# 程序代写案例-5SSPP232

5SSPP232 MATHEMATICS FOR ECONOMICS
1. Suppose that a prot maximising rm can sell as many pencils as it likes at a price
p > 0: Its prot function is gi
ven 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 rms prot.
(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,
d (q)
dw
=
@ (q)
@q
dq
dw
+
@ (q)
@w
=
@ (q)
@w
= q2 = p
2
4w2
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
f 0(x) = 3
p
x2 +
2
3
(x 1)x 13
= x
1
3 (x+
2
3
(x 1))
=
x
1
3
3
(5x 2) = 0;
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
f 00(x) = x
4
3
9
(5x 2) + 5
3
x
1
3 ;
which evaluated at x = 25 is f
00(25) = 2:262 > 0: Hence, x =
2
5 is a local minimum.
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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.
(a) Compute the limit from both above and below
lim
x!3
x2 4
x2 5x+ 6 =
5
0
= 1
lim
x!3
x2 4
x2 5x+ 6 =
5
0+
= +1
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
[email protected] 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
1 a
2 2

x1
x2

=

3
4

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

1 2
2 2

= 2

2 2
2 1

A1 =
1 1
1 0:5

So the solution to the system of equations is
x1
x2

=
1 1
1 0:5

3
4

=

1
1

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 =
[email protected] 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.
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