revenues accruing to the farm firm. Moreover, the
equations representing the first order conditions can be
divided by each other: Maximization in the Two-Input
Case 115 6.67 pf1/pf2 = v1/v2. Note that the output
price cancels in equation 6.67 such that 6.
) From Table
2.1, 160 pounds of nitrogen per acre will result in a corn
yield or TPP of 123 bushels per acre. The concept of a
function has a good deal of impact on the basic
assumptions underlying the economics of agricultural
production. Another possibl
problem for the mathematician. The saddle is level in the
middle, but it slopes upward at both ends and downward
at both sides. A saddle looks like neither a hill nor a
valley, but is a combination of both. So an approach
needs to be taken that will separ
On the expansion path: VMPx1/v1 = VMPx2/v2 = 0 Global
output maximization VMPx1/v1 = VMPx2/v2 < 0 Stage III
for both inputs; the profit-maximizing farmer would not
operate here VMPx1/v1 = VMPx2/v2 = 0 < . < 1 Between
profit and output maximum; farmer woul
< 0 6.86 pf11pf22 ! pf12pf21 > 0 6.87 p2 (f11f22 !
f12f21) > 0 Since p2 is positive, the required signs on the
second-order conditions are the same for both profit and
yield maximization. 6.8 Concluding Comments This
chapter has developed some of the fund
was the rapid growth in the use of the computer as a
device for estimating or measuring relationships within
an economy. Economists now routinely use techniques
for estimating models in which the computational
requirements would have been considered impos
producers. Production of hogs and cattle in the United
States is often closer to a purely competitive environment
in which a large number of farm firms take prices
generated by overall supply and demand for hogs and
cattle. However, there are a relatively
at the top of the hill is zero in all directions. It is not
possible to distinguish the bottom of a valley from the
top of a hill simply by looking at the slope at that point,
because the slope for both is zero. Much of the
mathematics of maximization and
bundle of the two inputs x1 and x2. Suppose that the
proportion of each input contained in the bundle is
defined by the expansion path. If the expansion path has
a constant slope, then as one moves up the expansion
path, the proportion of x1 and x2 does n
remainder of the text. Rather, reliance will be placed on
the purely competitive model as the starting point for
much of our analysis, with modifications made as needed
to meet the particular features of the problem. 1.13
Concluding Comments The purely co
the maximum output given the budgeted dollars C or
subject to the budget constraint. Another approach is to
think of the amount of output represented by a particular
isoquant as being fixed. Then the point of tangency
between the isoquant and the iso-outl
rationale, it seems unlikely that 140 pounds of nitrogen
would produce a yield of 119 bushels, and a more likely
guess might be 120 or 121 bushels. These are only
guesses. In reality no information about the behavior of
the function is available at nitrog
regard to what to produce given available land, labor,
machinery, and equipment. The manager must not only
decide how much of each particular commodity to be
produced, but also how available resources are to be
allocated among alternative commodities. The
economy. Economists rely heavily on what is sometimes
called comparative statics. The economic relationships
are often represented by a graph: for example, a graph
showing a supply curve and a demand curve. An event or
shock affecting demand or supply is
variable, but nature presents other challenges. Cattle
develop diseases and die, and crops are affected by
insects and disease. Most farmers would scoff at
economic theory that assumes that a production function
is known with certainty. Although farmers m
for a maximum are 6.11 f1 = 10 ! 2x1 = 0 6.12 x1 = 5
6.13 f2 = 10 ! 2x2 = 0 6.14 x2 = 5 The critical values for
a function is a point where the slope of the function is
equal to zero. The critical values for this function occur at
the point where x1 = 5,
indicated earlier, the data contained in Table 2.1 assume
that there is some residual nitrogen in the soil on which
the corn is grown. The nitrogen is in the soil because of
decaying organic material and leftover nitrogen from
fertilizers applied in years
competition, the consumer has complete knowledge with
respect to prices. Most importantly, with perfect
competition the producer is assumed to have complete
knowledge of the production process or function that
transforms inputs or resources into outputs o
Like the earlier saddle points, a minimum exists in one
direction and a maximum in another direction at a value
for x1 and x2 of 0.25, but the saddle no longer is parallel
to one of the axes, but rather lies along a line running
between the two axes. This
) Production
with One Variable Input 17 The corn yields (TPP)
generated by the production function in Table 2.2 are not
the same as those presented in Table 2.1. There is no
reason for both functions to generate the same yields. A
continuous function that
conclusions is reassuring. The use of mathematics as a
tool for presenting production theory does not mean that
the marginal principles change. Rather, the mathematics
provides further insight as to why the rules developed in
Section 7.8 work the way they
of x produces more and more additional y. Hence the law
of diminishing returns does not hold here either. Notice
that as the use of input x is increased, x becomes more
productive, producing more and more additional y. An
example of a function that would
nitrogen produces 123 bushels of corn, the yield at 140
pounds might be (115 + 123)/2 or 119 bushels per acre.
However, incremental increases in nitrogen application
do not provide equal incremental increases in corn
production throughout the domain of th
Production 14 Agricultural Production Economics 2.1
What Is a Production Function? A production function
describes the technical relationship that transforms
inputs (resources) into outputs (commodities). A
mathematician defines a function as a rule for a
Budget Constraints 121 7.3 ($100/$3.00)@($100/$5.00)
= ($100/$3.00)@($5.00/$100) = $5.00/$3.00 = 1.67 In
equation 7.3 , the budget constraint has a constant
slope of 5/3 or 1.67. Under the assumption of fixed input
prices, the budget constraint will alway
assuming that the second input is fixed. This is not the
global point of profit maximization, since only one input
is allowed to vary. For input x1, this is where pMPPx1/v1
= 1, assuming that x2 is fixed at x*. For input 2 x2, this is
where pMPPx2/v2 = 1,
fixed or "sunk" once they have been applied. At the start
of harvest, the only variable input is the labor, fuel, and
repairs to run the harvesting equipment and to move the
grain to market. This view treats the input categories as a
continuum rather than
what a negative quantity of an input might be. Functions
might be expressed in other ways. The following is an
example: If x = 10, then y = 25. If x = 20, then y = 50. If x =
30, then y = 60. If x = 40, then y = 65. If x = 50, then y =
60. Notice again th
greatest. There is no other point more profitable (Figure
7.4). The global point of profit maximization, where
profits are greatest when both inputs can be varied, is at
once a point on the expansion path, a point of least cost
combination, and a point wh
mechanism is the same as that described for the synthesis of
2,4-dinitrophenylhydrazones (Topic J6). R R' C O R R' C H H R R' C
N NH2 NH2NH2 NaOH R' = H or alkyl hydrazone -N2 (g) Fig. 1.
WolffKishner reduction. However, the simple hydrazone
formed under
also too reactive to be used as protecting groups. CC bond
Aromatic carboxylic acids can be obtained by oxidation of alkyl
benzenes (Topic formation I7). It does not matter how large the
alkyl group is, since they are all oxidized to a benzoic acid
struct
bond. A positive charge on oxygen is not very stable and so the
second stage in the mechanism is the loss of a proton. Both
electrons in the OH bond move onto the oxygen to restore a
second lone pair of electrons and thus neutralize the charge.
Methanol c
the resonance shown (Fig. 2). Since two electrophilic centers
are present, there are two H3C H C O H3C CH3 C O a) b)
Fig. 1. (a) a, b-unsaturated aldehyde; (b) a, b-unsaturated
ketone. places where a nucleophile can react. In both
situations, an addition
compounds may be sensitive to the acid conditions used in this
reaction and an alternative way of carrying out the oxidation is
to use a basic solution of silver oxide (Fig. 4b). Both reactions
involve the nucleophilic addition of water to form a 1,1-diol