where C is consumer expenditure, Yd is disposable income, Y is national income, I is investment, t is the tax rate and G is government expenditure. What is the marginal propensity to consume out of Y? What is the value of the govern-ment expenditure multiplier? How much does government expenditure need to be increased to achieve a national income of 700?
10.3 Second-order partial derivatives
Second-order partial derivatives are found by differentiating the ﬁrst-order partial derivatives of a function.
When a function has two independent variables there will be four second-order partial derivatives. Take, for example, the production function
Q = 25K0.4L0.3
There are two ﬁrst-order partial derivatives
∂Q = 10K−0.6L0.3 ∂Q = 7.5K0.4L−0.7
These represent the marginal product functions for K and L. Differentiating these functions a second time we get
∂K2 = −6K−1.6L0.3 ∂2Q = −5.25K0.4L−1.7
These second-order partial derivatives represent the rate of change of the marginal product functions. In this example we can see that the slope of MPL (i.e. ∂2Q/∂L2) will always be negative (assuming positive values of K and L) and as L increases, ceteris paribus, the absolute value of this slope diminishes.
We can also ﬁnd the rate of change of ∂Q/∂K with respect to changes in L and the rate of change of ∂Q/∂L with respect to K. These will be
∂2Q −0.6 −0.7 ∂K∂L
∂2Q −0.6 −0.7 ∂L∂K
and are known as ‘cross partial derivatives’. They show how the rate of change of Q with respect to one input alters when the other input changes. In this example, the cross partial derivative ∂2Q/∂L∂K tells us that the rate of change of MPL with respect to changes in K will be positive and will fall in value as K increases.
You will also have noted in this example that
∂2Q ∂2Q ∂K∂L ∂L∂K
In fact, matched pairs of cross partial derivatives will always be equal to each other. Thus,foranycontinuoustwo-variablefunctiony = f(x,z),therewillbefoursecond-order
partial derivatives:
2 (i) ∂x2
(ii) ∂zy
2 (iii) ∂x∂z
2 (iv) ∂z∂x
© 1993, 2003 Mike Rosser
with the cross partial derivatives (iii) and (iv) always being equal, i.e.
∂2y ∂2y ∂x∂z ∂z∂x
Example 10.12
Derive the four second-order partial derivatives for the production function
Q = 6K +0.3K2L+1.2L2
and interpret their meaning.
Solution
The two ﬁrst-order partial derivatives are
∂Q = 6 +0.6KL ∂Q = 0.3K2 +2.4L
and these represent the marginal product functions MPK and MPL. The four second-order partial derivatives are as follows:
2
(i) ∂K2 = 0.6L
This represents the slope of the MPK function. It tells us that the MPK function will have a constant slope along its length (i.e. it is linear) for any given value of L, but an increase in L will cause an increase in this slope
2
(ii) ∂L2 = 2.4
This represents the slope of the MPL function and tells us that MPL is a straight line with slope 2.4. This slope does not depend on the value of K.
2
(iii) ∂K∂L = 0.6K
This tells us that MPK increases if L is increased. The rate at which MPK rises as L is increased will depend on the value of K.
2
(iv) ∂L∂K = 0.6K
This tells us that MPL will increase if K is increased and that the rate of this increase will depend on the value of K. Thus, although the slope of the MPL schedule will always be 2.4, from (ii) above, its actual position will depend on the amount of K used.
Some other applications of second-order partial derivatives are given below.
© 1993, 2003 Mike Rosser
Example 10.13
A ﬁrm sells two competing products whose demand schedules are
q1 = 120 −0.8p1 +0.5p2 q2 = 160 +0.4p1 −12p2
How will the price of good 2 affect the marginal revenue of good 1?
Solution
To ﬁnd the total revenue function for good 1 (TR ) in terms of q1 we ﬁrst need to derive the inverse demand function p1 = f(q1). Thus, given
q1 = 120 −0.8p1 +0.5p2 0.8p1 = 120 +0.5p2 −q1
p1 = 150 +0.625p2 −1.25q1 TR1 = p1q1
= (150 +0.625p2 −1.25q1)q1 = 150q1 +0.625p2q1 −1.25q2
Thus
MR1 = ∂TR1 = 150 +0.625p2 −2.5q1 1
This marginal revenue function will have a constant slope of −2.5 regardless of the value of p2 or the amount of q1 sold.
The effect of a change in p2 on MR1 is shown by the cross partial derivative
∂q1∂p2 = 0.625
Thus an increase in p2 of one unit will cause an increase in the marginal revenue from good 1 of 0.625, i.e. although the slope of the MR1 schedule remains constant at −2.5, its position shifts upward if p2 rises. (Note that in order to answer this question, we have formulated the total revenue for good 1 as a function of one price and one quantity, i.e. TR1 = f(q1,p2).)
Example 10.14
A ﬁrm operates with the production function Q = 820K0.3L0.2 and can buy inputs K and L at £65 and £40 respectively per unit. If it can sell its output at a ﬁxed price of £12 per unit, what is the relationship between increases in L and total proﬁt? Will a change in K affect the extra proﬁt derived from marginal increases in L?
© 1993, 2003 Mike Rosser
Solution
TR = PQ = 12(820K0.3L0.2) TC = PKK +PLL = 65K +40L
Therefore proﬁt will be
π = TR −TC
= 12(820K0.3L0.2)−(65K +40L) = 9,840K0.3L0.2 −65K −40L
The effect of an increase in L on proﬁt is shown by the ﬁrst-order partial derivative:
∂π = 1,968K0.3L−0.8 −40 (1)
This effect will be positive as long as
1,968K0.3L−0.8 > 40
However, if L is continually increased while K is held constant, the value of the term 1,968K0.3L−0.8 will eventually fall below 40 and so ∂π/∂L will become negative.
To determine the effect of a change inK on the marginal proﬁt function with respect to L, we need to differentiate (1) with respect to K, giving
∂L∂K = 0.3(1,968K−0.7L−0.8) = 590.4K−0.7L−0.8
This cross partial derivative will be positive as long as K and L are positive. This is what we would expect and so an increase in K will have a positive effect on the extra proﬁt generated by marginal increases in L. The magnitude of this impact will depend on the values of K and L.
Second-order and cross partial derivatives can also be derived for functions with three or more independent variables. For a function with three independent variables, such as y = f(w,x,z) there will be the three second-order partial derivatives
∂2y ∂2y ∂2y ∂w2 ∂x2 ∂z2
plus the six cross partial derivatives
∂2y ∂2y
∂w∂x ∂x∂w
∂2y ∂2y
∂x∂z ∂z∂x
∂2y ∂2y
∂w∂z ∂z∂w
These are arranged in pairs because, as with the two-variable case, cross partial derivatives will be equal if the two stages of differentiation involve the same two variables.
© 1993, 2003 Mike Rosser
Example 10.15
For the production function Q = 32K0.5L0.25R0.4 derive all the second-order and cross partial derivatives and show that the cross partial derivatives with respect to each possible pair of independent variables will be equal to each other.
Solution
The three ﬁrst-order partial derivatives will be
∂Q = 16K−0.5L0.25R0.4 ∂Q = 8K0.5L−0.75R0.4
∂Q = 12.8K0.5L0.25R−0.6
The second-order partial derivatives will be
∂K2 = −8K−1.5L0.25R0.4 ∂2Q = −6K0.5L−1.75R0.4
∂2Q = −7.68K0.5L0.25R−1.6
plus the six cross partial derivatives:
∂2Q −0.5 −0.75 0.4 ∂2Q ∂K∂L ∂L∂K
∂2Q 0.5 −0.75 −0.6 ∂2Q ∂L∂R ∂R∂L
∂2Q −0.5 0.25 −0.6 ∂2Q ∂R∂K ∂K∂R
Second-order derivatives for multi-variable functions are needed to check second-order conditions for optimization, as explained in the next section.
Test Yourself, Exercise 10.3
1. For the production function Q = 8K0.6L0.5 derive a function for the slope of the marginal product of L. What effect will a marginal increase in K have upon this MPL function?
2. Deriveallthesecond-orderandcrosspartialderivativesfortheproductionfunction Q = 35KL+1.4LK2 +3.2L2 and interpret their meaning.
© 1993, 2003 Mike Rosser
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