)
3. (−2, 4, 1, 1)
4. (2, 4, 1, 1)
7
Additional Exercises: Solutions
Hessian Matrix:

2x2
2x1 + 2x2 0
0


 2x1 + 2x2 2x1 + 2x2 0
0



0
0
−1 0

0
0
0 −2
Hessian Matrix

0


 4


 0

0
8








at (2,0,1,1):
4
0
0



4 0
0 


0 −1 0 

0 0 −2
det H1 = 0 → undefined
Hessian Matrix at (-2,0,1,1):


0 −4 0
0




 −4 −4 0
0 




 0
0 −1 0 


0
0
0 −2
det H1 = 0 → undefined
Hessian Matrix at (-2,4,1,1):


8 4 0
0




 4 4 0
0 




 0 0 −1 0 


0 0 0 −2
det H1 = 8 > 0
det H2 = 16 > 0
det H3 = −16 < 0 → undefined
Additional Exercises: Solutions
Hessian Matrix at (2,-4,1,1):

−8 −4 0
0


 −4 −4 0
0


 0
0 −1 0

0
0
0 −2
9








det H1 = −8
det H2 = 16
det H3 = −16
det H4 = 32
→ negative definite → Maximum
Problem 4



grad f = 

∂f
∂x
∂f
∂y
∂f
∂z


2x
 
 
4
2
 =  −4y 3 e−y +z + 2y
 
4
2
2ze−y +z





second-order derivatives:
∂ 2f
=2
∂x2
∂ 2f
=0
∂x∂y
∂ 2f
=0
∂x∂z
∂ 2f
=0
∂y∂x
∂ 2f
3 −y 4 +z 2
3
2 −y 4 +z 2
+
−4y
e
·
−4y
+2
=
−4
·
3y
e
∂y 2
∂ 2f
4
2
= −4y 3 e−y +z · 2z
∂y∂z
∂ 2f
=0
∂z∂x
∂ 2f
4
2
= 2ze−y +z · −4y 3
∂z∂y
∂ 2f
4
2
4
2
= 2e−y +z + 2ze−y +z · 2z
2
∂z
Additional Exercises: Solutions
Hessian Matrix:

2
0
0


4
2
4
2
4
2
 0 −12y 2 e−y +z + 16y 6 e−y +z + 2
−8zy 3 e−y +z

4
2
4
2
4
2
2e−y +z + 4z 2 e−y +z
0
−8zy 3 e−y +z
10





Hessian Matrix for point (0, 0, 0):


2 0 0




 0 2 0 


0 0 2
Finding the determinants:
det H1 = 2
det H2 = 4
det H3 = 8
The Hessian Matrix is positive definite. The function has a minimum at point 0, 0, 0
Problem 5
a)
U (p1 , p2 , p3 ) = p1 x1 + p2 x2 + p3 x3
= (120 − 20p1 + 2p2 + 3p3 )p1 + (80 + 8p1 − 15p2 + 2p3 )p2
+ (100 + 7p1 + 8p2 − 30p3 )p3
= −20p21 − 15p22 − 30p23 + 10p1 p2 + 10p1 p3 + 10p2 p3
+ 120p1 + 80p2 + 100p3
b)
objective function: sales → max.
−20p21 − 15p22 − 30p23 + 10p1 p2 + 10p1 p3 + 10p2 p3 + 120p1 + 80p2 + 100p3
I
II
III
∂U
= −40p1 + 10p2 + 10p3 + 120 = 0
∂p1
∂U
= 10p1 − 30p2 + 10p3 + 80 = 0
∂p2
∂U
= 10p1 + 10p2 − 60p3 + 100 = 0
∂p3
Additional Exercises: Solutions
11
p1
p2
p3
−40
10
10
−120
10
−30
10
−80
·4
10
10
−60
−100
·4
−40
10
10
−120
40
−120
40
−320
II+I
40
40
−240
−400
II+I
−40
10
10
−120
0
−110
50
−440
·5
0
50
−230
−520
·11
−40
10
10
−120
0
−550
250
−2200
0
550
−40
10
10
−120
0
−550
250
−2200
0
0
−2530 −5720
III+II
−2280 −7920
−2280p3 = −7920
66
p3 =
19
−550p2 + 250 ·
−40p1 + 10 ·
66
= −2200
19
106
p2 =
19
106
66
+ 10 ·
= −120
19
19
100
p1 =
19
Additional Exercises: Solutions
12
Problem 6
objective function: area → max.
l·b
l = length
w = width
constraint: length of the fence
2l + 2w = P
2l + 2w − P = 0
Lagrange function:
L = l · w + λ (2l + 2w − P )
I
II
III
∂L
= w + 2λ = 0
∂l
∂L
= l + 2λ = 0
∂w
∂L
= (2l + 2w − P ) = 0
∂λ
I=II:
w=l
in III:
2l + 2w = P
2l + 2l = P
4l = P
l = 14 P
b = 14 P
⇒ w = −2λ
⇒ l = −2λ
Additional Exercises: Solutions
13
Problem 7
objective function: production → max
Y (F, H) = 120F − F 2 + 80H − 4H 2
constraint:
60F + 40H = 20000
Lagrange function:
L = 120F − F 2 + 80H − 4H 2 + λ (60F + 40H − 20000)
∂L
= 120 − 2F + 60λ = 0
∂F
∂L
= 80 − 8H + 40λ = 0
∂H
∂L
= 60F + 40H − 20000
∂λ
I and II in III:
60 (60 + 30λ) + 40 (10 + 5λ) = 20000
3600 − 1800λ + 400 + 200λ = 20000
2000λ = 16000
λ=8
F
F
H
H
= 60 + 30λ
= 300
= 10 + 5λ
= 50
→
→
F = 60 + 30λ
H = 10 + 5λ
Additional Exercises: Solutions
14
Problem 8
objective function: production → max
2
1
f (x, y) = 2x 3 y 3
constraint:
x + y = 3000
Lagrange function:
2
1
L = 2x 3 y 3 + λ (x + y − 3000)
∂L
4 1 1
= x− 3 y 3 + λ = 0
∂F
3
2 2 2
∂L
= x 3 y− 3 + λ = 0
∂H
3
∂L
= x + y − 3000
∂λ
I/II:
4 − 13 13
x y
3
2 32 − 23
x y
3
=1
2y
=1
x
x = 2y
in III:
2y + y = 3000
y = 1000
x = 2000
4 1 1
− λ = x− 3 y 3
3
2 2 −2
− λ = x3 y 3
3
Additional Exercises: Solutions
15
Problem 9
objective function: costs → min.
K(t, m) = 40t + 10m
constraint:
√ √
30 · t · m = 900
√ √
30 · t · m − 900 = 0
Lagrange function:
√ √
L = 40t + 10m + λ 30 · t · m − 900
I
II
III
√ m
∂L
= 40 + λ 15 √
=0
∂t
t
√ ∂L
t
= 10 + λ 15 √
=0
∂m
m
√ √
∂L
= 30 · t · m − 900 = 0
∂λ
I/II: 4 =
√
m
√
t
√
√t
m
m
t
m = 4t
4=
√ √
in III: 30 t · 4t = 900
30 · 2t = 900
t = 15
m = 60
→
→
√
m
40 = −15λ √
t
√
t
10 = −15λ √
m
Additional Exercises: Solutions
16
Problem 10
objective function: K = 4a2 − 16a + 5b2 − 30b + 70 → min
constraint: 40a + 20b = 380
Lagrange function: 4a2 − 16a + 5b2 − 30b + 70 + λ (40a + 20b − 380)
∂L
= 8a − 16 + 40λ = 0
∂a
∂L
= 10b − 30 + 20λ = 0
∂b
∂L
= 40a + 20b − 380 = 0
∂c
4a − 8 = 10b − 30
5
11
a= b−
2
2
11
5
b−
+ 20b = 380
40
2
2
100b − 220 + 20b = 380
b=5
5
11
a= b−
=7
2
2
→
→
−20λ = 4a − 8
−20λ = 10b − 30