BEE1020 – Basic Mathematical Economics
Dieter Balkenborg, Iannis Krassas
Class Exercises - Solutions
Department of Economics
Week 16 13-17/02/2006
University of Exeter
Exercise 1
Use the Lagrangian method to maximize the function
f (x, y) = xy
subject to the constraint
x + 2y = 200
Solution 1
L (x, y) = xy − λ (x + 2y − 200)
The solution must satisfy the three conditions
∂L
= y−λ=0⇔y =λ
∂x
∂L
= x − 2λ = 0 ⇔ x = 2λ
∂y
x + 2y = 200
The first two equations yield x = 2y and hence the last gives 4y = 200. Overall λ = y =
50, x = 100. (Notice that the “area” is xy = 50 × 100 = 5000.)
Exercise 2
A consumer has the following utility when he consumes x units of apples and y units of oranges:
2
2
u (x, y) = −x + 4x − y + 16y
Suppose the consumer has a budget of £3.20 to be spend on oranges and apples. Each apple and each orange costs £0.40. Use the method of
Lagrange to find the optimal consumption bundle:
a) Write down the budget constraint and the Lagrangian.
b) Write down the first-order conditions for a critical point for the Lagrangian. Find a condition for a critical point that does not involve the
Lagrange multiplier.
c) Use the latter condition and the budget constraint to find a candidate for the optimum. (One can show that it is indeed the optimum.)
Solution 2 a) With the given prices and budget the budget constraint is
0.4x + 0.4y = 3.2.
The Lagrangian for the constrained utility maximization problem is hence
L (x, y) = u (x, y) − λ [0.4x + 0.4y − 3.2]
= −x2 + 4x − y 2 + 16y − λ [0.4x + 0.4y − 3.2]
b) In the constrained utility maximum the two partial derivatives of the Lagrangian must
be zero:
∂L
= −2x + 4 − λ [0.4] = 0
∂x
∂L
= −2y + 16 − λ [0.4] = 0
∂x
or
or
− 2x + 4 = 0.4λ
− 2y + 16 = 0.4λ
Division of the two equations on the right yields
0.4λ
−2x + 4
=
=1
−2y + 16
0.4λ
−2x − 12 = −2y
or
− 2x + 4 = −2y + 16
or
or
y =x+6
Thus y = x + 6 must hold at the constrained utility maximum.
c) In addition, the constraint
0.4x + 0.4y = 3.2
or
x+y =
3.2
=8
0.4
or
y = 8−x
must hold at the optimum. Overall y = x + 6 and y = 8 − x must hold, so
x+6=8−x
or
2x = 2
or
x=1
Moreover, x = 1 implies y = 8 − x = 7. Thus the Lagrangian approach suggests that it is
optimal for the consumer to buy one apple and seven oranges.
Exercise 3
2z
∂ 2 z , ∂ 2 z and ∂ 2 z of the function
a) Find all second derivatives ∂ 2
, ∂y∂x
∂y∂x
∂x
∂y 2
2
2
z = f (x, y) = 2x − 2xy − 16x + 5y + 2y + 34
b) Calculate the determinant of the Hessian matrix
2
det H = det 4
∂2 z
∂x2
∂2 z
∂y∂x
∂2 z
∂y∂x
∂2 z
∂y 2
3
5
= ∂2 z
∂x2
∂2 z
∂y∂x
∂2 z
∂y∂x
∂2 z
∂y 2
=
∂2z ∂2z
∂x2
∂y 2
∂2z
−
!2
∂y∂x
Is it positive or negative?
c) Is the function convex, concave or saddle shaped? Does the function have a peak, a trough or a maximum?
Solution 3 a) The partial derivatives are
∂z
= 4x − 2y − 16
∂x
∂z
= −2x + 10y + 2
∂y
Hence the Hessian matrix with the four partial derivatives as entries is
" 2
# ∂ z
∂2z
4 −2
∂x2
∂y∂x
=
H=
∂2z
∂2z
−2 10
∂y∂x
∂y 2
b)
det H = 4 × 10 − (−2)2 = 36
2
∂ z
c) Because det H > 0 the functionis not saddle-shaped. Since ∂x
2 > 0 it is strictly convex.
1 2
The critical point (x, y) = 4 3 , 3 is hence a trough (and indeed an absolute minimum)
of the function.
2
Exercise 4
Find the critical point of the functions
a) z
=
f (x, y) = xy − 2x + 3y − 6
b) z
=
g (x, y) = 2x + 2xy − 6x + 5y − 6y + 5
c) z
2
2
2
2
h (x, y) = −2x − 2xy + 6x − 5y − 5
=
Determine whether they are troughs, peaks or saddlepoints.
Solution 4 a)
∂z
= y−2=0⇔y =2
∂x
∂z
= x + 3 = 0 ⇔ x = −3
∂y
The critical point is (x, y) = (3, 2).
det H =
0 1
1 0
= −1 < 0
The critical point is hence a saddle point.
b)
∂z
= 4x + 2y − 6 = 0
∂x
∂z
= 2x + 10y − 6 = 0| × 2
∂y
4x + 2y = 6| −
4x + 20y = 12| +
−18y = −6
Hence y = 31 , 4x = 6 −
2
3
x = 34 . The critical point is (x, y) =
4 2
det H =
= 40 − 4 = 36 > 0
2 10
=
16
,
3
2
4 1
,
3 3
. Since
∂ z
the critical point is not a saddle point. Since ∂x
2 = 4 > 0 it is a trough. (In fact, the
function is strictly convex and the critical point a global minimum.)
c)
∂z
= −4x − 2y + 6 = 0
∂x
∂z
= −2x − 10y = 0| × (−2)
∂y
−4x − 2y = −6| +
4x + 20y = 0| +
18y = −6
3
Hence y = − 13 , 4x = 6 + 32 =
20
,
3
x = 53 . The critical point is again (x, y) =
−4 −2
det H =
= 40 − 4 = 36 > 0
−2 −10
5
, − 13
3
. Since
2
∂ z
the critical point is not a saddle point. Since ∂x
2 = −4 < 0 it is a peak. (In fact, the
function is strictly concave and the critical point a global maximum.)
Exercise 5
Suppose £5000 is invested at an annual interest rate of 10%. Compute the balance after 10 years if interest is compounded
(a) annually, (b) quarterly, (c) monthly, (d) continuously.
Solution 5
a) P = 5000 (1 + 0.1)10 ≈ 12969
b) P = 5000 (1 + 0.025)40 ≈ 13425
120
1
c) P = 5000 1 +
≈ 13535
120
d) P = 5000e10×0.1 ≈ 13591
Exercise 6
How much should be invested today at 7% compounded quarterly so that it will be worth £5000 in five years?
Solution 6
20
0.07
= 1.4148P0
5000 = P0 1 +
4
5000
= 3534.1
P0 =
1.4148
£3535 must be invested.
4