3. TIC i esport. Una necessitat i un repte
3.3. Internet. Evolució i conceptes bàsics de la xarxa
3.3.3. Internet i societat: geografia i usos de la xarxa
In Chapter 4, we discussed how we should go about solving the maximisation (minimi-sation) problem for a single-variate function. What are the procedures if the function is bi-variate? Recalling that we used the first and second derivatives in Chapter 4, some of you may think about using the first- and second-order partial derivatives in the same way.
It’s great if you think like that, but the problem turns out to be much more complicated.
Forget about the utility maximisation problem for a while, and consider finding the maximum/minimum of a general function, f (x1, x2). This problem is referred to as the unconstrained optimisation problem because you can choose any (x1, x2) to max-imise/minimise the function; that is, you are not constrained to choose from particular combinations of (x1, x2). In contrast, in the utility maximisation problem, which we come back to later, you can’t choose any (x1, x2) because you have a limited budget. For now, however, you are allowed to choose any combination of (x1, x2).
Look at Figure 6.10. As you can see, the maximum of the function is achieved at Point A, and (x1∗, x2∗) is the optimal combination of x1 and x2. Let us think about the mathematical properties at Point A. What has to be true at Point A?
Firstly, it has to be the case that if we draw a plane (the shaded parallelogram) that is tangent to the function, it has to be parallel to the (x1, x2) plane (also shaded). This is analogous to the single-variable case where we could draw a horizontal line at the maximum. In the mathematics context, it means the following.
Allow x1and x2to move simultaneously by an infinitesimally small amount from Point A(both directions as the arrows suggest on the diagram). It corresponds to taking the total differential of the function:
df = ∂f(x1, x2)
∂x1 dx1+ ∂f(x1, x2)
∂x2 dx2, (6.27)
165 6.6 Maxima and minima revisited
x2 x1
f (x1, x2)
B f (x1*, x2*)
x1* x2*
Figure 6.11 Local minimum.
where dx1 and dx2 correspond to the changes in x1 and x2. For Point A to be the local maximum, the total differential df must be zero, for any small movements dx1 and dx2
(because it is on the top; not going up or down). It follows that both the partial derivatives in Equation (6.27) must be zero.
However, as we can guess, this property is not enough. As in the one-variable case, we can find the exact opposite case, where we find the local minimum of a function by using the same condition. That situation is depicted in Figure 6.11, where Point B is the local minimum of the function.
In any case, we find one of the conditions for the local maximum/minimum of a bi-variate function:
∂f(x1, x2)
∂x1 = ∂f(x1, x2)
∂x2 = 0. (6.28)
At the local maximum/minimum, the above set of equations has to hold. It constitutes the first-order conditions for the local maximum/minimum. Now, go back to Figure 6.10, where the local maximum is depicted. The second thing that has to be true at Point A is that the function is at the local maximum in both directions x1 and x2. It is the same as saying that both of the second-order partial derivatives of the function are negative:
∂2f(x1, x2)
∂x12 <0, ∂2f(x1, x2)
∂x22 <0. (6.29)
Likewise, in Figure 6.11, the opposite has to hold at the local minimum, Point B:
∂2f(x1, x2)
∂x12 >0, ∂2f(x1, x2)
∂x22 >0. (6.30)
They are parts of the second-order conditions for the local maximum and local mini-mum, respectively.
In passing, I have depicted in Figure 6.12 a special case, where the first-order conditions are met but, for the second-order conditions (those so far covered), one of the partial
x2 x1
0
C
x1*
x2* f (x1, x2)
Figure 6.12 A saddle point.
derivatives is positive, and the other is negative. Point C corresponds to such a case, where it is the local minimum in the direction of x1, but is the local maximum in the direction of x2. Mathematically,
∂2f(x1, x2)
∂x12 >0, ∂2f(x1, x2)
∂x22 <0. (6.31)
Such a point is referred to as a saddle point, for obvious reasons.
Let us return to the second-order conditions for the local minimum, i.e. (6.30). I have said that they are parts of the conditions. Indeed, we can find a counter-example in which these conditions are satisfied at a point, yet the point is neither a local maximum nor a local minimum. I will demonstrate this case in the following.
Consider the following function:
y= f (x1, x2)= x12+ 10x1x2+ x22. (6.32) The first-order conditions are:
∂f(x1, x2)
∂x1
= 2x1+ 10x2= 0, (6.33)
∂f(x1, x2)
∂x2 = 10x1+ 2x2 = 0. (6.34)
The only combination of x1 and x2 that simultaneously satisfies Equations (6.33) and (6.34) is (x1∗, x2∗)= (0, 0). To check whether this point is the local maximum, local minimum or a saddle point, we investigate the second-order conditions:
∂2f(x1, x2)
∂x21 = 2 > 0, ∂2f(x1, x2)
∂x22 = 2 > 0. (6.35) Equation (6.35) implies that (x1∗, x2∗)= (0, 0) is the local minimum (see (6.30)), and the value of the function at that point is y = f (0, 0) = 0. However, consider a real number a and take a point (x1, x2)= (−a, a). Substituting this point into Equation (6.32) yields the
167 6.6 Maxima and minima revisited
following:
f(−a, a) = a2− 10a2+ a2 = −8a2. (6.36) Note that lim
a→0f(−a, a) = 0, which means f (−a, a) approaches 0 as a becomes closer to zero. But, more importantly, f (−a, a) approaches 0 from negative values. Looking at it from the other way, it implies that a movement from (0, 0) to (−a, a) will lower the value of the function, which suffices in showing that (0, 0) is not the local minimum. It is not a local maximum either, in passing (try using (a, a) instead of (−a, a) to verify this claim).
It turns out that, for the point we find using the first- and second-order conditions (that have been introduced so far) to be either the local maximum or the local minimum, the following relationship has to be satisfied at the point in question:
∂2f(x1, x2)
∂x12 ·∂2f(x1, x2)
∂x22 >
∂2f(x1, x2)
∂x2∂x1
2
, (6.37)
or, more concisely,
f11· f22> f122. (6.38) This expression completes the second-order conditions. The derivation of this condi-tion will not be given here, to maintain the flow of our discussion. If you are interested in the derivation, see Appendix A. In the meantime, we will just check that this condition is not met for the problem we just investigated. For f (x1, x2)= x12+ 10x1x2+ x22:
∂2f(x1, x2)
∂x2∂x1 = 10. (6.39)
The RHS of (6.37) is 100, whereas the LHS is 4. Hence (6.37) (or (6.38)) is violated.
Question (The profit maximisation problem with two inputs) Think about applying what we have learnt to the profit maximisation problem. Although we maintain the assumption of the competitive input and output markets, the setup here is slightly different from that in Chapter 4. That is, we have two inputs of production, instead of one. Denoting the level of production by q, the production function of the firm can be expressed as follows:
q = f (K, L), (6.40)
where L is labour and K is the number of robots. Since there are two inputs the firm want to choose the optimal K and L, not just the output, q. In the case where K does not exist (as in Chapter 4), controlling L meant the same as controlling q. But here, the firm can control K as well as L, and different combinations of K and L might result in producing the same amount of q. In that case, the firm would want to choose the most efficient combination of K and L to produce that amount. Hence, we consider the firm that tries maximising the profits by choosing the amount of both robots (K) and labour (L). Let us denote the price of the good, a robot and labour by p > 0, r > 0, and w > 0,
respectively. Then the profits of the firm can be expressed as:
π(K, L)= pf (K, L) − rK − wL. (6.41)
Notice that π (K, L) is expressed as a function of K and L: but not π (q) as it was in Chapter 4. Also we are assuming that the fixed cost of production is zero. Now we can mathematically represent the profit maximisation problem as follows:
Max
K,L π(K, L). (6.42)
Now, the question is the following: if the production function is given as (6.43) and inputs prices are given as r= 2 and w = 1, solve the profit maximisation problem and obtain the amounts of the inputs that the firm hires:
f(K, L)= K13L13. (6.43)
Solution
MaxK,L π(K, L)= Max
K,L [pf (K, L)− 2K − L] . The first-order conditions are:
p∂f(K, L)
∂K = 2,
p∂f(K, L)
∂L = 1.
These equations imply the following:
1
3pK−23L13 = 2, 1
3pK13L−23 = 1.
Solving the above for K and L, the input combination that satisfies the first-order condi-tions is,
K∗, L∗
=
p3 108,p3
54
.
To check this combination is the true local (and global) maximum, we check the second-order conditions:
∂2π
∂K2 <0,
∂2π
∂L2 <0,
∂2π
∂K2 ·∂2π
∂L2 −
∂2π
∂L∂K
2
>0.
169 6.7 The utility maximisation problem: constrained optimisation
Calculating the LHS of these, we have:
∂2π
Therefore, given p > 0, the second-order conditions are satisfied for any K > 0 and L > 0
and so of course they are satisfied at the point of concern (K∗, L∗)=p3
Exercise 6.6 The profit maximisation with two inputs.