1 Production with two inputs

1.1 Introduction

In the previous lecture you were introduced to the basics of a production function and we looked at a production process where you produce an output using one varialble inpunt and number of fixed inputs. Now we will look at a case where we produce an output using two variable inputs and a number of fixed inputs.

\[ y = f(x) =TPP= f(x_1, x_2 |x_3...x_n) \] Lets take the example of a farmer who is producing corn (maize) using two variable inputs as \(x_1\): potash(K) and \(x_2\): phosphate(P), with nitrogen (\(x_3\)) as the only fixed input with an application rate of 180 pounds per acre. Therefore the production process is as follow.

\[ y = f(x) =TPP= f(x_1, x_2 |x_3) \\[10pt] corn= f(potash, phosphate |nitrogen) \\ \]

Table 1 below shows the hypothetical production relationship of this production process. Notice that the two inputs are complimentary since applying potash only is not very productive (looking horisontally where phosphate = 0). In fact, the maximum yield of only applying potash is a mere 99 bushels per acre at a rate of 20 to 30 pounds per acre. Hereafter production starts to decrease and thus Stage III of the production process is reached. Applying only phosphate instead of potash is a bit more productive since a maximum yield of a 104 bushels per acre is reached between 40 and 50 pounds per acre (looking vertically where potash = 0). Note that each one of the rows of the table represent a production function for potash fertiliser since both the nitrogen and phosphate levels are kept constant. The same would also apply for phosphate where each column would represent a production function.

Table 1: Hypothetical Corn Response to Phosphate and Potash Fertilizer. Source: Debertin (2012,83)

Table 1: Hypothetical Corn Response to Phosphate and Potash Fertilizer. Source: Debertin (2012,83)

Looking between rows and columns it is clear that the productivity of potash is mostly increased by adding more phosphate and visa versa, but that it also possible to apply too much of both. Fore example, if you are applying 80 pounds per acre of both, you could increase yield by dialling both back by one increment to 70 pounds per acre which would result in an increase in yield from 134 to 136 bushels per acre. These relationships speak to agronomical principles, specifically Von Liebig’s “Law of the Minimum”. Due to the biology of crop growth, a synergistic effect is present. This means that the presence of ample amounts of phosphate makes the productivity of potash greater. Ample amounts of potash makes the productivity of phosphate greater. The two fertilizers, taken together, result in productivity gains in terms of increased yields greater than would be expected by looking at yields resulting from the application of only one type of fertilizer. Ok, so lets look at the data graphically. Lets get the data into R and plot it out using plotly.

Figure 1 shows the 3D surface plot of the yield response to potash and phosphate. Note that you can interact with the plot by changing the zoom level and the orientation. Since we have three variables as potash, phosphate and maize yield the plot is three dimensional with potash on the \(x_1\) axis, phosphate on the \(x_2(y)\) axis and maize yield on the \(z\) axis. The white lines shows constant applications of potash and the grey lines constant applications of phosphate. If you orientate your figure such that maize yield is on the vertical axis and phosphate is on the horisontal axis, then the you will see the respective production functions for phosphate at different potash levels. Now hover your cursor over the figure. Not only does it show you the levels of potash, phosphate and yield (the black lines), but it also shows the amounts of potash and phosphate applied that would give you the same yield (the red line). This red line is similar to a contour on an elevation map of the word, only difference is that the contour shows areas with the same hight whereas these contours show the same yield. If you plot some of the contours on the phosphate and potash plain (i.e. 2D), as I do in this figure and you look closely (and orientate the map correctly) then you will recognise their shape as isoquants (“iso” comes from the Greek “isos” meaning equal, and “quant” short for quantity). Therefore, whenever you hear the term isoquant or see them, you must always think contour! And remind yourself of the fact that the isoquant that you’re seeing is just one of many. Fun fact, the very top isoquant goes all the way arround, i.e it is closed. Go ahead, check it out.

Figure 1: Production response to potash and phosphate, 3D surface plot

1.2 The marginal rate of substitution (MRS)

The slope of the isoquant is the marginal rate of substitution (MRS) and represents the rate at which the one input can be substituted for the other whist keeping output the same. As you saw in Table 1 there are limitations to this especially if you move to very low levels of either inputs. This speaks to the shape of the isoquant which exhibits a decreasing marginal rate of substitution meaning as you apply additional units of potash, it replaces fewer and fewer units of phosphate. This is illustrated in Figure 2 where \(x_1\) represents potash and \(x_2\) represents phosphate. Note that the same increments of additional potash is. By convention we write this as \(MRS_{x_1x_2} = \frac{\Delta x_2}{\Delta x_1}\)

Table 2: Hypothetical Corn Response to Phosphate and Potash Fertilizer. Source: Debertin (2012,83)

Table 2: Hypothetical Corn Response to Phosphate and Potash Fertilizer. Source: Debertin (2012,83)

Note that the shape of the isoquant is a factor of the underlying substitution relationship between the inputs and thus does not always exhibit a decreasing marginal rate of substitution. THis is illustrated in Figure 3 where diagrams E and F shows an production process where two of the inputs exhibits an increasing marginal rate of substitution.

Figure 3: Some Possible Production Surfaces and Isoquant Maps. Source: Debertin (2012,91)

Figure 3: Some Possible Production Surfaces and Isoquant Maps. Source: Debertin (2012,91)

1.3 MRS and Marginal Physical Products

The slope of an isoquant (MRS) and the underlying productivity of the production functions used to derive an isoquant map are closely related. An algebraic relationship can be derived between the MRS and the marginal products of the underlying production functions.

Suppose that we want to determine the change in output (\(\Delta y\)) that would result if we changed the use of \(x_1\) by some small amount (called \(\Delta x_1\)) and the use of \(x_2\) were also changed by some small amount (called \(\Delta x_2\)). To determine the resulting change in output (\(\Delta y\)), two pieces of information would be needed. First, the exact magnitude of the changes in the use of each of the inputs \(x_1\) and \(x_2\). It is not possible to determine the change in output by merely summing the respective change in the use of the two inputs. The second piece of information needed is the rate at which each input can be transformed into output. This rate is the marginal physical product of each input \(x_1\) and \(x_2\) (\(MPP_{x_1}\) and \(MPP_{x_2}\)). Therefore:

\[ \Delta y = MPPx_1\Delta x_1 + MPPx_2\Delta x_2 \] But since we’re on a isoquant the the quantity of \(y\) will stay the same, thus \(\Delta y = y'-y'' = 0\) therefore,

\[ \begin{aligned} 0 &= MPPx_1\Delta x_1 + MPPx_2\Delta x_2 \\[10pt] MPPx_2\Delta x_2 &= -MPPx_1\Delta x_1 \\[10pt] \frac{\Delta x_2}{\Delta x_1} &= \frac{-MPPx_1}{MPPx_2} \\[10pt] MRSx_1x_2 &= -\frac{MPPx_1}{MPPx_2} \end{aligned} \] From the above it is clear that the marginal rate of substitution between a pair of inputs is equal to the negative ratio of the marginal products. Thus the slope of an isoquant at any point is equal to the negative ratio of the marginal products at that point, and if the marginal products for both inputs are positive at a point, the slope of the isoquant will be negative at that point.

1.4 Partial and Total Derivatives and the Marginal Rate of Substitution

You will recall that \(MPP\) is the rate of change of the production function and therefore represents the first derivative of the production function. But how do we proceed if we have more than one input? Lets consider the production function \(y = f(x_1,x_2)\). The only way to obtain the marginal product of \(x_1\) is by making an assumption about the level of \(x_2\). The same is also true the other way around. To calculate the \(MPP_{x_1}\) we have to assume that \(x_2\) stays constant (represented like this \(x^*_2\)) so that we can take the partial derivative of the function \(f\). Therefore:

\[MPP_{x_1}= \frac{\partial f}{\partial x_1}|x_2=x^*_2\] Note that the partial derivative is represented by the \(\partial\) symbol and the \(|\) symbol stands for “such that” or “it is true that”. Therefore the equation reads as follow: the \(MPP_{x_1}\) is equal to the partial derivative of the function \(f\) with respect to \(x_1\) given it is true that \(x_2\) is constant. Similarly the \(MPP_{x_2}\) is represented by

\[MPP_{x_2}= \frac{\partial f}{\partial x_2}|x_1=x^*_1\] To illustrate partial derivatives in action consider these two examples:
\[y = x_{1}^{0.5}x_{2}^{0.5}\] Then \[MPP_{x_1} = \frac{\partial y}{\partial x_1}=0.5x_{1}^{-0.5}x_{2}^{0.5}\] Note that we treat \(x_2\) as a constant and the derivative of an constant is a constant. Also note that the production function is a product of \(x_1\) and \(x_2\) the \(MPP_{x_1}\) is a function of both the values of \(x_1\) and \(x_2\).

Now consider \[y = x_{1}^{0.5}+x_{2}^{0.5}\] Then \[MPP_{x_1} = \frac{\partial y}{\partial x_1}=0.5x_{1}^{-0.5}\] Note that we still treat \(x_2\) as constant but since its not multiplied with our variable it falls away.

Now lets consider again the expression representing the total change in output.

\[\Delta y = MPPx_1\Delta x_1 + MPPx_2\Delta x_2\] If we make the changes in \(x_1\) and \(x_2\) infinitesimally small, then we can rewrite the function as \[ \Delta y = \frac{\partial y}{\partial x_1}\Delta x_1 + \frac{\partial y}{\partial x_2}\Delta x_2 \] And since we are on a isoquant the value of y stays the same and therefore \(\Delta y =0\), then we can rewrite the function as \[ \begin{aligned} 0 &= \frac{\partial y}{\partial x_1}\Delta x_1 + \frac{\partial y}{\partial x_2}\Delta x_2 \\[10pt] \frac{\partial y}{\partial x_2}\Delta x_2 &= -\frac{\partial y}{\partial x_1}\Delta x_1 \\[10pt] \frac{\Delta x_2}{\Delta x_1} &= -\frac{\partial y}{\partial x_1}/\frac{\partial y}{\partial x_2} \\[10pt] MRS_{x_1x_2} &= -\frac{\partial y}{\partial x_1}/\frac{\partial y}{\partial x_2} \\[10pt] \end{aligned} \]

1.5 Conclusion

This section evaluated the physical and technical relationships underlying production when two inputs are used in the production of a single output. An isoquant is a line connecting points of equal output on a graph with the axes represented by the two inputs. The slope of an isoquant is referred to as a marginal rate of substitution (MRS). It indicates the extent to which one input substitutes for another as we move from one point to another along an isoquant (representing constant output). The marginal rate of substitution is usually diminishing i.e., when output is maintained at a constant level and additional units of input \(x_1\) are added, then each additional unit of \(x_1\) that is added replaces a smaller and smaller quantity of \(x_2\).

A diminishing marginal rate of substitution between two inputs normally occurs if the production function exhibits positive but decreasing marginal product with respect to incremental increases in the use of each input, a condition normally found in stage II of production. Therefore the marginal rate of substitution is closely linked to the marginal product functions for the inputs. This section has illustrated how the marginal rate of substitution can be calculated if the marginal products for the inputs are known.

2 Cost minimisation with input-input relationships

Now suppose that that we face a budget constraint which limits the amount of inputs that we can buy, how do we determine the optimal combination of the two inputs? Now we now the physical production relationship between the inputs what is their cost relationship. To know that we have to look at the total cost function which can be expressed as

\[ TC = P_{x_1}x_1 + P_{x_2}x_2 \] Similar to an isoquant the total cost stays the same as we move along the total cost line, also called the isocost line. Therefore if we change the amounts of inputs \(x_1\) and \(x_2\) used provided that \(TC\) stays constant, then it can expressed as follow

\[ \begin{aligned} \Delta TC = 0 &= P_{x_1}\Delta x_1 + P_{x_2}\Delta x_2\\[10pt] P_{x_2}\Delta x_2 &= -P_{x_1}\Delta x_1 \\[10pt] \frac{\Delta x_2}{\Delta x_2} &= -\frac{P_{x_1}}{P_{x_2}} \end{aligned} \]

This represents the so called inverse input price ratio that you saw during your undergraduate studies. From it you will recall that cost is minimised where the slope of the isocost line is equal to that of the isoquant, therefore where the inverse input price ratio is equal to the marginal rate of subsititution. From the above this can be expressed as follow (note that I drop the (\(-\)) since its negative of both sides). \[ \begin{aligned} \frac{MPPx_1}{MPPx_2}&=\frac{P_{x_1}}{P_{x_2}}\\[10pt] \frac{\partial y}{\partial x_1}/\frac{\partial y}{\partial x_2} &=\frac{P_{x_1}}{P_{x_2}} \end{aligned} \]

3 More on substitution among inputs

Whilst using the marginal rate of substitution (MRS) to measure the substitution amongst inputs is perfectly legitimate, it suffers from one major drawback in the sense that it magnitude depends on the units of measure. If capital and labour is used in the production process for example, and the one is measured in 1 000s of dollars and the other simply in dollars, then the MRS will differ by a factor of a 1000 relative to situation where both were in dollars. Hence, changing the units of inputs used would also change the isoquant. To overcome this we can use a measure of substitution that is independent of the units used.

3.1 Hicks Elasticity of Substitution

The Hicks elasticity of substitution (\(\sigma\)) between two inputs is defined as the percentage change in the input ratio for each one percent change in their MPP ratios (Hicks, 1932). This is defined as follow:

\[ \sigma = \frac{\%\Delta[\frac{x_1}{x_2}]}{\%\Delta[\frac{MPP_{x_1}}{MPP_{x_2}}]} \] The shape of the isoquant is determined by the values of \(\sigma\). As shown in the first case of Figure 4 if \(\sigma=0\) then the isoquant is L-shaped given that the two inputs are used in fixed proportions, thus the inputs cannot be substituted for one another. When baking a bread for example, flour cannot be substituted for yeast since both has to be used in the right proportions. The second case of Figure 4 shows a value for \(\sigma\) of between \(0\) and \(\infty\) yields a isoquant that exhibits a diminishing marginal rate of substitution between the two inputs and they are imperfect substitutes. As the value of \(\sigma\) approaches \(\infty\) the two inputs become perfect substitutes which means that they can be substituted at fixed proportions (not necessarily 1 to 1). This hardly every happens in reality.

Figure 4: Hicks elasticities of substitution. Source: Peterson (1999,25)

Figure 4: Hicks elasticities of substitution. Source: Peterson (1999,25)

As agricultural economists we are most interested in the values for \(0 <\sigma<\infty\). In order to think of inputs in fixed proportions one has to take a narrow short run view on the production process since over the longer term alternatives might be developed. For example if two inputs are used in fixed proportions but one of them becomes relatively expensive, companies will be incentivised to develop alternatives. The same would also hold for imperfect substitutes. Take the example of capital and labour, if the price of labour increases relative to labour, farmers will be incentivised to devise or adopt labour saving machines. In the case of harvesting grapes for example, farmers could replace workers with mechanised harvesters (the more machines scenario) or if they have done so already, replace two smaller machines with a bigger one (the better machines scenario). This process might take time since the necessary machines will have to be developed or the farmer has to change their trellising system to allow for mechanical harvesting. Therefore we can say that the Hicks elasticity of substitutions increases over time in most cases.

3.2 Other measures of elasticity of substitution

The Hicks elasticity of substitution only holds for the two input case. Ways to estimate the elasticity of substitution with more than two inputs have been devised, examples include Allen and Morishima elasticities of substitution, but these fall beyond the scope of this course.

4 Substitutes versus Compliments

People often confuse the concept of the elasticity of substitution as measured by \(\sigma\) and the notion of substitutes versus compliments. As illustrated above the elasticity of substitution (\(\sigma\)) deals with the shape of the isoquant, whereas substitutes versus compliments has to do with the direction of the shift in the \(MPP\) curve of one input if the quantity of the the other changes. Two inputs are substitutes if the \(MPP\) of one decreases if more of the other is used. For example maize and sorghum in a dairy feed ration is substitutes since both of them are energy rich feed items that can be substituted for each other. Two inputs are compliments when the \(MPP\) of one increases if more of the other is used. Green pastures for example is high in protein but low in energy, therefore if you feed your cattle some additional maize they will be able to utilise the pasture better thereby raising the \(MPP\) thereof as shown by the second graph of Figure 5.

Figure 4: Substitutes versus Compliments. Source: Peterson (1999,30)

Figure 4: Substitutes versus Compliments. Source: Peterson (1999,30)

Note that \(x^1_2\) means more of input \(x_1\) relative to level \(x^0_2\) and the same applies for \(x^1_4\) versus \(x^0_4\).