BEE1020 — Basic Mathematical Economics Dieter Balkenborg
Week 12, Lecture Thursday 19.01.06
Department of Economics
Functions in two variables
University of Exeter
1. Objective
Functions in two variables.
Partial differentiation ←→ partial analysis in economics
level curves ←→ indifference curves or isoquants
unconstrained extrema, first order conditions
The lecture should enable you to calculate partial derivatives
and marginal rates of substitution and to determine the first order
conditions for a maximum.
2. Functions in two variables
A function z = f (x, y) or simply z (x, y) in two independent
variables with one dependent variable assigns to each pair (x, y)
of (decimal) numbers from a certain domain D in the two-dimensional
plane IR2 a number z = f (x, y).
x and y are hereby the independent variables
z is the dependent variable.
The graph of f is the surface in 3-dimensional space consisting of
all points (x, y, f (x, y)) with (x, y) in D.
2
Example: The cubic polynomial
z = f (x, y) = x3 − 3x2 − y 2
6
4
2
-12
2
1 x
3
0
-2
y
-4
-6
-8
Exercise: Evaluate
z = f (2, 1) , z = f (3, 0) , z = f (4, −4) , z = f (4, 4)
Could the information fit with the graph?
3
Example: production function:
√
√
1 1
6
Q = K L = K 6 L2
capital K ≥ 0, labour L ≥ 0, output Q ≥ 0
6
4
2
0
K
14
5
0
2
4
6
4
L
16
18
20
Example: profit function. Assume that the firm is a price
taker in the product market and in both factor markets.
P is the price of output
r the interest rate (= the price of capital)
w the wage rate (= the price of labour)
total profit of this firm:
Π (K, L) = P Q − rK − wL
=
1 1
P K 6 L2
P = 12, r = 1, w = 3:
Π (K, L) =
− rK − wL
1 1
12K 6 L 2
5
− K − 3L
20
15
10
5
0
-5
-10
K
20
0
L
5
Profits is maximized at K = L = 8.
6
15
20
The level curve of the function z = f (x, y) for the level c is the
solution set to the equation
f (x, y) = c
where c is a given constant.
Geometrically, a level curve is obtained by intersecting the graph
of f with a horizontal plane z = c and then projecting into the
(x, y)-plane. This is illustrated here for the cubic polynomial discussed above:
2
-1
1
5
2
1 x
1
2
3
y
y
1 x
-1
-2
-2
7
In the case of a production function the level curves are called
isoquants. An isoquant shows for a given output level capitallabour combinations which yield the same output.
20
18
16
14
12
6
4
2
0
K
K
8
6
4
2
6 8 L
0 0 2 4
20
14 16 18
5
0
2
4
6
L
8
20
16 18
Finally, the linear function
z = 3x + 4y
has the graph and the level curves:
5
4
30
3
y
2
20
10
y
0
0
2
x
3
0
4
0
5
1
2
x
3
4
5
The level curves of a linear function form a family of parallel lines:
c 3
c = 3x + 4y ⇔ 4y = c − 3x ⇔ y = − y
4 4
slope − 34 , variable intercept 4c .
9
Exercise: Describe the isoquant of the production function
Q = KL
for the quantity Q = 2.
Exercise: Describe the isoquant of the production function
√
Q = KL
for the quantity Q = 4.
10
Remark: The exercises illustrate the following general principle:
If h (z) is an increasing (or decreasing) function in one variable,
then the composite function h (f (x, y)) has the same level curves
as the given function f (x, y) . However, they correspond to different levels.
25
5
20
4
15
3
10
4
L
2
L
2
5
0
2
4
1
2
K
3
4
5
1
0
√
Q = KL
Q = KL
11
1
2
K
3
4
5
3. Partial derivatives
Exercise: What is the derivative of
z (x) = a3x2
with respect to x when a is a given constant?
Exercise: What is the derivative of
z (y) = y 3b2
with respect to y when b is a given constant?
12
Consider function z = f (x, y). fix y = y0, vary only x: z =
F (x) = f (x, y0).
The derivative of this function F (x) at x = x0 is called
the partial derivative of f with respect to x and de∂z
dF
dF (x ).
noted by ∂x
=
=
dx |x=x
dx 0
|x=x ,y=y
0
0
0
Notation: “d”=“dee”, “δ”=“delta”, “∂”=“del”
It suffices to think of y and all expressions containing only y as
exogenously fixed constants. We can then use the familiar rules
∂z .
for differentiating functions in one variable in order to obtain ∂x
∂f
Other common notations for partial derivatives are ∂f
,
∂x ∂y or fx,
fy .
13
Example: Let
Then
z (x, y) = y 3x2
∂z
= 3y 2x2
∂y
∂z
= 2y 3x
∂x
14
Example: Let
z = x3 + x2y 2 + y 4.
Setting e.g. y = 1 we obtain z = x3 + x2 + 1 and hence
∂z
= 3x2 + 2x
∂x |y=1
∂z
=5
∂x |x=1,y=1
15
For fixed, but arbitrary, y we obtain
∂z
= 3x2 + 2xy 2
∂x
as follows:
We can differentiate the sum x3 + x2y 2 + y 4 with respect to x
term-by-term.
Differentiating x3 yields 2x2, differentiating x2y 2 yields 2xy 2 because we think now of y 2 as a constant and
2
d ax
= 2ax
dx
holds for any constant a.
Finally, the derivative of any constant term is zero, so the derivative
of y 4 with respect to x is zero.
16
Similarly considering x as fixed and y variable we obtain
∂z
= 2x2y + 4x3
∂y
∂q
∂q
Example: The partial derivatives ∂K
and ∂L
of a production
function q = f (K, L) are called the marginal product of capital and (respectively) labour. They describe approximately by
how much output increases if the input of capital (respectively
labour) is increased by a small unit.
17
Fix K = 64, then q =
1 1
K 6 L2
=
1
2L 2
which has the graph
L
3
4
3
Q
2
1
0
1
2
18
4
5
This graph is obtained from the graph of the function in two variables by intersecting the latter with a vertical plane parallel to
L-q-axes.
6
4
2
0
K
14
5
0
2
4
6
16
18
20
L
∂q
∂q
The partial derivatives ∂K
and ∂L
describe geometrically the slope
of the function in the K- and, respectively, the L- direction.
19
Diminishing productivity of labour: The more labour is used,
the less is the increase in output when one more unit of labour is
employed. Algebraically:
√
∂q 1 1 − 1 1 6 K
> 0,
= K 6L 2 = √
2
∂L 2
2 L
√
2
∂ q
1 1 −3
∂ ∂q
1 6K
= − K 6L 2 = − √
=
< 0,
2
2
∂L ∂L
4
4 L3
∂L
Exercise: Find the partial derivatives of
z = x2 + 2x y 3 − y 2 + 10x + 3y
20
4. The tangent plane
df
For a function y = f (x) in one variable the derivative dx
at
|x=x0
x0 is the slope of the tangent to the graph of f through the point
(x0, f (x0)).
4
3
2
1
0
0.2
0.4
0.6
0.8
1x
1.2
1.4
1.6
1.8
2
-1
The tangent of f (x) = x2 at x = 1.
This tangent is the graph of the linear function
df
y = l (x) = f (x0) +
(x − x0) .
dx |x=x0
21
This is so because l (x0) = f (x0) , so the graph of l goes through
the point (x0, f (x0)).
Moreover,
y = l (x) =
df
f (x0) −
x0
dx0 |x=x0
df
+
x,
dx |x=x0
so l (x) is indeed a linear function in x with intercept
df
x0
f (x0) −
dx0 |x=x0
df
and the “right” slope dx
.
|x=x0
22
The graph of a function in one variable is a curve and its tangent
is a line. By analogy, the graph of a function z = f (x, y) in two
variables is a surface and its tangent at a point (x0, y0) is a plane.
5
4
3
2
1
0
2
1
1
3
2
4
3
4
23
Non-vertical planes are the graphs of linear functions z = ax +
by + c. The tangent plane of the function z = f (x, y) in (x0, y0)
is, correspondingly, the graph of the linear function
z = l (x, y)
(1)
∂f
∂f
= f (x0, y0) +
(x − x0) +
(y − y(2)
0)
∂x |x=x0,y=y0
∂y |x=x0,y=y0
since the graph of this linear function contains the point
(x0, y0, f (x0, y0))
and has the right slopes in the x- and y-directions.
24
Exercise: This is the example shown
in the above graph. The
√
√
tangent plane of z = f (x, y) = x + y at the point (1, 1) is
obtained as follows: We have
∂z 1 1
∂z 1 1
and
= √
= √
∂x 2 x
∂y 2 y
and hence
∂z
1
=
∂x |x=1,y=1 2
and
∂z
1
= .
∂y |x=1,y=1 2
Since f (1, 1) = 2 the tangent plane is the graph of the linear
function
1
1
1
1
z = l (x, y) = 2 + (x − 1) + (y − 1) = 1 + x + y
2
2
2
2
25
Using the abbreviations dx = x−x0, dy = y −y0 and dz = z −z0
the formula for the tangent can be neatly rewritten as
∂z
∂z
dz = dx + dy
(3)
∂x
∂y
an expression which is called the total differential.
26
5. The slope of level curves
Suppose (x0, y0) is a point on the level curve f (x, y) = c. Then
the slope of tangent to the level curve in this point in the x-y-plane
is
dy
∂z
∂z
=−
(4)
dx |x=x0,y=y0
∂x |x=x0,y=y0 ∂y |x=x0,y=y0
∂z
provided ∂y
|x=x0,y=y0
= 0. This is known as the rule for implicit
differentiation: If the function y = y (x) is implicitly defined by
dy
is given by the
the equation f (x, y (x)) = c, then its derivative dx
above formula.
27
Example: A linear function
y = ax + by + c
∂z = a and ∂z = b.
has the partial derivatives ∂x
∂y
According to the above formula its level curves have the slope
a
∂z ∂z
=− .
−
∂x ∂y
b
Indeed, its level curves are the solutions to the equations
c̄ = ax + by + c
with some constant c̄. Solving the equation for y we obtain
a
c̄ − c
y =− x+
b
b
which is a linear function with slope − ab .
28
1 1
K 6 L2
Example: An isoquant of the production function Q =
has according to the above formula the slope
dK
∂Q ∂Q
1 −5 1
1 1 −1
=−
= − K 6L 2
K 6 L2
dL
∂L ∂K
2
6
K
6 1 5 −1 −1
6
6
2
2
= − K K L L = −3 .
2
L
in the point (K, L).
29
Indeed, an isoquant is the set of solutions to the equation q̄ =
1 1
K 6 L 2 for fixed q̄. Solving for K we see that the isoquant is the
graph of the function
6
1
K = q̄L− 2 = q̄ 6L−3.
Differentiation yields
dK
= −3q̄ 6L−4
dL
and since (K, L) is on the isoquant
1 1 6
dK
K
−4
= −3 K 6 L 2 L = −3 .
dL
L
30
Economically, − dK
dL is the marginal rate of substitution of
labour for capital: If labour is reduced by one unit, how much more
capital is (approximately) needed to produce the same output?
Just reducing labour by one unit reduces output by the marginal
∂q
∂q
product of labour ∂L
. Each additional unit of capital produces ∂K
— the marginal product of capital
— more units. Therefore, if now
∂q
∂q
capital input is increased by ∂L
∂K units, output changes overall
by
∂q
∂q
∂q ∂q
−
+
= 0.
∂L
∂L ∂K ∂K
31
Remark: A quick way to memorize the formula (4) is to use the
total differential as follows: In order to stay on the level curve one
must have dz = z − z0 = 0, so
∂z
∂z
0 = dz = dx + dy
∂x
∂y
hence
∂z ∂z
dy = −
dx
∂x ∂y
or
dy
∂z ∂z
=−
.
dx
∂x ∂y
32
Remark: The implicit function theorem states that if
∂z
= 0
∂y |x=x0,y=y0
for some point (x0, y0) on a level curve f (x, y) = c, then one
can find a function y = g (x) defined near x = x0 such that
the level curve is locally the graph of this function, i.e. one has
f (x, g (x)) = c for x near x0.
Exercise: For the production function
√
Q = KL
calculate the marginal rate of substitution for K = 3 and L = 4.
33
6. The chain rule
Suppose that z = f (x, y) is a function in two variables x, y.
Suppose that x depends on the variable t via x = g (t) and that
y depends on the variable t via y = h (t). Then we can define
the composite function in one variable z = F (t) = f (x (t) , y (t))
which has t as the independent and z as the dependent variable.
The derivative of this function, if it exists, is calculated as
dz ∂z dx ∂z dy
=
+
.
dt ∂x dt ∂y dt
Notice that this is just the total differential divided by dt.
34
Example: Suppose z = xy, x = t3 + 2t, y = t2 − t. Then,
according to the chain rule,
∂z dx ∂z dy
dz
=
+
= y × 3t2 + 2 + x × (2t − 1)
dt
∂x dt ∂y
dt = t2 − t 3t2 + 2 + (2t − 1) t3 + 2t
3
2
Since z = xy = t + 2t t − t we get the same result using
the product rule:
dz 2
= 3t + 2 t2 − t + t3 + 2t (2t − 1) .
dt
35