ECON7030 Microeconomic Analysis
Lecture 5: Expenditure Minimisation
University of Queensland
Semester 1, 2022
Outline
1 Expenditure Minimisation
• Tangency Method
• Lagrangian Method
2 Duality: The link between Utility Maximisation and
Expenditure Minimisation
Part I
Expenditure Minimisation
Utility Maximisation (Last Lecture)
• Given income and prices, what is the consumption bundle that
maximises an individual’s utility among all the affordable
bundles?
• Tangency Method
• Lagrangian Method
• Corner solutions and non-differentiable utility functions
This Lecture: Expenditure Minimisation
• But we can turn the question around: Given prices and a
target utility level, what is the least cost bundle that achieves
the targeted utility?
• This is known as Expenditure Minimisation
Why Expenditure Minimisation?
• It turns out to be a good auxiliary tool for analysing demand
(we will see this in Lecture 6)
• It helps answering welfare-related question: How much do we
have to give to a consumer for her to achieve certain utility?
• It turns out to be very similar to cost minimisation in
production — which we will see later in this course
Expenditure Function
• The expenditure associated with a bundle (x , y) is the cost of
buying it:
E (x , y) = Pxx + Pyy .
• Again, since E is a two-variable function, we are typically
interested in its contour lines on the x-y plane
• But this is easy: the collection of (x , y) such that E (x , y) is
constant (let’s call it k) is simply the budget line with income
k
• The closer the line is to the origin, the smaller the expenditure
Contour Lines of the Expenditure Function
x
y
k
Py
Pxx + Pyy = k
k
Px
k ′
Py
Pxx + Pyy = k
′, k ′ < k
k ′
Px
Expenditure Minimisation on a Graph: 1
x
y
UE (x , y) = k
A
Key:
Achieves U
Cost less
than A
• Bundle A is not optimal — there are bundles achieving at
least U which cost less than A
Expenditure Minimisation on a Graph: Part 2
x
y
UE (x , y) = k
A
Key:
Achieves U
Costs strictly
less than A
• Bundle A is optimal — any bundle that costs strictly less than
A cannot achieve U
Expenditure Minimisation on a Graph: Part 3
• In other words, if bundle A is expenditure minimising subject
to achieving at least U, then:
1 A must achieve at least U AND
2 The set of bundles that costs strictly less than A is separated
from the set of bundles that are weakly preferred to A
The Tangency Condition
x
y
BL U1
A
Key:
Affordable
Strictly
Prefers to A
• When
1 The utility function (hence the indifference curves) is
differentiable; and
2 The optimal bundle does not occur on either axes,
the previous condition implies that the expenditure minimising
bundle is where the budget line passing through the bundle is
tangent to the targeted indifference curve
Tangency Condition, I know it!
• As in Utility Maximisation, tangency of the budget line and
the indifference curve at A means
−Px
Py
= −MUx(A)
MUy (A)
Or, the absolute value of the MRS at A is equal to the
relative price
Tangency Condition: Interpretation
x
y
BL
BL∗ U
A
Willing to give up Y for X exactly as
required, no further adjustment
More willing to give
up Y for X than
required, give up
more Y for X
Less willing to give
up Y for X than
required, give up
more X for Y
• MRS measures how many units of good Y a consumer is
willing to give up for an additional unit of good X
• Relative price measures how many units of good Y a
consumer needs to give up for an additional unit of good X
Finding the Expenditure-Minimising Bundle
• The tangency condition gives us
MUx(x , y)
MUy (x , y)
=
Px
Py
• The constraint is now that (x , y) has to achieve U (rather
than being affordable as in the utility maximisation problem)
U(x , y) = U
• This gives us two equations in two unknowns (x and y)
• We can solve for the optimal bundle by solving these two
equations simultaneously
Finding the Expenditure-Minimising Bundle
Numerical Example
• Recall the utility function that we have been looking at:
U(x , y) = x
1
2 y
1
2
• Suppose
Px = 2
Py = 3
U =
√
6
• Goal: Find the expenditure-minimising bundle
Finding the Expenditure-Minimising Bundle
Numerical Example
• Recall that the MRS of this utility function is −y/x
• Hence the tangency condition gives us
y
x
=
Px
Py
=
2
3
• Meanwhile, the utility constraint is
x
1
2 y
1
2 = U
x
1
2 y
1
2 =
√
6
Finding the Optimal Bundle: Numerical Example
• We are solving for x and y in
y
x
=
2
3
(1)
x
1
2 y
1
2 =
√
6 (2)
• Rearranging Equation (1)
y =
2
3
x
• Substitute this into Equation (2)
x
1
2
(
2
3
x
) 1
2
=
√
6
x =
√
6× 3
2
= 3
Finding the Optimal Bundle: Numerical Example
• Finally back out y :
y =
2x
3
=
2× 3
3
= 2.
• The expenditure-minimising bundle is (3, 2)
A More General Example
• Consider a general Cobb-Douglas utility function:
U(x , y) = xαyβ
• The expenditure function is Pxx + Pyy
• And we have the target utility constraint
U(x , y) = U
• Goal: Find the optimal consumption bundle (x , y) (In terms
of α, β, Px , Py and U)
Step 1: Find MRS
• Since I only need the MRS, I am going to use the trick in last
lecture and take natural log of the utility function. Let
V (x , y) = lnU(x , y) = α ln x + β ln y
• This gives me
MUVx =
∂V
∂x
= α
1
x
MUVy =
∂V
∂y
= β
1
y
• Hence we have
MRS = −MU
V
x
MUVy
=
αy
βx
Step 2: Equate MRS with Relative Price
• Next I am going to use the tangency condition:
MRS = −Px
Py
αy
βx
=
Px
Py
αPyy = βPxx
y =
βPx
αPy
x
Step 3: Solve the Simultaneous Equations (part 1)
• Now we have two equations (tangency condition and budget
line) in two unknowns (x and y):
y =
βPx
αPy
x
α ln x + β ln y = lnU
• Write ln y as a function of ln x using the tangency condition:
y =
Pxβ
Pyα
x
ln y = ln x + ln
(
Pxβ
Pyα
)
Step 3: Solve the Simultaneous Equations (part 2)
• Substitute the expression of ln y in terms of ln x into the
target utility constraint:
α ln x + β ln x + β ln
(
Pxβ
Pyα
)
= lnU
(α + β) ln x = lnU − β ln
(
Pxβ
Pyα
)
ln x =
1
α + β
lnU − β
α + β
ln
(
Pxβ
Pyα
)
Step 3: Solve the Simultaneous Equations (part 3)
• You could have taken exponentials on both sides of the
equation on the last slide and recover x (instead of ln x), but
I’d rather leave it in ln x for a while until I find ln y :
ln y =
1
α + β
lnU − β
α + β
ln
(
Pxβ
Pyα
)
+ ln
(
Pxβ
Pyα
)
=
1
α + β
lnU +
α
α + β
ln
(
Pxβ
Pyα
)
Step 4: Loose Ends
• Mathematically, if I had found ln x and ln y , I have found x
and y
• But you’d probably want to see just x and y , so here they are:
ln x =
1
α + β
lnU +
β
α + β
ln
(
Pyα
Pxβ
)
ln y =
1
α + β
lnU +
α
α + β
ln
(
Pxβ
Pyα
)
• Taking exponentials on both sides:
x = U
1
α+β
(
Pyα
Pxβ
) β
α+β
y = U
1
α+β
(
Pxβ
Pyα
) α
α+β
Why is this different from the U-Max Solution?
1 The most fundamental answer: Because they are functions of
different variables
• Optimal bundle under U-Max is a function of Px , Py and M
• Optimal bundle under E-Min is a function of Px , Py and U
2 Procedurally, the tangency condition is substituted into the
• Budget Constraint in U-Max
• Target Utility Constraint in E-Min
3 The two optimal bundles may be numerically the same given a
certain M and U, but they respond differently to (even the
same) changes in Px and Py
Corner Solutions and Kinked Indifference Curves
• As in utility maximisation, we can have corner solutions or
kinked indifference curves
• The ideas are the same so I won’t repeat them
• They are also less important for our purposes of introducing
expenditure minimisation
Lagrangian
• As in utility maximisation, we can also use the Lagrangian for
the expenditure minimisation problem
• With two goods, the Lagrangian does not offer much more
advantage over the tangency method, but it allows extensions
to more than two goods
• The method is very similar to what we have learnt last week,
except that
1 We have a minimisation instead of maximisation
2 The constraint is different
Expenditure Minimisation
min
x≥0,y≥0
Pxx + Pyy
s.t. U(x , y) ≥ U
• The above is read as:
“minimise Pxx + Pyy by choosing x and y (both of which
must be weakly positive) subject to the utility constraint”
• Some terminology:
Objective function The function to be minimise
Constraints (In)equalities that the solution must satisfy
Choice variables Variables to be solved, also known as
endogenous variables
Parameters Variables given, also known as exogenous
variables
The Lagrangian
min
x ,y ,µ
M = Pxx + Pyy + µ
(
U − U(x , y))
• As in utility maximisation, we form the Lagrangian (M)
• The added variable µ is the Lagrange Multiplier — it
measures how “tight” the constraint is
• If the constraint is not binding (“loose”), µ is zero
• If the constraint is binding (“tight”), U − U(x , y) is zero
• (I use M and µ just to distinguish this problem from the
Lagrangian for utility maximisation. There is no deep meaning
to the choice of notations.)
Sign of the Lagrangian
• You might be wondering whether it should be U(x , y)− U or
U − U(x , y)
• Mathematically this only changes the sign of µ and makes no
real differences
• Lagrangian Multipliers (LM) and constraint setup:
LM Constraint
Maximisation ≥ 0 ≥ 0
≤ 0 ≤ 0
Minimisation ≥ 0 ≤ 0
≤ 0 ≥ 0
• The ”positive” LM will have easy to interpret economic
meanings.
Lagrangian: First Order Conditions (FOC)
min
x ,y ,µ
M = Pxx + Pyy + µ
(
U − U(x , y))
• We then (partially) differentiate the Lagrangian with respect
to each of the choice variables and set the derivatives to
zeroes to get the First Order Conditions (FOC):
[x ] :
∂M
∂x
= Px − µUx(x , y) = 0
[y ] :
∂M
∂y
= Py − µUy (x , y) = 0
[µ] :
∂M
∂µ
= U − U(x , y) = 0
How about the Second Order Conditions (SOC)?
• Just as in a single-variable unconstrained maximisation, the
Second Order Condition (SOC) guarantees that the FOC is
characterising a local minimum, rather than a local maximum
or an inflection point
• As in utility maximisation, the SOC involves a Hessian Matrix
• However, you would expect that the sign requirements on the
SOC are different from that for a maximisation problem
Back to the FOC
• We have 3 equations:
Px − µUx(x , y) = 0
Py − µUy (x , y) = 0
U − U(x , y) = 0
in 3 unknowns: x , y , µ
• Unless the Hessian Matrix is singular (don’t worry about what
this is), this system of equations can be solved
• Economists are often more interested in knowing that there is
a solution, rather than finding the solution, so we are happy
now
Another Look at the FOC
• The 3 equations can be written as
Px = µUx(x , y)
Py = µUy (x , y)
U − U(x , y) = 0
• If I divide the first two equations I get:
Px
Py
=
Ux(x , y)
Uy (x , y)
which is the tangency condition!
• Needless to say, the third equation is the utility constraint
• So we have got exactly the same equations as in the tangency
condition approach
µ as Marginal Cost of Utility
• Moreover, I can also write the first two equations as
Px
Ux(x , y)
= µ
Py
Uy (x , y)
= µ
• The two fractions on the LHS are the extra expenditure the
consumer has to pay to get the last unit of utility from x and
y , respectively
• Hence µ can be interpreted as the marginal cost of utility —
how much extra expenditure the consumer is required to pay
if he wants to increase his utility by one unit
You may be wondering
• Both U-Max and E-Min have the tangency condition
• I can see that the constraints are different, but. . .
• Suppose I choose the M and U “deliberately” so that the M
would correspond to the minimal expenditure solved under
some U, wouldn’t I get the same bundle?
• This idea is correct — but let’s make it more precise
Utility Maximisation
max
x ,y
U(x , y) s.t. Pxx + Pyy ≤ M
• Given Px , Py and M, we can solve for the optimal bundle for
this problem
• So write
xM(Px ,Py ,M) the optimal amount of x given Px , Py and M
yM(Px ,Py ,M) the optimal amount of y given Px , Py and M
U∗(Px ,Py ,M) the maximal utility achieved given Px , Py and
M
(Note: The “M” superscript stands for “Marshallian” — we
will see this in next lecture)
Expenditure Minimisation
min
x ,y
Pxx + Pyy s.t. U(x , y) ≥ U
• Given Px , Py and U, we can solve for the optimal bundle for
this problem
• So write
xH(Px ,Py ,U) the optimal amount of x given Px , Py and U
yH(Px ,Py ,U) the optimal amount of y given Px , Py and U
E ∗(Px ,Py ,U) the minimal expenditure required given Px , Py
and U
(Note: The “H” superscript stands for “Hicksian” — we will
see this in next lecture)
Duality: Graphical Approach
x
y
U = U∗
E ∗/PyM/Py
A
Duality: In Words
• Fix some Px , Py
• Suppose we start with some U, do the expenditure
minimisation, and find xH(Px ,Py ,U), y
H(Px ,Py ,U) and
E ∗(Px ,Py ,U)
• Now let’s run a utility maximisation with Px , Py and
M = E ∗(Px ,Py ,U)
• Then we should get the same bundle
• Moreover, the maximal utility that can be achieved should be
U
Duality
• Therefore,
xH(Px ,Py ,U) ≡ xM(Px ,Py ,E ∗(Px ,Py ,U))
yH(Px ,Py ,U) = ≡ yM(Px ,Py ,E ∗(Px ,Py ,U))
U∗(Px ,Py ,E ∗(Px ,Py ,U)) = ≡ U
• Notice that the above holds for any Px , Py and U — they are
identities
• This result is known as Duality, and it will become a very
handy tool next lecture
Can we do it the other way?
• Just now I start with U, run expenditure min, get E ∗, and
then use it as the income in a utility max
• You may wish to start with M, run utility max, get U∗ and
then use it as the utility in an expenditure min
• Nothing prohibits you from doing so, just that the result
won’t be as useful