# 5SSPP232辅导、辅导Java/Python编程语言

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1. Suppose that a prot maximising rm can sell as many pencils as it likes at a price

p > 0: Its prot function is given by:

(q) = qp wq2

where w > 0.

(a) How much output will the rm produce?

The FOC is p 2wq = 0: Hence q = p2w : The SOC is 2wq which ensures

that q indeed maximizes the prot.

(b) How much pro?t will the rm make?

(q) = qp wq2 = p24w :

(c) Now suppose that w increases. How does this a¤ect the optimal choice of

output and the ?rm?s pro?t?

For the optimal choice of output just compute dq

dw = p2w2 : Hencet, output is

decreasing in w: For the pro?t, use the envelope theorem to obtain,

Alternatively, you could just compute directly d(q

)

dw = p

2

4w2

: Hence, pro?t is

also decreasing in w:

2. Find the stationary points of the following function, and determine whether each

corresponds to a max, a min, or an in?ection point.

f(x) = (x 1) 3

p

x2

Any stationary point must satisfy

which only has one solution at x = 25 : Note that f

0(0) = 20 = 1; so x = 0 is

not a stationary point. The second derivative is

which evaluated at x = 25 is f

00(25) = 2:262 > 0: Hence, x =

2

5 is a local minimum.

1

3. According to the O￠ ce of National Statistics, 100 GBP in 1970 were equivalent to

1,457.38 GBP in 2017.

(a) Using this data, ?nd the anual increase in prices (i.e. in?ation rate). In other

words, at which rate prices had to increase annually so 100 GBP in 1970

would buy the same basket of goods (i.e. have the same value) as 1,457.38

GBP in 2017.

Call the in?ation rate I: We are thus looking to ?nd

100(1 + I)47 = 1457:38

which yields I = 0:0587 or, in other words, a 5.87% annual in?ation rate.

(b) Find the anual rate of in?ation assuming now that prices change monthly.

Now we are looking to ?nd

100(1 +

I

12

)4712 = 1457:38;

which yields I = 0:05714, or, in other words, a 5.714% annual in?ation rate.

4. Calculate the following limits. Please show your work.

(a) Compute the limit from both above and below

Hence the limit does not exist :

(b) limx!a x

3a3

x2a2 =

3

2a

(c) limx!1

p

x+2px+1p

x

= 0

5. Find the value of the following integrals

(a)

R 4

1

x2+x1p

x

dx = 22615 15:06

(b)

R

x sin(x)dx = sinx x cosx+ C

6. Determine whether the following matrices are invertible:

(a) det

1 2

1 1

= 1 so it?s invertible

(b) det

0@ 1 2 11 3 4

2 4 2

1A = 0 so it?s not invertible

2

7. Consider the following system of equations

x1 + ax2 = 3

2x1 + 2x2 = 4

(a) Write these equations in matrix form Ax = b

(b) Suppose that a = 2. Find the solution to this system of equations by inverting

So the solution to the system of equations is

8. Consider the following consumer problem:

max

x;y

u(x; y) = 60x+ 90y 2x2 3y2

s.t. 2x+ 4y = 68

(a) Use the Lagrange method to ?nd the stationary point(s) of this function.

The Lagrangian is

L(x; y; ) = 60x+ 90y 2x2 3y2 (2x+ 4y 68)

The ?rst order conditions are

@L(x; y; )

@x

= 60 4x 2 = 0

@L(x; y; )

@y

= 90 6y 4 = 0

@L(x; y; )

@

= 2x 4y + 68 = 0

3

Use the ?rst two FOCs to establish that

30 2x

45 3y =

1

2

This together with the third FOC yields (x; y) = (12; 11):

(b) Find the bordered Hessian matrix and show whether this point(s) is a maxi-

mum or a minimum

HB =

0@ 0 2 42 4 0

4 0 6

1A

The minors

det(HB1 ) = 4 < 0

det(HB) = 88 > 0

Hence (x; y) = (12; 11) is a maximum.

(c) Find the value of the Lagrange multiplier . What is its economic interpre-

tation?

Substituting (x; y) = (12; 11) in one of the ?rst two FOCs yields = 6. This

is how much u(x; y) increases when the income of the consumer moves away

from 68.

9. Consider the function

f(x; y) = x2 + y2 3xy

a. Find the Hessian matrix and ?nd the conditions for f(x; y) to be concave/convex

H =

2 3

3 2

det(H1) = 2 > 0

det(H2) = 5

The Hessian is inde?nite. The function is neither concave or convex.

b. Find the maximum, minimum and saddle points (if any) of this function.

The stationary points are given by the FOCs

@f(x; y)

@x

= 2x 3y = 0

@f(x; y)

@y

= 2y 3x = 0;

which has only one solution (x; y) = (0; 0): This must be a saddle point because

the Hessian is inde?nite.

4

10. Use the Kuhn-Tucker method to solve the following maximisation problem

max

x;y

3x y

s.t. y ex

x 1

See solution to mid-term problem 5, which was the same problem.

1. Suppose that a prot maximising rm can sell as many pencils as it likes at a price

p > 0: Its prot function is given by:

(q) = qp wq2

where w > 0.

(a) How much output will the rm produce?

The FOC is p 2wq = 0: Hence q = p2w : The SOC is 2wq which ensures

that q indeed maximizes the prot.

(b) How much pro?t will the rm make?

(q) = qp wq2 = p24w :

(c) Now suppose that w increases. How does this a¤ect the optimal choice of

output and the ?rm?s pro?t?

For the optimal choice of output just compute dq

dw = p2w2 : Hencet, output is

decreasing in w: For the pro?t, use the envelope theorem to obtain,

Alternatively, you could just compute directly d(q

)

dw = p

2

4w2

: Hence, pro?t is

also decreasing in w:

2. Find the stationary points of the following function, and determine whether each

corresponds to a max, a min, or an in?ection point.

f(x) = (x 1) 3

p

x2

Any stationary point must satisfy

which only has one solution at x = 25 : Note that f

0(0) = 20 = 1; so x = 0 is

not a stationary point. The second derivative is

which evaluated at x = 25 is f

00(25) = 2:262 > 0: Hence, x =

2

5 is a local minimum.

1

3. According to the O￠ ce of National Statistics, 100 GBP in 1970 were equivalent to

1,457.38 GBP in 2017.

(a) Using this data, ?nd the anual increase in prices (i.e. in?ation rate). In other

words, at which rate prices had to increase annually so 100 GBP in 1970

would buy the same basket of goods (i.e. have the same value) as 1,457.38

GBP in 2017.

Call the in?ation rate I: We are thus looking to ?nd

100(1 + I)47 = 1457:38

which yields I = 0:0587 or, in other words, a 5.87% annual in?ation rate.

(b) Find the anual rate of in?ation assuming now that prices change monthly.

Now we are looking to ?nd

100(1 +

I

12

)4712 = 1457:38;

which yields I = 0:05714, or, in other words, a 5.714% annual in?ation rate.

4. Calculate the following limits. Please show your work.

(a) Compute the limit from both above and below

Hence the limit does not exist :

(b) limx!a x

3a3

x2a2 =

3

2a

(c) limx!1

p

x+2px+1p

x

= 0

5. Find the value of the following integrals

(a)

R 4

1

x2+x1p

x

dx = 22615 15:06

(b)

R

x sin(x)dx = sinx x cosx+ C

6. Determine whether the following matrices are invertible:

(a) det

1 2

1 1

= 1 so it?s invertible

(b) det

0@ 1 2 11 3 4

2 4 2

1A = 0 so it?s not invertible

2

7. Consider the following system of equations

x1 + ax2 = 3

2x1 + 2x2 = 4

(a) Write these equations in matrix form Ax = b

(b) Suppose that a = 2. Find the solution to this system of equations by inverting

So the solution to the system of equations is

8. Consider the following consumer problem:

max

x;y

u(x; y) = 60x+ 90y 2x2 3y2

s.t. 2x+ 4y = 68

(a) Use the Lagrange method to ?nd the stationary point(s) of this function.

The Lagrangian is

L(x; y; ) = 60x+ 90y 2x2 3y2 (2x+ 4y 68)

The ?rst order conditions are

@L(x; y; )

@x

= 60 4x 2 = 0

@L(x; y; )

@y

= 90 6y 4 = 0

@L(x; y; )

@

= 2x 4y + 68 = 0

3

Use the ?rst two FOCs to establish that

30 2x

45 3y =

1

2

This together with the third FOC yields (x; y) = (12; 11):

(b) Find the bordered Hessian matrix and show whether this point(s) is a maxi-

mum or a minimum

HB =

0@ 0 2 42 4 0

4 0 6

1A

The minors

det(HB1 ) = 4 < 0

det(HB) = 88 > 0

Hence (x; y) = (12; 11) is a maximum.

(c) Find the value of the Lagrange multiplier . What is its economic interpre-

tation?

Substituting (x; y) = (12; 11) in one of the ?rst two FOCs yields = 6. This

is how much u(x; y) increases when the income of the consumer moves away

from 68.

9. Consider the function

f(x; y) = x2 + y2 3xy

a. Find the Hessian matrix and ?nd the conditions for f(x; y) to be concave/convex

H =

2 3

3 2

det(H1) = 2 > 0

det(H2) = 5

The Hessian is inde?nite. The function is neither concave or convex.

b. Find the maximum, minimum and saddle points (if any) of this function.

The stationary points are given by the FOCs

@f(x; y)

@x

= 2x 3y = 0

@f(x; y)

@y

= 2y 3x = 0;

which has only one solution (x; y) = (0; 0): This must be a saddle point because

the Hessian is inde?nite.

4

10. Use the Kuhn-Tucker method to solve the following maximisation problem

max

x;y

3x y

s.t. y ex

x 1

See solution to mid-term problem 5, which was the same problem.