Solutions to Problem Set 4: Constrained Optimisation
1. Draw a function that is quasi-concave but not concave and explain why.
Lots of possibilities. The dotted line shows that the curve drawn is not concave.
More to the point you need to show that your curve satisfies the definition of quasiconcave.
Quasi-convexity requires
f(x″) = f(λx + (1-λ)x‘ ) <= max(f(x), f(x’) )
(compare with convexity f(x″) < λf(x) + (1-λ)f(x‘) )
Quasi-concavity requires
f(x″) = f(λx + (1-λ)x‘ )>= min(f(x), f(x’) )
2. Sketch the function 2x3
convex
-
and work out if the function is quasi-concave or quasi-
Note that this function is both concave and convex around a point of inflection
at x = 0
Consider f(x) =2x3 and f(x’) = 2x’3 and let x’ > x
and compare with f(λx + (1-λ)x‘ )= [λ2x + (1-λ)2x‘]3
If let x’ > x then 2x’3 >= 2x3 so f(x’) >= f(x)
And since 0 < λ < 1 it must be that 2x’3 >= [λ2x + (1-λ)2x‘]3 >= 2x3
f(x’) > f(λx + (1-λ)x‘ ) > f(x)
So this function is either quasi-concave or quasi-convex
Lagrangeans
1. Optimise the following function subject to the given constraint
Z = 4x2 – 2xy + 6y2
st
x+ y = 72
Lagrangrean is then
L = 4x2 – 2xy + 6y2 + λ(72 - x – y)
FOC
Lx = 8x – 2y – λ = 0
Ly = -2x +12y – λ = 0
L λ = 72 - x –y = 0
(1) –(2) gives
10x-14y =0
(1)
(2)
(3)
so x = 1.4y
Sub,. X = 1.4y into (3) gives
2.4y=72
So y = 30
and x = 42 and so λ = 8(42) – 2(30) = 276
2.
For each case, identify the objectives and the constraints. Write down the
constraint function and indicate whether it is a constraint that must hold with equality
or not. Finally, write down the Lagrangean.
Example:
A new business employs L workers and wishes to maximize
profits 2L-L2. There are only 12 workers available for hire
Answer:
The objective function is 2L-L2. There are two constraints: L ≥ 0
and 12 –L ≥ 0. Neither needs to hold with equality. The
Lagrangean is 2L-L2 + λ(12-L) + µL whereµ and λ are the
Lagrangean multipliers
(i)
A government wishes to choose its spending on health (H) and
education (E) expenditure levels to maximize its popularity P = E + 2H
+ EH2. Its annual budget is £40bn and it is not allowed to borrow
money
(ii)
A consumer's utility function is: u = x0.25y0.75
where x and y are two
goods. Total income is £10,000 and the prices of the two goods are £4
and £6 respectively
1.
(i)
Objective function: P = E + 2H + EH2.
Constraints are 40 – E – H ≥ 0 and that E≥ 0 and H ≥ 0.
Ignoring these last two constraints,
Lagrangrean is then just, L = E + 2H + EH2 + λ(40- H-E).
(ii)
Objective function is x0.25y0.75
Constraint is 10000-4x-6y ≥ 0
and
that x and y must both be non-negative.
Ignoring these last two constraints, Lagrangrean is then
L = x0.25y0.75 + λ(10000-4x-6y)
3.
A monopolist faces a demand curve of q = 20 – p, where q is output and p is
its price. Total costs are given by 4q
(i).
(ii).
Write down an expression for the monopolist’s profits as a function of its
output
Find the profit-maximizing level of output and the associated level of
profits
(iii).
A fire in one of its factories means that the maximum output for the firm
is constrained to 6. Use constrained optimization to find the profit
maximizing output for the firm under these conditions
(iv).
What are the firm’s profits?
(v).
What is the value of the lagrangean multiplier?
(vi).
The firm can build extra production capacity at a cost of 5. Should it do
so?
.
(i)
(ii)
(iii)
20q-q2 -4q or 16q-q2
FOC says differentiate wrt q and set equal to zero
Gives q = 8, so profits = 64
Max π = 16q-q2 st q<=6
Set up lagrangean L = 16q-q2 + λ(6-q)
With
L
 2q 16    0
q
complementary slackness condition
g  0 and   0 and g  0
λ(6-q) = 0 and λ ≥ 0 and 6-q ≥ 0
Option 1: g =0 and λ ≥ 0
Gives q = 6 and λ ≥ 0
From FOC we have λ = 16-2q which is > 0 if q==6
Option 2: g>0 and λ= 0
With q = 6 g = 6 -6 =0 so not consistent
Hence option 1 must hold
q=6
constraint is binding – would prefer to produce more output
(compare with unconstrained solution)
(iv)
(v)
(vi)
profits =16q-q2 =16(6)-62 = 60
4
If the firm can build a fraction of a unit of capacity then yes,
We know that and λ = the marginal gain in profit of an increase in output
below the constraint
-
From the FOC of the lagrangean below the constraint
L
C q  0

because the marginal gain in profits

 2q 16  0
q
and the FOC of the lagrangean wrt q is
L
 2q 16    0
q
Then λ equals the marginal gain in profits = 6 at q = 5
and the question says that the marginal cost of capacity is 5 at any level
of output
If it can only build whole units of capacity then the firm should not bother
because with a capacity of 7 its profits would only be 63 – an increase
of only 3 and therefore not enough to outweigh the extra cost
Linear & Non-Linear Programming
1. Draw a diagram to show the constraints and the constraint set.
(i)
(ii)
(iii)
x ≥ 0, y ≥ 0, x + 2y ≤ 10
x ≥ 0, y ≥ 0, 2x + y ≤ 16, y + x ≥ 4
consumption of x is not negative, consumption of y is not negative, the
price of x is 4, the price of y is 6 and the consumer has 600 pounds to
spend. The consumer cannot buy more than 30 units of y.
y
5
10
(i)
(ii)
x
x ≥ 0, y ≥ 0, x + 2y ≤ 10. The constraint set is shaded.
x ≥ 0, y ≥ 0, 2x + y ≤ 16, y + x ≥ 4
y
16
4
4
8
x
(iii)
consumption of x is not negative, consumption of y is not negative, the price of
x is 4, the price of y is 6 and the consumer has 600 pounds to spend. The
consumer cannot buy more than 30 units of y.
y
100
150
2.
2 x1  3 x 2
x
Maximise the following objective function using linear programming
s.t.
x1  4 x 2  160
3 x1  x 2  135
x1  x 2  50
x1  0
x2  0
First 3 constraints
are “resource constraints”
Last two are non-negativity constraints
Graphically
3x1+x2 =135
x1+x2=50
x1+4x2=160
Constraints intercept x and y axes at
(160, 40)
(45, 135)
(50,50)
Respectively
Constraints form the feasible set
Optimal solution lies within extreme points – on the nodes of the boundary of the set
bounded by
A= (0, 40)
B = (34.5, 31.4)
from constraint 1
with x1 =0
from intersection of constraint 1 and constraint 2
X1+4x2 =160
3x1 + x2 =135
Gives x1 = 34.5, x2 =31.4
C= (42.5, 7.5)
from intersection of constraints 2 & 3
3x1 + x2 =135
x1 + x2 = 50
D = (45,0)
from constraint 2 with x2 =0
Note that all other constraints are not feasible
Now consider the output level at each of these points given by 2x1 + 3x2
D = (45,0) = 90
C= (42.5 7.5) = 2(42.5) +3(7.5) =107.5
A = (0,40) = 3(4) =120
B = (34.5, 31.4) = 2(34.5) +3(31.4) = 163.2
So B is the preferred solution
3.
A consumer's utility function is:
u = x0.25y0.75 where x and y are two goods.
(i)
(ii)
(iii)
(iv)
(v)
Suppose total income is £10,000 and the prices of the two goods are
£4 and £6 respectively. Use constrained optimisation to find the
consumer's demand for both goods.
Now replace the price of the second good with p. Find a formula for the
consumer’s demand for this good. Draw the demand curve and
comment on its properties
What is the own-price elasticity of demand for the first good?
What happens to the demand for the first good if income rises to
£12,000?
What can you say about the income elasticity of demand for the first
good?
.
(i)
(ii)
(iii)
(iv)
(v)
x = 625, y = 1,250
y = 7,500/p. points to note are:
a. it’s downward sloping,
b. definitely not linear,
c. but a hyperbola instead,
d. that the curve never touches the axes,
e. so that as price falls towards zero demand goes to infinity
f. and as price goes to infinity demand goes towards zero
g. and the derivative is -7500/p2.
1 (note: to answer this question you need to find a formula for x, in the
same way you obtained a formula for y.)
It rises to 750
That’s 1 as well. To answer this you need first to find the formula for
the demand curve, which is x = 0.25m/px where m is income and px is
the price of x. Then use the definition of elasticity to get an answer.
p
y
6. Optimise the function
z = 4x2 + 3xy +6y2
subject to
x + y = 56
using the bordered Hessian to check the second order conditions
Form the Lagrangean L= 4x2 + 3xy +6y2 +λ(56-x-y)
FOC
Lx = 8x + 3y – λ = 0
Ly = 3x + 12y – λ = 0
L λ = 56 – x – y = 0
(1)
(2)
(3)
(1)-(2) gives
5x -9y = 0
Sub. X = 1.8y into (3)
56 -1.8y – y = 0
For SOC form the border Hessian
 2L
2L
2L 



2
x1
xn 
 
 2L
2L
2L 



H  x 
x12
x1xn 
 1


 
 
 2L
2L
2L 
 x  x x  x 2 
n
1
n
 n

so x = 1.8y
so y=20 and x = 36

 0

 g
H   x1

 
 g
 x
 n
g
x1
2L
x12

2
 L
xn x1
g 

xn 
2L 


x1xn 

 
2
 L 

xn2 

Which in this case is
0

H  g x
g y

_
gx
f xx
f yx
g y   0  1  1

f xy    1 8
3 
f yy   1 3 12 
For this to be a maximum we need leading principal minors to alternate in sign
D0  0, D1  0; D2  0; D3  0...
And for a minimum the determinants of the bordered Hessian should be all nonpositive
(and 1st determinant will always be zero so don’t need to check and second
2
determinant D1 will always be
<=0 so don’t need to check )
 g 

 
 x1 
So sign depends on
_
H2
 0  1((1.12)  (1.3))  1((1.3)  (8.  1))  9  5  14  0
So determinant of bordered Hessian is negative which means the optimisation at
y =20 x = 36 is a minimum
7. Find the second order conditions that guarantee maximisation of the following
utility function
U = q1q2
Subject to the constraint Y = p1q1+p2q2
Where Y = income and U = utility
Form the Lagrangean L= q1q2 +λ(Y - p1q1 - p2q2)
FOC
L1 = q2 – λp1 = 0
L2 = q1 – λp2 = 0
L λ = Y - p1q1 - p2q2 = 0
(1)
(2)
(3)
(1)/(2) gives
Q2/p1 = q1/p2
so q2 = q1p1/p2
Sub. q2 = q1p1/p2
Y= p1q1 + p2 q1p1/p2
and q1 = q2p2/p1
into (3)
so q1 = Y/2p1 and q2 = Y/2p2
(these are the Marshallian demand functions for each good which maximise utility)
For SOC form the bordered Hessian
0
H   g1
 g 2
_
g1
f11
f 21
g2   0
f12    p1
f 22   p 2
p1
0
1
p2 
1 
0 
For this to be a maximum we need leading principal minors to alternate in sign
D0  0, D1  0; D2  0; D3  0...
(and 1st determinant will always be zero so don’t need to check and second
2
determinant D1 will always be
<=0 so don’t need to check )
 g 

 
 x1 
So sign depends on
_
H2
 0  p1 ((0  (1. p 2 ))  p 2 ( p1  0)  2 p1 p 2  0
(assuming non-negative prices)
So the determinant of the bordered Hessian is positive
optimisation of the utility function is a maximum
which means the