Intermediate Mathematics for Economics

This document provides solutions to questions related to convex functions, partial and total derivatives, symmetric functions, production functions, and isoquants in the context of intermediate mathematics for economics. It covers topics such as strict convexity, marginal rate of technical substitution, and Euler's theorem.

EC206 Intermediate Mathematics for Economics
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[DATE]

Question 1
The given function f(x) is strictly convex. The definition of a convex function is highlighted
below.
When its domain is also a convex set and is true for all x, y and for
The condition for strict convexity of the given function is shown below.
It is an imperative use of the strictly convexity that the obtained optimal solution is itself
unique.
In other words, the strict convexity is to ensure the uniqueness of the optimal solution.
1

Question 2
(a) Partial and total derivative of z with respective to x and y
z=ln ( x2 y )
x + y=1
Partial derivative of z with respect to x
∂ z
∂ x = ∂ ln ( x2 y )
∂ x
Let u=x2 y
∂ z
∂ x = ∂ ln ( u )
∂ x . ∂ ( x2 y )
∂ x
∂ z
∂ x = 1
u∗2 y∗x2 y−1= 1
x2 y ∗2 y x2 y−1=2 y x2 y−1+2 y =2 y x−1= 2 y
x
Now ,
With respect to y
∂ z
∂ y = ∂ ln ( x2 y )
∂ y
Let u=x2 y
∂ z
∂ y = ∂ ln ( u )
∂ y . ∂ ( x2 y )
∂ y
From exponent rule: x2 y=e2 yln(x)
∂ z
∂ y = ∂ ln ( u )
∂ y . ∂ ( e2 yln (x ) )
∂ y =1
u .2 x2 y ln ( x )
∂ z
∂ y = 1
x2 y . 2 x2 y ln ( x ) =1.2 ln ( x )=2 ln (x )
Hence, the partial derivatives would be 2 y
x ,2 ln (x)
Total derivative of z
dz= ∂ z
∂ x dx + ∂ z
∂ y dy
dz= 2 y
x dx +2 ln( x )dy
Here,
x + y=1
2

dx +dy =0
dy =−dx
Thus,
dz= 2 y
x dx +2 ln ( x ) (−dx )
dz= ( 2 y
x −2 ln ( x ) )dx
dz= 2
x ( y −x ln ( x ) ) dx
(b) Partial and total derivative of z with respective to x and y
z=x + y
A xa yb =k
Where, a , b∧k are constants
Partial derivative of z with respect to x
∂ z
∂ x = ∂(x + y )
∂ x =1
With respect to y
∂ z
∂ y = ∂( x+ y )
∂ y =1
Total derivative of z
dz= ∂ z
∂ x dx + ∂ z
∂ y dy
Divide by dx
dz
dx = ∂ z
∂ x + ∂ z
∂ y
dy
dx
dz
dx =1+1. dy
dx
Now,
A xa yb =k
x= ( k
y b A )1
a
Similarly,
3

y= ( k
xa A )1
b
dz
dx =1+1.
d ( ( k
xa A ) 1
b
)
dx =1+ d
dx ( k
1
b
x
a
b . A
1
b )
dz
dx =1+ ( k
A )1
b
( d
dx (x
a
b ) )=1+ ( k
A )1
b
( a
b )x− ( a
b +1 )
Further,
dz= ∂ z
∂ x dx + ∂ z
∂ y dy
Divide by dy
dz
dy = ∂ z
∂ y + ∂ z
∂ x
dx
dy
dz
dy =1+1. dx
dy
dz
dy =1+1.
d ( ( k
yb A ) 1
a
)
dy =1+ d
dy ( k
1
a
y
b
a . A
1
a ) =1+ ( k
A ) 1
a
( d
dx ( y
−b
a ) )=1− ( k
A ) 1
a
( b
a ) y−( b
a +1 )
Question 3
(a) The given function would be termed as symmetric in its arguments when the reversal of
the arguments would not make any change in the function.
z=f (x1 , x2)
z= A (∝1 x1
−β +∝2 x2
−β )−1
β
Where, A>0 ,∝>0 , i=1 , 2∧β >−1
Swapping of the arguments
Original: z=f (x1 , x2)=A ( ∝1 x1
− β +∝2 x2
− β ) −1
β
Swapping: z=f (x2 , x1)=A ( ∝1 x2
− β +∝2 x1
− β ) −1
β
It is apparent that ∝1∧∝2are constant and hence, can be assumed to be equal ∝1=∝2
4

Therefore, it can be concluded that swapping of the arguments would not change the original
function.
z=f ( x1 , x2)=f (x2 , x1 )
(b)
(i) ∂ z
∂ x1
Function of x1 for x2
∂ z
∂ x1
= ∂
∂ x1
{ A ( ∝1 x1
− β +∝2 x2
− β
)
−1
β }
∂ z
∂ x1
= A ( −1
β ) ( ∝1 x1
−β +∝2 x2
−β ) −1
β −1 ∂
∂ x1
( ∝1 x1
− β +∝2 x2
− β )
∂ z
∂ x1
= A ( −1
β ) ( ∝1 x1
−β +∝2 x2
−β ) −1
β −1
. ∝1 ( −β ) x1
−β −1
∂ z
∂ x1
= A ( −1
β ) ( −β ) ( ∝1 x1
− β +∝2 x2
− β ) −1
β −1
. ∝1 x1
−β −1
∂ z
∂ x1
= A ∝1 x1
− β−1 ( ∝1 x1
− β +∝2 x2
− β )−1
β −1
(ii) ∂ z
∂ x1
Function of x1 for x2
∂ z
∂ x1
= A ∝1 x1
− β−1 ( ∝1 x1
− β +∝2 x2
− β )−1
β −1
The given function is symmetric in its arguments when the reversal of the arguments has not
made any change in the function.
z=f ( x1 , x2)=f (x2 , x1 )
Thus,
Function of x1 for x2
∂ z
∂ x2
= A ∝1 x2
− β−1 ( ∝1 x1
− β +∝2 x2
− β )−1
β −1
5

(c) ∂ z
∂ x1
i=1 , 2
x1
∂ z
∂ x1
+ x2
∂ z
∂ x2
=z
Here, put the values of
∂ z
∂ x1
∧∂ z
∂ x2
in the above function
LHS ¿ x1
∂ z
∂ x1
+ x2
∂ z
∂ x2
¿ x1 . {A ∝1 x1
−β−1
(∝1 x1
− β +∝2 x2
− β
)−1
β −1
}+ x2 . {A ∝2 x2
−β−1
(∝1 x1
− β +∝2 x2
− β
)−1
β −1
}
¿ A (∝1 x1
−β +∝2 x2
−β )−1
β −1
(∝1 x1
−β −1 . x1+∝2 x2
−β −1 x2)
¿ A (∝1 x1
−β +∝2 x2
−β )−1
β −1
(∝1 x1
−β −1+1 +∝2 x2
− β−1+1 )
¿ A (∝1 x1
−β +∝2 x2
−β )−1
β −1
(∝1 x1
−β + ∝2 x2
−β )
¿ A (∝1 x1
−β +∝2 x2
−β )−1
β −1+1
¿ A (∝1 x1
−β +∝2 x2
−β )−1
β
¿ z
RHS = z
Hence, proved LHS =RHS
x1
∂ z
∂ x1
+ x2
∂ z
∂ x2
=z
(d) Euler’s theorem holds this function or not?
It can be seen that the degree of the function is 1.
Hence,
∑
i=1
2
xi
df
d xi
=z=f
The function holds Euler’s theorem.
6

Question 4
(a) Production function
Q=a ln ( x1 )+ b ln(¿ x2)¿
Where, a , b=Positive numbers, x1∧x2=Quantity inputs
Marginal rate expression
Total derivative of output Q w. r. t. x1
dQ
d x1
= d
d x1
¿
Total derivative of output Q w. r. t. x2
dQ
d x2
= d
d x2
¿
Hence, Marginal rate of technical substitution=
a
x1
b
x2
= a x2
b x1
(b) Equation of an isoquant x2 for a fixed Q in terms of x1
¿
ln ( x1
a ) +ln (¿ x2
b )=Q ¿
ln (¿ x2
b )=Q−ln ( x1
a ) ¿
Take the exponential of both the sides
x2
b=eQ −x1
a
x2=¿ ¿
Equation of an isoquant x2 for a fixed Q in terms of x1 would be x2=¿ ¿
(c) Marginal rate of technical substitution of x1for x2 , for x2 not in terms of Q and hence, first
task is to eliminate Q
Expression from part b
x2=¿ ¿
Here,
x2
b=eQ −x1
a
7

eQ=x2
b + x1
a
Take natural log
Q=ln ¿ ¿
Q=ln( x2
b∗x1
a )
Now, Again
eQ=x2
b∗x1
a
Substitute the eQ=x2
b∗x1
a in eq. 1
x2=¿
x2={x1
a
( x2
b – 1 ) }
1
b
x2= ( x1
a
b ) { ( x2
b – 1 ) }
1
b
Now,
∆ x2
∆ x1
= ∆
∆ x1
(x1
a
b ){( x2
b – 1 ) }
1
b
∆ x2
∆ x1
={( x2
b – 1 ) }
1
b
( a
b ) ( x1
a
b −1
)
∆ x2
∆ x1
={( x2
b – 1 ) }
1
b
( a
b ) ( x1
a−b
b )
∆ x2
∆ x1
= {( x2 ) ( a
b ) ( x1
a ) ( x1
−1 ) }= ( a
b ) ( x2
x1 )x1
a
Or
∆ x2
∆ x1
= ( a
b ) ( x2
x1 )
Question 5
(a) The aim of the manager is to maximize the profit (π )
x , y are the outputs of the two products manufactured by the fir,
Profit function
Maximize π=50 x−2 x2−xy−3 y2 +95 y
x + y=25
8

Lagrangean for the given case to show the Lagrange multiplier
∂
∂ x ∅ =π + λ ( 25−x− y )
∂∅
∂ x = ∂
∂ x { ( 50 x−2 x2−xy−3 y2 +95 y ) + λ ( 25−x− y ) }
(b) First order conditions
∂∅
∂ x = ∂
∂ x { ( 50 x−2 x2−xy−3 y2 +95 y ) + λ ( 25−x− y ) }
∂∅
∂ x =50−4 x− y−0+ 0+ ( −λ )
∂∅
∂ x =50−4 x− y−λ
Put ∂∅
∂ x =0
50−4 x− y −λ=0 … … … … ( 1 )
Now,
∂∅
∂ y = ∂
∂ y { ( 50 x−2 x2−xy−3 y2 +95 y ) + λ ( 25−x− y ) }
∂∅
∂ y =−x −6 y+ 95+ (−λ )
Put
∂∅
∂ y =0
∂∅
∂ y =95−6 y −x−λ … … … ..(2)
Similarly,
∂∅
∂ λ = ∂
∂ λ { ( 50 x−2 x2 −xy−3 y2+ 95 y ) + λ ( 25−x− y ) }
Put
∂∅
∂ λ =0
∂∅
∂ λ =25−x− y=0 … … … … .(3)
9

(c) Solve the three equation to find the value of first order conditions which is x*, y* λ*
50−4 x− y −λ=95−6 y−x−λ
−45−3 x=−5 y
y=9+( 3
5 ) x
25−x− y=0
25−x− {9+( 3
5 ) x }=0
25−x−9−3
5 x=0
16−8 x
5 =0
80−8 x=0
x=10 or say x∗¿ 10
y=9+( 3
5 ) x +9+ (3
5 )10
y∨ y∗¿ 15
50−4 x− y −λ=0
50−40−15−λ=0
−5−λ=0
λ∗¿−5
Now, profit
π=50 x−2 x2−xy−3 y2 +95 y
π=50∗10−2¿ 102−10∗15−3 152+ 95∗15
π∗¿ 1925−1025=900
(d) Show that the solution is maximum
The first step is to find the second derivative of the function
∂2 L
∂ x2 =0 (For maximization , second order ) (w.r.t.x)
∂2 L
∂ x2 = ∂2
∂ x2 { ( 50 x−2 x2 −xy−3 y2+95 y ) +λ ( 25−x− y ) }
10

¿ ∂
∂ x ¿
¿ ∂
∂ x {50−4 x− y +5 }
¿−4
−4<0
Similarly,
∂2 L
∂ y2 =0( For maximization , second order) (w.r.t.y)
∂2 L
∂ y2 = ∂2
∂ y2 { ( 50 x−2 x2−xy−3 y2 +95 y ) + λ (25−x− y ) }
¿ ∂
∂ y ¿
¿ ∂
∂ y { −x−6 y +95+5 }
¿−6
−6<0
Both,
∂2 L
∂ x2 ∧∂2 L
∂ y2 <0
Hence, it can be sad that the solution is a maximum because the second derivative comes out
to be lower than zero.
(e) Interpretation
50−4 x− y −λ=0
50−40−15−λ=0
−5−λ=0
λ∗¿−5
The economical interpretation of the Lagrange multiplier represents that for a given utility
maximizing problem is the marginal utility of the time. The λ∗¿ indicates the rate of change in the
extreme value with respect to the change in the constraints. The optimal value of the λ∗¿ indicates
the measures of the marginal utility of the income which means the rate of increase in maximized
utility increases as the corresponding income increases.
(f) Approximate change in the maximized profit when the output constraints reduced from 25 to 24.
11

Here,
x + y=24
x=24− y
Now,
−45−3 x=−5 y
−45−3 ( 24− y )=−5 y
−45−72+3 y +5 y=0
8 y=117
y= 117
8
And x=24− y =24−( 117
8 )=75
8
Thus, profit through profit function
Maximize π=50 x−2 x2−xy−3 y2 +95 y
π=50 ( 75
8 )−2 ( 75
8 )2
− ( 75
8 )∗( 117
8 )−3 ( 117
8 )2
+95 (117
8 )=903.5625
The profit has increased by (903.5625 -900) = 3.56 when the output constraint has reduced
from 25 to 24. Thus, the approximate change in the maximized profit is 3.5625.
Question 6
Utility function
U =f ( x , y )
(a) (i) Objective function and constraints and variables
Objective is to maximize the utility function.
Let the unit price of goods x=Px
Let the unit price of goods y=P y
Suppose, x goods are to be produced and the total amount spend in goods x would be Px(x )
Whereas, the total amount spend in goods y would be Py ( y )
Consumer spends all of his income M to purchase the two goods.
Total price would be Px ( x ) + P y ( y )
Objective function
12

U =f ( x , y )
Constraints
Px ( x ) + P y ( y ) ≤ M (Budget constraints)
Choice variables: x and y (Independent variables)
(ii) Lagrangean multiplier (λ) to obtain the Lagrangean of this given problem
L ( x , y , λ ) =f ( x , y ) +λ g ( x , y )
Here, g ( x , y ) =Px ( x ) +Py ( y ) −M
Now,
g( x , y)=0
Px ( x ) + P y ( y ) =M
f x + λ gx=0
f x + λ Px=0 ;
f y +λ gy=0
f y +λ Py=0 ;
Hence, the final Lagrangean of the given problem
L=f ( x , y ) + λ ( Px ( x ) +Py ( y ) −M )
(b) (i) First order condition of the given problem in terms of marginal utilities
∂ L
∂ x =0
∂ f
∂ x −L Px=0
Let the Marginal Utilities is (MU)
M U x= ∂ f
∂ x =L Px … … … … .(1)
Similarly,
∂ L
∂ y =0
∂ f
∂ y −L Py=0
M U y= ∂ f
∂ y =L Py … … … … .(2)
13

(ii) First order equation represents the ratio of marginal utilities which is equal to the price
ratio
Based on the above two equations
M U x=L Px ……………. (3)
And
M U y=L P y………………. (4)
Divide equation (3)/(4) (Ratio of marginal utilities)
M U x
M U y
= L Px
L P y
M U x
M U y
= Px
Py
Thus, proved that ratio of marginal utilities is equal to the price ratio.
(iii) Interpretation of the tangency condition
Slopeof the budget line= Px
P y
Further,
Slopeof the indifferene curve= M U x
M U y
Here, in order to maximize the utility the consumer should consume both x and y goods or
else the bundle of goods x and y at which the budget line is tangent to the respective
indifferences curve. Further, higher income would result higher utilities. Thus, it can be said
that M should be high.
(c) The given condition that must be satisfy by the problem so as to say that the solution to
the first order condition is a maximum is shown below.
The problem is related to the equality constraints problem, for which the constraints were
exactly satisfied (Active) at a solution and there was a slack which is strictly positive then the
optimal condition of first order problem would easily be replaced by the optimal conditions
by considering equality case.
Equality cases
14

Further, the above sets can be expressed in terms of the Lagrange multiplier.
The sign of Lagrange multiplier is an important factor to decide whether the first order
condition for the maximization is satisfied or not. The sign of Lagrange multiplier must be
negative to satisfy the first order conditions for maximization.
(d) Let the given utility function
U =5 x0.8 y0.2
Unit prices
Px=Py=1
(i) Maximize the problem in this case
U =5 x0.8 y0.2
L ( x , y , λ ) =5 x0.8 y0.2+ λ ( Px ( x ) + Py ( y ) −M )
∂ L
∂ x =0(For maximization)
∂ L
∂ x = ∂
∂ x (5 x0.8 y0.2+ λ ( 1 x +1 y−M ) )
4.0 (x¿ ¿−0.2) y0.2 + λ=0 ¿
4 ( y
x )
0.2
+λ=¿…….. (1)
Further , ∂ L
∂ x =0
( 1 ) 0.8 y−0.8 +λ=0
( x
y )
0.8
+ λ=0 … … … … … (2)
Solve both the equation
4 ( y
x )
0.2
+ λ= ( x
y )
0.8
+ λ
4 ( y
x )0.2
=( x
y )0.8
4 y=x
15

y∗¿ x
4
Now, x + y−M=0
x +( x
4 )−M=0
M = 5
4 x∨¿
x∗¿ 4 M
5
Hence, λ∗¿− ( 4 )0.8 M
Therefore, U∗¿ f ¿
(ii) Second order condition for maximum satisfied in the present case
∂2 L
∂ x2 =0 (For maximization , second order )
∂2 L
∂ x2 = ∂2
∂ x2 ( 5 x0.8 y0.2 + λ ( x− M ) )
¿ ∂
∂ x ¿
¿ ∂
∂ x { 4 ( y
x )
0.2
+λ }
¿ 4 y0.2 x−0.2 −1 +0
¿−4 ¿ 0.2 y0.2 x−1.2
¿−0.8 y0.2 x1.2
∂2 L
∂ x2 =−0.8 y0.2
x1.2
∂2 L
∂ x2 =0
−0.8 y0.2
x1.2 =0
−0.8 y0.2
x1.2 < 0
Hence, it can be sad that for a second order condition for being maximum is satisfied in the
present case because the second derivative comes out to be lower than zero.
16

(iii) The maximized utility from the above is highlighted below.
U∗¿ f ¿
Here, the income of the consumer has increased of 2 units and hence,
U∗¿ ( 40.8 ) ( M +2 )
Approximate change in the utility ¿ 2 ( 40.8 ) =6.06286
The approximate change in the maximized utility when the income has increased by 2 units would be
6.06286.
17