BEE1020 — Basic Mathematical Economics
Dieter Balkenborg, Giovanni Caggiano
Axel Dreher, Iannis Krassas
Class Exercises - Solutions
03-04/02/2004
Week 14
Exercise 1
Department of Economics
University of Exeter
Find the critical point of the function
2
2
z = f (x, y) = 2x − 2xy − 16x + 5y + 2y + 34
Solution 1 The first-order conditions for a critical point are
∂z
= 4x − 2y − 16 = 0
∂x
∂z
= −2x + 10y + 2 = 0
∂y
Adding twice the second equation to the first equation yields
18y = 12 or y =
2
3
Substituting this into the second equation yields
2x = 10 ×
Exercise 2
2
20 6 26
13
1
+ 2 or 2x =
+ =
or x =
=4
3
3
3
3
3
3
A T-shirt shop carries two competing shirts, one endorsed by Michael Jordan and the other by Shaq O’Neal. The owner
of the store can obtain both at a cost of $2 per shirt and estimates that if Jordan shirts are sold for x dollars apiece and O’Neal shirts for y
dollars apiece, consumers will buy approximately 40 − 50x + 40y Jordan shirts and 20 + 60x − 70y O’Neil shirts each day.
a) Express as functions of x and y: i) the revenue from selling Jordan shirts, ii) the revenue from selling O’Neal shirts iii) the costs for
shirts and iv) the overall profit.
b) Find the critical point of the profit function.
c) How should the owner price the shirts in order to generate the largest possible profit?
d) Calculate he Hessian matrix for this problem and its determinant. Is the solution found in b) indeed an absolute maximum?
Solution 2 a) i) revenue from selling Jordan shirts (price times quantity sold):
(40 − 50x + 40y) x
ii) revenue from selling O’Neil shirts (price times quantity sold):
(20 + 60x − 70y) y
iii) costs from buying both types of shirts (buying price times total quantity bought):
2 ((40 − 50x + 40y) + (20 + 60x − 70y)) = 2 (60 + 10x − 30y) = 120 + 20x − 60y
iv) total profit
Π (x, y) = (40 − 50x + 40y) x + (20 + 60x − 70y) y − (120 + 20x − 60y)
= (40 − 50x + 40y) x + (20 + 60x − 70y) y − 120 − 20x + 60y
b) I use the product rule and the multiplicative constant rule because it is faster. Those
enjoying calculations can first expand and then differentiate.
0=
∂Π
= −50x + 40 − 50x + 40y + 60y − 20 = 20 − 100x + 100y
∂x
0=
∂Π
= 40x − 70y + 20 + 60x − 70y + 60 = 100x + 80 − 140y
∂y
c) We must solve the simultaneous linear system of equations
100x − 100y = 20
100x − 140y = −80
Subtraction of the two equations yields
40y = 100
10
100
=
= 2.5
y =
40
4
The first equation yields therefore
100x − 100 × 2.5
100x − 250
100x
x
=
=
=
=
20
20
270
2.7
Thus Jordan T-shirts should be sold for $2.70 and O’Neal shirts for $2.50.
d) The Hessian matrix is
# ·
"
¸
∂Π2
∂Π2
−100 100
2
∂x
∂x∂y
=
H=
∂Π2
∂Π2
100 −140
∂y∂x
∂y2
and its determinant is
det H = (−100) (−140) − (100)2 = 14000 − 10000 = 4000 > 0.
2
< 0 and det
The function has a relative maximum at the critical point found because ∂Π
∂x2
H > 0. Since the function is defined on the entire plane (a convex set) and since there is
only one relative maximum, this is also an absolute maximum.
Exercise 3
The highway department is planning to build a picnic area for motorists along a major highway. It is to be rectangular
with an area of 5, 000 square yards and is to be fenced off on the three sides not adjacent to the highway. What is the least amount of fencing
that will be needed to complete the job.
a) Identify this problem as a constraint optimization problem. What objective function f (x, y) is to be maximized / minimized subject
to what constraint g (x, y) ≥ c?
b) Write down the Lagrangian L (x, y) for this problem.
c) Find the solution to the three equations a) ∂L
= 0 b) ∂L
= 0 c) g (x, y) ≥ c.
∂x
∂y
2
Solution 3 a) Let x denote the length of the side of the rectangle adjacent to the highway
and let y denote the length of the other side. Then we have to minimize the amount of
fencing z = x +2y susbject of the constraint that the area xy is at least 5000 (xy ≥ 5000).
b) The Lagrangian is
L (x, y) = x + 2y − λ (xy − 5000)
c)
∂L
= 1 − λy = 0
∂x
∂L
= 2 − λx = 0
∂y
5000 = xy
or
1 = λy
or
2 = λx
λy
Division of the two equations on the right yields 12 = λx
= xy or x = 2y. Hence 5000 =
xy = 2y 2 , y 2 = 2500, y = 50 (since the solution −50 does not make sense). We obtain
x = 100 and y = 50.
Exercise 4
1
1
A consumer with utility function u (x, y) = x 2 y 4 and a budget of b = $1, 200 to spend on two commodities wishes to
maximize his utility. The first commodity has a price of $10 per unit, the second a price of $20. In order to find how much he should buy of
each commodity
a) write down the Lagrangian. L (x, y) for this problem.
2
∂L
c) find the solution to the three equations a) ∂L
∂x = 0 b) ∂y = 0 c) g (x, y) ≥ c. (Remark: You find the same solution using v (x, y) = x y
instead of u (x, y).)
d) What is the Lagrange multiplier? (harder:) Does the Lagrangian L (x, y) indeed have a maximum at the solution found in c)?
Solution 4 a)
1
b)
1
L (x, y) = x 2 y 4 − λ (10x + 20y − 1200)
∂L
1 −1 1
=
x 2 y 4 − 10λ = 0
∂x
2
∂L
1 1 −3
=
x 2 y 4 − 20λ = 0
∂y
4
10x + 20y = 1200
or
or
1 −1 1
x 2 y 4 = 10λ
2
1 1 −3
x 2 y 4 = 20λ
4
Division of the equations on the right yields
1 − 12 14
x y
2
1 12 − 34
x y
4
1
1
1
=
3
10λ
20λ
2x− 2 x− 2 y 4 y 4 = 2x−1 y = 2
y
1
=
x 2
x = 4y
The budget equation gives 10x + 20y = 40y + 20y = 60y = 1200, so y = 20 and x = 4y =
80.
3
1
1
1
d) The Lagrange multiplier is λ = 20
(80)− 2 (20) 4 = 18.22. The Hessian matrix for
the Lagrangian is
# ·
"
¸
∂L2
∂L2
1 − 32 14
1 − 12 − 34
−
x
y
x
y
2
∂x
∂x∂y
= 1 4− 1 − 3 8 3 1 − 7
H=
∂L2
∂L2
x 2 y 4 − 16 x 2 y 4
2
∂y∂x
∂y
8
and has the determinant
µ
¶µ
¶ µ
¶2
1 −3 1
3 1 −7
1 −1 −3
2
4
2
4
2
4
det H =
− x y
− x y
−
x y
4
16
8
6
3
3 −1 − 6
1
2
x y 4 − x−1 y − 4 = x−1 y − 2
=
64
64
64
2
which is positive everywhere, especially at the solution found. Since ∂L
< 0 and det H >
∂x2
0 the Lagrangian has a relative maximum at this solution found. Because the Lagrangian
is defined for all positive pairs of numbers (x, y), which form a convex set, and since it
has a unique critical point, we have an absolute maximum of the Lagrangian. Hence we
have also an absolute constrained maximum of the utility function.
4