Mathematics for Economists Assignment 2

This document contains solutions for Mathematics for Economists Assignment 2. It includes questions on functions, demand function, production function, implicit function, and more. Subject: Mathematics, Course Code: N/A, Course Name: N/A, College/University: N/A

MATHEMATICS FOR ECONOMISTS
ASSIGNMENT 2
[DATE]

Question 1
1.1 f ( x )=2 x5− 4
3 x3− 7
x +2
df ( x )
dx = d
dx (2 x5− 4
3 x3− 7
x +2 )
df ( x )
dx =10 x4 −4 x2 + 7
x2 +0
df ( x )
dx =10 x4 −4 x2 + 7
x2
1.2 z ( λ )=7 ln λ− 6
λ +x5
d z ( λ )
d λ = d
d λ (7 ln λ− 6
λ + x5
)
d z ( λ )
d λ = 7
λ + 6
λ2 + 0
d z ( λ )
d λ = 7
λ + 6
λ2
1.3 g ( t )=( 2
t + t5
) ( t3 +1 )
d g ( t )
d t = d
d t ( ( 2
t +t5
) ( t3 +1 ) )
d g ( t )
d t = ( t3+1 ) d
dt ( 2
t + t5
)+( 2
t + t5
) d
dt (t3 +1)
d g ( t )
d t = ( t3+1 ) ( −2
t 2 +5 t4
)+ ( 2
t +t5
) .( 3t2)
d g ( t )
d t =−2 t +5 t7− 2
t2 +5t4 +6t +3 t 7
1

d g ( t )
dt =8 t7 +5 t4 − 2
t2 ++4 t
1.4 y ( m ) =em ( 3
√ m )
d y ( m )
d t = d
d t ( em ( 3
√ m ) )
d y ( m )
d t = ( em ) d
dt ( 3
√ m ) + ( 3
√ m ) d
dt ( em )
d y ( m )
d t =em 1
3 m
2
3
+ 3
√m em
d y ( m )
d t =
( 3
√ m+ 1
3 m
2
3 ) em
1.5 Y = ex2−1
x2−1
dY
dx = d
dx ( ex2−1
x2−1 )
dY
dx =
( x2 −1 ) d
dx ( ex2−1 ) − ( ex2−1 ) d
dx ( x2−1 )
( x2−1 )
2
dY
dx = ( x2 −1 ) ( 2 x . ex2−1 ) −ex2−1∗2 x
( x2−1 ) 2
dY
dx = ( x2 −1 ) ( 2 x . ex2−1 )−2 x ex2−1
( x2−1 )2
dY
dx = 2 x3 ex2−1 −2 x . ex2−1 −2 x ex2−1
( x2 −1 )
2
dY
dx = 2 x3 ex2−1 −4 x . ex2−1
( x2−1 )2
2

1.6 d3 y
d x3
y=−x2 +2 √ x3
dy
dx =−2 x+ 2∗3 x2
√ x3
d2 y
d x2 =−2+ 3 x
2 √x3
d3 y
d x3 =0− 3
4 √ x3
d3 y
d x3 = −3
4 √ x3
Question 2
2.1 Function
y=f ( x )=e2 x−6 x2 +16
At x = 1
dy
dx = df ( x)
dx = d
dx ( e2 x−6 x2 +16 )
dy
dx = df ( x)
dx =2 e2 x−12 x+ 0
Now,
d
dx ( dy
dx )= d f ( x)
dx ( 2 e2 x−12 x )
d2 y
d x2 = d2 f (x )
d x2 =4 e2 x−12
Now,
3

At x=1
d2 y
d x2 =4 e2−12=17.556
It can be seen that d2 y
d x2 comes out to be positive and hence, it can be said that the function is
convex.
2.2 Demand function p = 100 - 0.01 x
Cost function C(x) = 50x +10000
2.2.1 Average cost function (AC)
AC= C ( x )
x
AC= ( 50 x+10000 )
x
Average cost function ( AC ) =50+( 10000
x )
2.2.2 Profit function (π)
Revenue=x∗p=x∗(100−0.01 x )=100 x – 0.01 x2
Cost function C ( x )=50 x+10000
Hence,
Profit function ( π )= ( 100 x – 0.01 x2 ) − ( 50 x+10000 )
π=100 x−0.01 x2−50 x−10000
π=−0.01 x2 +50 x −10000
4

2.2.3 Value of x for which the profit is maximum
First derivative of profit function =0 (Maximum profit)
dπ
dx = d
dx (−0.01 x2+50 x−10000 ) =−0.02 x+ 50
−0.02 x+50=0
x= 50
0.02 =2500
x=2500
2.2.4 Maximum profit would be at x = 2500
π=−0.01 x2 +50 x −10000
π=−0.01 ( 2500 )2 + ( 50∗2500 )−10000
π=525,00
2.2.5 Price for this level of production
p = 100 - 0.01 x
x = 2500
p=100− ( 0.01∗2500 )
p=75
Question 3
3.1 Production function
Q= 3
√ 3 K2 +2 L3
Where, L=Labour ( work−hours )
K=cost of capital ( $ )
Q=Daily production
5

3.1.1 Marginal productivity of capital
dQ
dK = d
dK ( 3
√ 3 K2 +2 L3 )
Let u=3 K 2+2 L3
f =3
√u
dQ
dK = d
dK ( 3
√u ) . d
dK ( 3 K2 +2 L3 )
dQ
dK = 1
3u
2
3
.6 K = 2 K
u
2
3
Now,
dQ
dK = 2 K
( 3 K2+ 2 L3 )
2
3
Marginal productivity of capital
2 K
( 3 K2+ 2 L3 )
2
3
.
Marginal productivity of labour
dQ
dL = d
dL ( 3
√ 3 K2 +2 L3 )
Let u=3 K 2+2 L3
f =3
√u
dQ
dL = d
dL ( 3
√ u ) . d
dL ( 3 K2+ 2 L3 )
dQ
dL = 1
3u
2
3
.6 L2= 2 L2
u
2
3
Now,
dQ
dL = 2 L2
( 3 K 2+2 L3 )
2
3
6

Marginal productivity of labour
2 L2
( 3 K2+ 2 L3 )
2
3
3.1.2 MRTS of productions of shoes
MRTS= Marginal productivity of labour
Marginal productivity of capital
MRTS=
2 K
( 3 K2 +2 L3 )
2
3
2 L2
( 3 K2 +2 L3 )
2
3
MRTS= 2 K∗( 3 K 2+2 L3 )
2
3
( 3 K 2+2 L3 )
2
3∗2 L2
MRTS= K
L2
MRTS when labour work hours L = 8 hours per day
Cost of capital K = 4
Now,
MRTS= 4
82 =0.0625
3.2 Implicit function
x2+ √ y +5 y =5 x2 y2+ 4 x3
Differentiation w.r.t.x
d
dx ( x2+ √ y +5 y ) = d
dx ( 5 x2 y2+4 x3 )
2 x+ 1
2 √ y
dy
dx +5 dy
dx =5 ( 2 x y2 +2 y dy
dx x2
) ++12 x2
7

Put dy
dx = y '
2 x+ 1
2 √ y y'+5 y' =5 ( 2 x y2 +2 y y' x2 ) ++12 x2
1
2 √ y y' +5 y'=5 ( 2 x y2+ 2 y y ' x2 ) +12 x2−2 x
Multiply both side with 2y
( 1
2 √ y y' +5 y'
) 2 y=10 y ( 2 x y2+2 y y' x2 ) +24 y x2−4 xy
( √ y∗√ y
√ y y' +10 y y'
) =10 y ( 2 x y2+2 y y' x2 ) +24 y x2−4 xy
√ y y' +10 y y'=20 y3 x +24 y x2 −4 xy
√ y y' +10 y y'−20 y2 x2 y ' =20 y3 x +24 y x2 −4 xy
y' ( √ y +10 y−20 y2 x2 ) =20 y3 x +24 y x2−4 xy
y' = 20 y3 x +24 y x2 −4 xy
√ y +10 y −20 y2 x2
Hence,
dy
dx =20 y3 x+ 24 y x2−4 xy
√ y +10 y−20 y2 x2
Slope of the tangent line at (1,0)
Slope= dy
dx =20. 03 .0+24 .1 . 02−4.1.0
√1+10 .1−20 02 12 =0
Question 4
4.1 Firm spends on fixed costs = $650
MC=82−16 y +1.8 y2
8

Total cost function =?
TC=∫82−16 y +1.8 y2 dx
TC=82 y− 16 y2
2 +1.8 y3
3 +C
TC=82 y−8 y2 +0.6 y3 +C
At y=0 ,TC =$ 650
650=C
Hence,
Total cost fucntion TC=82 y −8 y2+ 0.6 y3+ 65
4.2 Demand function p=600−6 q0.5
Marginal revenue MR=600−9 q0.5
4.2.1 Change in TR when q has increased from 2025 to 2500
¿ ∫
2025
2500
MR dq
¿ ∫
2025
2500
600−9 q0.5 dq
¿ [600 q ]2025
2500− [ 9 q1.5 ]2025
2500
+ C
¿ 750000−668250
¿ 81,750
4.2.2 The value of consumer surplus when q = 2500
9

¿ ∫
0
2500
600−6 q0.5 dq
¿ [ 600 q ]0
2500− [ 4 q1.5 ]0
2500
+C
¿ 1000000
Now,
TR=p∗q= p∗( 600−6 q0.5 ) =600 q−6 q1.5
TR= ( 600∗2500 )− ( 6∗25001.5 )=750,000
4.3 Young’s theorem through Z=3 x3 y−x y3−5 y2
According to Young’s theorem, Z would be a valued function defined in such a way that both
the first order partial derivatives Z(x) and Z(y) would be differentiable and then,
Z ( xy ) =Z ( yx )
∂ Z
∂ x = ∂
∂ x ( 3 x3 y−x y3 −5 y2 )
f ( x )= ( 9 x2 y− y3 ) …… … ….. ( 1 )
∂ Z
∂ y = ∂
∂ y ( 3 x3 y−x y3−5 y2 )
f ( y ) =3 x3 −3 x y2−10 y2 …… … .(2)
f ( xy ) = ∂ f ( x )
∂ y = ∂
∂ y ( 9 x2 y − y3 ) =9 x2−3 y2 … … … … … ..(A )
f ( yx ) = ∂ f ( y )
∂ x = ∂
∂ x ( 3 x3−3 x y2−10 y2 ) =9 x2−3 y2 … … … … … .( B)
f ( xy ) =f ( yx )
Proved!!
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