PRELIMINARY AND INCOMPLETE - DO NOT CIRCULATE
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11.2 Cost Minimization
If a firm has multiple variable inputs, it faces a cost minimization problem: what is the least-costly way of producing a given level of output? That is, if we view a firm’s isoquant as a “menu” of options for producing a certain level of output, what is the cheapest “item” (i.e. combination of inputs) on that menu?
Isocost Lines
Let’s start by thinking about how much any combination of labor and capital costs. If the firm has to pay $w$ for each unit of labor and $r$ for each unit of capital, then the total cost of some combination of inputs $(L,K)$ is $$c(L,K) = wL + rK$$ Conversely, the set of all combinations of labor and capital that cost some amount $c$ may be given by the equation $$wL + rK = c$$ If we plot this line in a graph with labor on the horizontal axis and capital on the vertical axis, we get what we might call an isocost line : that is, all combinations of labor and capital that cost the same amount:
The cost minimization problem
The goal of the firm’s cost minimization problem is to produce a given quantity at the lowest possible cost: that is, find the point along an isoquant which is along the lowest possible isocost line.
The key thing here is that we’re treating the amount of output as fixed; that is, some target amount $q$. In other words, ⊕ In fact, this is identical to the cost minimization problem we used when finding the Hicks decomposition bundle for a consumer. Just as the utility function maps goods into utility, the production function maps inputs into the quantity of goods produced. In each case, the agent takes the prices of the goods being bought as given, and the agent is trying to minimize the amount spent. the firm faces a constrained optimization problem to produce this amount $q$ at the lowest possible cost. We can write this problem as
$$\begin{aligned} \min_{L,K} \ \ & wL + rK\\ \text{s.t. } \ & f(L,K) = q \end{aligned}$$
If the firm is solving this problem correctly, then it is choosing a point along its isoquant that costs as much or less than every other point along an isoquant.
Visually, this occurs at a point along the isoquant where the relevant isocost line never crosses the isoquant: that is, every other point along the isoquant costs more than that point:
The “gravitational pull” along an isoquant
Suppose a firm is not currently solving the problem correctly: it’s either using too much labor (and too little capital) or too much capital (and too little labor). If that’s the case, then it could save money by shifting its production mix around.
In particular, suppose the firm is currently using some production combination $(L,K)$ to produce $q$ units of output, and is considering using a bit more labor and a bit less capital. In particular, let’s assume that if it used some amount $\Delta L$ more labor and $\Delta K$ less capital, it would stay along the isoquant. If it did so, its labor expenditures would increase by $w \times \Delta L$, and its capital expenditures would decrease by $r \times \Delta K$. It should do this if the amount it saved on capital would be greater than the amount it cost in additional labor: $$r \times \Delta K > w \times \Delta L$$ or $${\Delta K \over \Delta L} > {w \over r}$$ By construction, since they’re staying along the isoquant, the left-hand side of this is just the marginal rate of technical substitution (MRTS), or the magnitude of the slope of the isoquant; so this is just another way of saying that at such a point, the isoquant would be steeper than the isocost line: $$MRTS > {w \over r}$$ Intuitively, if the slope of the isoquant is not equal to the slope of the isocost line passing through that point, then there is an area of overlap between the set of input combinations that cost less, and the set of input combinations that produce more output.
When calculus works
The “objective” of the firm is to minimize the expenditure $wL + rK$; the “constraint” is that it wishes to produce $q$ units of output. Therefore, if calculus works to find the solution, relevant Lagrangian for this problem is
$$\mathcal{L}(L,K,\lambda) = wL + rK + \lambda [q - f(L,K)]$$
which has the three first-order conditions
$$\begin{aligned} \frac{\partial \mathcal{L}}{\partial L} &= w - \lambda \times MP_L = 0 \Rightarrow \lambda = \frac{w}{MP_L}\\ \frac{\partial \mathcal{L}}{\partial K} &= r - \lambda \times MP_K = 0 \Rightarrow \lambda = \frac{r}{MP_K}\\ \frac{\partial \mathcal{L}}{\partial \lambda} &= q - f(L,K) = 0 \Rightarrow q = f(L,K) \end{aligned}$$
Setting the $\lambda$ for the first two conditions equal to each other gives us the condition
$$\frac{MP_L}{MP_K} = \frac{w}{r}$$
The left-hand side of this is just the expression for the MRTS. Therefore, this represents a tangency condition between the isoquant constraint and the isocost lines defined by the price ratio $w/r$.
What is our interpretation of the Lagrange multiplier $\lambda$? As always, the Lagrange multiplier represents the effect on the objective function of relaxing the constraint by one unit. In this case, the constraint is defined by the quantity $q$, and the objective function is the cost of producing $q$ units; so $\lambda$ represents the marginal cost of producing an additional unit.
Intuitively, both $w/MP_L$ and $r/MP_K$ are such marginal costs: the first is the marginal cost of producing another unit using labor, and the second is the marginal cost of producing another unit using capital. More specifically, $1 / MP_L$ is the amount of labor required to produce an additional unit of output by increasing labor; multiplying that by $w$ gives the cost of producing that unit with labor. Likewise, $1 / MP_K$ is the amonut of labor required to produce an additional unit of output by increasing capital; multiplying that by $r$ gives the cost of producing that unit with capital.
Kinks and corners
For production functions that don’t have a smoothly decreasing MRTS, the Lagrange method will not work. For example, if we have the Leontief production function $f(L,K) = \min\{2L, 3K\}$, the cost-minimizing way to produce any quantity of output will be to produce at the base of the L-shaped isoquant: that is, $q = 2L = 3K$. In this case the conditional demands for labor and capital will be $L(q) = {1 \over 2}q$ and $K(q) = {1 \over 3}q$. Likewise, if we have the linear production function $f(L,K) = 2L + 3K$, firm will cost minimize by producing either entirely with labor or entirely with capital (or be indifferent between the two). All of the techniques we derived in Unit I apply here.
Worked example
For this chapter, we’re going to use the production function
$$f(L,K) = \sqrt{LK}$$
Let’s look at this algebraically and visually first, and then get to the analytical solution. The graph below shows five different ways of producing $q = 16$ units of output given our production function $f(L,K) = \sqrt{LK}$. The lines show green isocost lines given the wage rate $w$ and the rental rate of capital $r$; the table evaluates the cost of each of the possible input combinations. Note that the lowest-cost combination of output corresponds to the farthest-in isocost line. For $w = 8$ and $r = 2$, this combination is $L = 8, K = 32$:
We can see that, in this case, the lowest-cost combination of labor and capital is the point along the isoquant where the isoquant is tangent to the isocost line. For the production function $q = \sqrt{LK}$, the $MRTS = K/L$, so the tangency condition is
$$\frac{K}{L} = \frac{w}{r} \Rightarrow K = \frac{w}{r}L$$
Plugging this into the constraint (i.e., the isoquant) gives us
$$\begin{aligned} q &= \sqrt{L\times\left[\frac{w}{r}L\right]}\\ q &= \sqrt{\frac{w}{r}} \times L\\ L^\star &= \sqrt \frac{r}{w} \times q \end{aligned}$$
and therefore
$$K^\star = \frac{w}{r}L^\star = \sqrt \frac{w}{r} \times q$$
For example, if $w = 8$, $r = 2$, and $q = 16$, the cost-minimizing combination of labor and capital is
$$\begin{aligned} L^\star &= \sqrt \frac{r}{w} \times q^2 = \frac{1}{2} \times 16 = 8\\ K^\star &= \sqrt \frac{w}{r} \times q^2 = 2 \times 16 = 32 \end{aligned}$$
as we found before. Visually, we can see that this occurs at the intersection of the line representing the tangency condition $K = \frac{w}{r}L$ and the isoquant for $q = 16$:
Next, let’s see how this optimal bundle changes as prices and target output change, and how we can use the solution to the cost minimization problem to derive the firm’s long-run total cost of production.
7 Minimizing Costs
The policy question will an increase in the minimum wage decrease employment.
Recently, a much-discussed policy topic in the United States is the idea of increasing minimum wages as a response to poverty and growing income inequality. There are many questions of debate about minimum wages as an effective policy tool to tackle these problems, but one common criticism of minimum wages is that they increase unemployment. In this chapter, we will study how firms decide how much of each input to employ in their production of a good or service. This knowledge will allow us to address the question of whether firms are likely to reduce the amount of labor they employ if the minimum wage is increased.
In order to provide goods and services to the marketplace, firms use inputs. These inputs are costly, so firms must be smart about how they use labor, capital, and other inputs to achieve a certain level of output. The goal of any profit maximizing firm is to produce any level of output at the minimum cost. Doing anything else cannot be a profit maximizing strategy. This chapter studies the cost minimization problem for firms: how to most efficiently use inputs to produce output.
Exploring the Policy Question
Read the Labor Day Turns Attention Back to Minimum-Wage Debate article and answer the following questions:
- What potential benefits to raising the minimum wage are identified in the article?
- What potential disadvantages to raising the minimum wage are identified?
Learning Objectives
7.1 the economic concept of cost.
Learning Objective 7.1 : Explain fixed and variable costs, opportunity cost, sunk cost, and depreciation.
7.2 Short-Run Cost Minimization
Learning Objective 7.2 : Describe the solution to the cost minimization problem in the short run.
7.3 Long-Run Cost Minimization
Learning Objective 7.3 : Describe the solution to the cost minimization problem in the long run.
7.4 When Input Costs and Output Change
Learning Objective 7.4 : Analyze the effect of changes in prices or output on total cost.
7.5 Policy Example Will an Increase in the Minimum Wage Decrease Employment?
Learning Objective 7.5 : Apply the concept of cost minimization to a minimum-wage policy.
From the isoquants described in chapter 6 , we know that firms have many choices of input combinations to produce the same amount of output. Each point on an isoquant represents a different combination of inputs that produces the same amount of output.
Of course, inputs are not free: the firm must pay workers for their labor, buy raw materials, and buy or rent machines, all of which are costly. So the key question for firms is, Which point on an isoquant is the best choice? The answer is the point that represents the lowest cost. The topics of this chapter will help us locate that point.
This chapter studies production costs—that is, how costs are related to output. In order to draw a cost curve that shows a single cost for each output amount, we have to understand how firms make the decision about which set of possible inputs to use to be as efficient as possible. To be as efficient as possible means that the firm wants to produce output at the lowest possible cost.
For now, we assume that firms want to produce as efficiently as possible—in other words, minimize costs. Later, in chapter 9 , “Profit Maximization and Supply,” we will see that producing at the lowest cost is what profit maximizing firms must do (otherwise, they cannot possibly be maximizing profit!).
A good way to think about the cost side of the firm is to consider a manager who is in charge of running a factory for a large company. She is responsible for producing a specific amount of output at the lowest possible cost. She must choose the mix of inputs the factory will use to achieve the production target. Her task, in other words, is to run her factory as efficiently as possible. She does not want to use any extra inputs, and she does not want to pick a mix of inputs that costs more than another mix of inputs that produces the same amount of output.
To make efficient or cost-minimizing decisions, it is important to understand some basic cost concepts, starting with fixed and variable costs as well as opportunity costs, sunk costs, and depreciation.
Fixed and Variable Costs
A fixed cost is a cost that does not change as output changes. For example, a firm might need to pay for the lights to be on in order for the workers to see what they are doing and for production to happen. But the lights are simply on or off, and the cost of powering them does not change when output changes.
A variable cost is a cost that changes as output changes. For example, a firm that wishes to produce more output might need to employ more labor hours by either hiring more workers or having existing workers work more hours. The cost of this labor is therefore a variable cost, as it changes as the output level changes.
Opportunity Costs
As we learned in chapter 3 , the opportunity cost of something is the value of the next best alternative given up in order to get it.
Suppose a firm has access to an input that it can use in production without paying a price for it. A simple example is a family farm. The farm uses land, water, seeds, fertilizer, labor, and farm machinery to produce a crop—let’s say corn—which it then sells in the marketplace. If the farm owns the land it uses to produce the corn, do we then say that the land component is not part of the firm’s costs? The answer, of course, is no. When the farm uses the land to produce corn, it forgoes any other use of the land; that is, it gives up the opportunity to use the resource for another purpose. In many cases, the opportunity cost is the market value of the input.
For example, suppose an alternative use for the land is to rent it to another farmer. The forgone rent from the decision to use the land to produce its own corn is the farm’s opportunity cost and should be factored into the production decision. If the opportunity cost, which in this case is the rental fee, is higher than the profit the farm will earn from producing corn, the most efficient economic decision is to rent out the land instead.
Now consider the more complex example of a farm manager who is told to produce a certain amount of corn. Suppose that the manager figures out that she can produce exactly that amount using a low-fertilizer variety of corn and all the available land. She also knows that another way to produce the same amount of corn is with a higher-yielding variety that requires a lot more fertilizer but uses only 75 percent of the land. The additional fertilizer for this higher-yielding corn will cost an extra $50,000. Which option should the farm manager choose?
Without considering the opportunity cost, she would use the low-fertilizer variety of corn and all of the land because it costs $50,000 less than the alternative method. But what if, under the alternative method, she could rent out for $60,000 the 25 percent of the land that would not be planted? In that case, the cost-minimizing decision is actually to use the higher-yielding corn variety and rent out the unused land.
Another classic example is that of a small business owner who runs, say, a coffee shop. The inputs into the coffee shop are the labor, the coffee, the electricity, the machines, and so on. But suppose the owner also works a lot in the shop. He does not pay himself a salary but simply pays himself from the shop’s excess revenues, or revenues in excess of the cost of the other inputs. The cost of his labor for the shop is not $0 but the amount he could earn working elsewhere instead. If, for example, he could work in the local bank for $4,000 a month, then the opportunity cost of his working at the coffee shop is $4,000, and if the excess revenues are less than $4,000, he should close the shop and work at the bank instead, assuming he likes both jobs equally well.
Some costs are recoverable, and some are not. An example of a recoverable cost is the money a farmer spends on a new tractor knowing that she can turn around and re-sell it for the same amount she paid.
A sunk cost is an expenditure that is not recoverable. An example of a sunk cost is the cost of the paint a business owner uses to paint the leased storefront of his coffee shop. Once the paint is on the wall, it has no resale value. Many inputs reflect both recoverable and sunk costs. A business that buys a car for an employee’s use for $30,000 and can resell it for $20,000 should consider $10,000 of its expenditure a sunk cost and $20,000 a recoverable cost.
Why do sunk costs matter in choosing inputs? Because after incurring sunk costs, a manager should not consider them in making subsequent decisions. Sunk costs have no bearing on such decisions. To see this, suppose you buy a $500 non-refundable and non-transferrable airline ticket to go to Florida during spring break. However, as spring break approaches, you are invited by friends to spend the break at a mountain cabin they have rented to which they will give you a ride in their car at no cost to you. You prefer to spend the break with your friends in the cabin, but you have already spent the $500 on the ticket, and you feel compelled to get your money’s worth by using it to go to Florida. Doing so would be the wrong decision. At the time of your decision, the $500 spent on the ticket is non-recoverable and therefore a sunk cost. Whether you go to Florida or not, you cannot get the $500 back, so you should do what makes you most happy, which is going to the cabin.
Depreciation
Depreciation is the loss of value of a durable good or asset over time. A durable good is a good that has a long usable life. Durable goods are things like vehicles, factory machines, or appliances that generally last many years. The difference between the beginning value of a durable good and its value sometime later is called depreciation. Most durable goods depreciate: machines wear out; newer, more advanced ones are produced, thus reducing the value of current ones; and so on.
Suppose you are a manager who runs a pencil-making factory that uses a large machine. How does the machine’s depreciation factor into the cost of using it?
The appropriate way to think about these costs is through the lens of opportunity cost. If your factory didn’t use the machine, you could rent it to another firm or sell it. Let’s consider the selling of the machine. Suppose it costs $100,000 to purchase the machine new, and it depreciates at a rate of $1,000 per month, meaning that each month, its resale value drops by $1,000. Note that the purchase cost of the machine is not sunk because it is recoverable—but each month, the recoverable amount diminishes with the rate of depreciation. So, for example, exactly one year after purchase, the machine is worth $88,000. At this point in time, the $12,000 depreciation has become a sunk cost. However, at the current point in time, you have a choice: you can sell the machine for $88,000 or use it for a month and sell it at the end of the month for $87,000. The opportunity cost of using the machine for this month is exactly $1,000 in depreciation.
Why does depreciation matter in choosing inputs? Well, suppose workers can do the same job as the machine, or in economics parlance, suppose you can substitute workers for capital. To produce the same amount of pencils as the machine, you need to add labor hours at a rate of $900 a month. A manager without good economics training might think that since the firm purchased the machine, the machine is free, and since the labor costs $900 a month, use the machine. But you—a manager well trained in economics—know that the monthly cost of using the machine is the $1,000 drop in resale value (the depreciation cost), and the monthly cost of the labor is $900. You will save the company $100 a month by using the labor and selling the machine.
In order to maximize profits, firms must minimize costs. Cost minimization simply implies that firms are maximizing their productivity or using the lowest cost amount of inputs to produce a specific output. In the short run, firms have fixed inputs, like capital, giving them less flexibility than in the long run. This lack of flexibility in the choice of inputs tends to result in higher costs.
In chapter 6 , we studied the short-run production function:
[latex]Q=f(L,\overline{K})[/latex],
where [latex]Q[/latex] is output, [latex]L[/latex] is the labor input, and [latex]\overline{K}[/latex] is capital. The bar over [latex]K[/latex] indicates that it is a fixed input. The short-run cost minimization problem is straightforward: since the only adjustable input is labor, the solution to the problem is to employ just enough labor to produce a given level of output.
Figure 7.2.1 illustrates the solution to the short-run cost minimization problem. Since capital is fixed, the decision about labor is to choose the amount that, combined with the available capital, enables the firm to produce the desired level of output given by the isoquant.
From figure 7.2.1 , it is clear that the only cost-minimizing level of the variable input, labor, is at [latex]L^*[/latex]. To see this, note that any level of labor below [latex]L^*[/latex] would yield a lower level of output, and any level of labor above [latex]L^*[/latex] would yield the desired level of output but would be more costly than [latex]L^*[/latex] because each additional unit of labor employed must be paid for.
Mathematically, this problem requires only that we solve the following production function for [latex]L[/latex]:
[latex]Q=f(\overline{K},L)[/latex]
Solving the production function requires that we invert it, which we can do only if the function is monotonic. This requirement is satisfied for our production functions because we assume that output always increases when inputs increase.
[latex]L^*=f^{−1}(\overline{K},q)[/latex]
Let’s consider a specific example of a Cobb-Douglas production function:
[latex]Q=10\overline{K}^{\frac{1}{2}}L^{\frac{1}{2}}[/latex]
To find the cost-minimizing level of labor input, [latex]L^*[/latex], we need to solve this equation for [latex]L^*[/latex]:
[latex]L^{\frac{1}{2}}=(\frac{q}{10\overline{K}^{\frac{1}{2}}})^2[/latex]
This simplifies to
[latex]L^*=\frac{q^2}{100\overline{K}}[/latex]
Note that this equation does not require a specific output target but rather gives us the cost-minimizing level of labor for every level of output. We call this an input-demand function : a function that describes the optimal factor input level for every possible level of output.
The long run, by definition, is a period of time when all inputs are variable. This gives the firm much more flexibility to adjust inputs to find the optimal mix based on their relative prices and relative marginal productivities. This means that the cost can be made as low as possible, and generally lower than in the short run.
Total Cost in the Long Run and the Isocost Line
For a long-run, two-input production function, [latex]Q=f(L,K)[/latex], the total cost of production is the sum of the cost of the labor input, [latex]L[/latex], and the capital input, [latex]K[/latex].
- The cost of labor is called the wage rate, [latex]w[/latex].
- The cost of capital is called the rental rate, [latex]r[/latex].
- The cost of the labor input is the wage rate multiplied by the amount of labor employed, [latex]wL[/latex].
- The cost of capital is the rental rate multiplied by the amount of capital, [latex]rK[/latex].
The total cost ([latex]C[/latex]), therefore, is
[latex]C(Q)=wL+rK[/latex]
If we hold total cost [latex]C(Q)[/latex] constant, we can use this equation to find isocost lines. An isocost line is a graph of every possible combination of inputs that yields the same cost of production. By picking a cost, and given wage rates, [latex]w[/latex], and rental rates, [latex]r[/latex], we can find all the combinations of [latex]L[/latex] and [latex]K[/latex] that solve the equation and graph the isocost line.
Consider the example of a pencil-making factory, where both capital in the form of pencil-making machines and labor to run those machines are utilized. Suppose the wage rate of labor for the pencil maker is $20 per hour and the rental rate of capital is $10 per hour. If the total cost of production is $200, the firm could be employing ten hours of labor and no capital, twenty hours of capital and no labor, five hours of labor and ten hours of capital, or any other combination of capital and labor for which the total cost is $200. Figure 7.3.1 illustrates this particular isocost line.
This figure represents the isocost line where total cost equals $200. But we can draw an isocost line that is associated with any total cost level. Notice that any combination of labor hours and capital that is less expensive than this particular isocost line will end up on a lower isocost line. For example, two hours of labor and five hours of capital will cost $90. Any combination of hours of labor and capital that are more expensive than this particular isocost line will end up on a higher isocost line. For example, twenty hours of labor and thirty hours of capital will cost $700.
Note that the slope of the isocost line is the ratio of the input prices, [latex]-w/r[/latex]. This tells us how much of one input (capital) we have to give up to get one more unit of the other input (labor) and maintain the same level of total cost. For example, if both labor and capital cost $10 an hour, the ratio would be −10/10 or −1. This is intuitive—if they cost the same amount, to get one more hour of labor, you need to give up one hour of capital. In our pencil-maker example, labor is $20 per hour, and capital is $10 per hour, so the ratio is −2: to get one more hour of labor input, you must give up two hours of capital in order to maintain the same total cost or remain on the same isocost line.
The solution to the long-run cost minimization problem is illustrated in figure 7.3.2 . The plant manager’s problem is to produce a given level of output at the lowest cost possible. A given level of output corresponds to a particular isoquant, so the cost minimization problem is to pick the point on the isoquant that is the lowest cost of production. This is the same as saying the point that places the firm on the lowest isocost line. We can see this by examining figure 7.3.2 and noting that the point on the isoquant that corresponds to the lowest isocost line is the one where the isocost is tangent to the isoquant.
From figure 7.3.2 , we can see that the optimal solution to the cost minimization problem is where the isocost and isoquant are tangent: the point at which they have the same slope. We just learned that the slope of the isocost is [latex]-w/r[/latex], and in chapter 6 , we learned that the slope of the isoquant is the marginal rate of technical substitution (MRTS), which is the ratio of the marginal product of labor and capital:
[latex]MRTS=-\frac{MP_L}{MP_K}[/latex]
So the solution to the long-run cost minimization problem is
[latex]MRTS=-\frac{w}{r}[/latex],
[latex]\frac{MP_L}{MP_K}=\frac{w}{r}[/latex] (7.1)
This can be rearranged to help with intuition:
[latex]\frac{MP_L}{w}=\frac{MP_K}{r}[/latex] (7.2)
Equation (7.2) says that at the cost-minimizing mix of inputs the marginal products per dollar must be equal. This conclusion makes sense if you think about what would happen if equation ([pb_glossay id=”752″]7.2[/pb_glossary]) did not hold. Suppose instead that the marginal product of capital per dollar was more than the marginal product of labor per dollar:
[latex]\frac{MP_L}{w}< \frac{MP_K}{r}[/latex]
This inequality tells us that this current use of labor and capital cannot be an optimal solution to the cost minimization problem. To understand why, consider the effect of taking a dollar spent on labor input away, thereby lowering the amount of labor input ( raising the [latex]MP_L[/latex]—remember the law of diminishing marginal returns), and spending that dollar instead on capital and increasing the capital input ( lowering the [latex]MP_K[/latex]). We know from the inequality that if we do this, overall output must increase because the additional output from the extra dollar spent on capital has to be greater than the lost output from the diminished labor. Therefore, the net effect is an increase in overall output. The same argument applies if the inequality were reversed.
An Example of Minimizing Costs in the Long Run
Calculus (long-run cost minimization problem).
Mathematically, we express the long-run cost minimization problem in the following way; we want to minimize total cost subject to an output target:
[latex]\frac{min}{(L,K)}{wL+rK}[/latex] (7.1C)
subject to [latex]Q=f=(L,K)[/latex] (7.2C)
We can proceed by defining a Lagrangian function:
(7.3C) [latex]\wedge (L,K,\lambda) = wL+rK-\lambda (f(L,K)-q)[/latex]
where [latex]\lambda[/latex] is the Lagrange multiplier. The first-order conditions for an interior solution (i.e., [latex]L \gt 0[/latex] and [latex]K \gt 0[/latex]) are as follows:
(7.4C) [latex]\frac{\partial \wedge}{\partial L}=0\Rightarrow w=\lambda \frac{\partial f(L,K)}{\partial L}[/latex]
(7.5C) [latex]\frac{\partial \wedge }{\partial K}=0\Rightarrow r=\lambda \frac{\partial f(L,K)}{\partial K}[/latex]
(7.6C) [latex]\frac{\partial \wedge }{\partial \lambda }=0\Rightarrow Q=f(L,K)[/latex]
From chapter 6 , we know that [latex]MP_L=\frac{\partial f(L,K)}{\partial L}[/latex] and [latex]MP_K=\frac{\partial f(L,K)}{\partial K}[/latex].
Substituting these in and combining 7.4C and 7.5C to get rid of the Lagrange multiplier yields expression (7.1):
(7.7C) [latex]\frac{MP_L}{MP_K}=\frac{w}{r}[/latex]
And 7.6C is the constraint:
(7.8C) [latex]Q=f(L,K)[/latex]
Equations (7.7C) and (7.8C) are two equations in two unknowns, [latex]L[/latex] and [latex]K[/latex], and can be solved by repeated substitution. Note that these are exactly the conditions that describe figure 7.3.2 . Equation (7.7C) is the mathematical expression of [latex]MRTS = −w/r[/latex], and equation (7.8C) pins us down to a specific isoquant, as [latex]MRTS = −w/r[/latex] holds for a potentially infinite number of isoquants and isocost lines depending on the [latex]Q[/latex] chosen.
Consider a specific example of a gourmet root beer producer whose labor cost is $20 an hour and whose capital cost is $5 an hour. Suppose the production function for a barrel of root beer, [latex]Q[/latex], is [latex]Q=10L^{\frac{1}{2}}K^{\frac{1}{2}}[/latex]. If the output target is one thousand barrels of root beer, they could, for example, utilize one hundred hours of labor, [latex]L[/latex], and one hundred hours of capital, [latex]K[/latex], to yield [latex]10(10)(10)=1,000[/latex] barrels of root beer. But is this the most cost-efficient way to do it? More generally, what is the most cost-effective mix of labor and capital to produce one thousand barrels of root beer?
To determine this, we must start with the marginal products of labor and capital, which for this production function are the following:
(7.3) [latex]MP_L=5L^{-\frac{1}{2}}K^{\frac{1}{2}}[/latex]
(7.4) [latex]MP_K=5L^{\frac{1}{2}}K^{-\frac{1}{2}}[/latex]
And thus the [latex]MRTS,-\frac{MP_L}{MP_K}[/latex] is [latex]-\frac{K}{L}[/latex].
The ratio [latex]-\frac{w}{r}[/latex] in this case is [latex]-\frac{20}{5}[/latex], or −4.
So the condition that characterizes the cost-minimizing level of input utilization is [latex]\frac{K}{L}=4[/latex], or [latex]K=4L[/latex]. That is, for every hour of labor employed, [latex]L[/latex], the firm should utilize four hours of capital. This makes sense when you think about the fact that labor is four times as expensive as capital. Now, what are the specific amounts? To find them, we substitute our ratio into the production function set at one thousand barrels:
(7.5) [latex]1,000=10L^{\frac{1}{2}}K^{\frac{1}{2}}[/latex]
If [latex]K = 4L[/latex], then [latex]1,000=10L^{\frac{1}{2}}(4L)^{\frac{1}{2}}[/latex], or [latex]L=\frac{100}{\sqrt{4}}[/latex], which equals fifty.
If [latex]L = 50[/latex], then [latex]K = 200[/latex]. So using fifty hours of labor and two hundred hours of capital is the most cost-effective way to produce one thousand barrels of root beer for this firm.
- Find the marginal product of labor.
- Find the marginal product of capital.
- Find the MRTS.
- Find the optimal amount of labor and capital inputs.
- For [latex]Q=5L^{\frac{1}{3}}K^{\frac{2}{3}}[/latex] and [latex]w[/latex] and [latex]r[/latex], solve for the input-demand functions using the Lagrangian method.
In the previous section, we determined the cost-minimizing combination of labor and capital to produce one thousand barrels of root beer. As long as the prices of labor and capital remain constant, this producer will continue to make the same choice for every one thousand barrels of root beer produced. But what happens when input prices change?
Suppose, for example, an increasing demand for the capital equipment used to make root beer drives the rental price up to $10 an hour. This means capital is more expensive than before not only in absolute terms but in relative terms as well. In other words, the opportunity cost of capital has increased. Before the price increase, for every extra hour of capital utilized, the root beer firm had to give up [latex]\frac{1}{4}[/latex] of an hour of labor. After the rental rate increase, the opportunity cost has increased to [latex]\frac{1}{2}[/latex] an hour of labor. A cost-minimizing firm should therefore adjust by utilizing less of the relatively more expensive input and more of the relatively less expensive input.
In this case, the ratio [latex]-\frac{w}{r}[/latex] is now [latex]-\frac{20}{10}[/latex], or −2. So the new condition that characterizes the cost-minimizing level of input utilization after the price change is [latex]\frac{K}{L}=2[/latex], or [latex]K=\,2\,L[/latex].
The production function for one thousand barrels has not changed:
[latex]1,000=10L^{\frac{1}{2}}K^{\frac{1}{2}}[/latex]
So if [latex]K=2L[/latex], then [latex]1,000=10L^\frac{1}{2}(2L)^\frac{1}{2}[/latex], or [latex]L=\frac{100}{\sqrt{2}}[/latex], which equals roughly 71. If [latex]L = 71[/latex], then [latex]K = 2L = 142[/latex]. As expected, the firm now uses more labor than it did prior to the price change and less capital.
We can also calculate and compare the total cost before and after the increase in the rental rate for capital. Total cost is [latex]C(Q)=wL+rK[/latex], so in the first case where [latex]w[/latex] is $20 and [latex]r[/latex] is $5, the total cost is
[latex]C(Q) = 20(L) + 5(K) = $20(50) + $5(200) = $1,000 + $1,000 = $2,000[/latex].
Now when capital rental rates increase to $10, total cost becomes
[latex]C(Q) = 20(L) + 10(K) = $20(71) + $10(142) = $1,420 + $1,420 = $2,840[/latex].
This new higher cost makes sense because the production function did not change, so the firm’s efficiency remained constant, wages remained constant, but rental rates increased. So overall, the firm saw a cost increase and no change in productivity, leading to an increase in production costs.
Expansion Path
A firm’s expansion path is a curve that shows the cost-minimizing amount of each input for every level of output. Let’s look at an example to see how the expansion path is derived.
Equation (7.5) describes the production function set to the specific production target of one thousand barrels of root beer. If we replace one thousand with the output level [latex]Q[/latex], we get the following expression:
(7.6) [latex]Q=10L^{\frac{1}{2}}K^{\frac{1}{2}}[/latex]
We use the [latex]K=4L[/latex] ratio of capital to labor that characterizes the cost-minimizing ratio when the wage rate for labor is $20 an hour and the rental rate for capital is $5 an hour:
- If [latex]K=4L[/latex], then [latex]Q=10L^{\frac{1}{2}}(4L)^{\frac{1}{2}}[/latex], or [latex]Q=20L[/latex] or [latex]L(q)={\frac{q}{20}}[/latex].
- If [latex]L(q)=\frac{q}{20’}[/latex], then [latex]K(q)=\frac{4q}{20}=\frac{q}{5}[/latex].
So using fifty hours of labor and two hundred hours of capital is the most cost-effective way to produce one thousand barrels of root beer for this firm.
Note that [latex]L(q)=\frac{q}{20}[/latex] and [latex]K(q)=\frac{4q}{20}=\frac{q}{5}[/latex] are both functions of output [latex]Q[/latex]. These are the input-demand functions.
Input-demand functions describe the optimal, or cost-minimizing, amount of a specific production input for every level of output. Note that when the output [latex]Q = 1,000[/latex], [latex]L(Q) = 50[/latex] and [latex]K(Q) = 200[/latex], just as we found before. But from these factor demands, we can immediately find the optimal amount of labor and capital for any output target at the given input prices. For example, suppose the factory wanted to increase output to two thousand or three thousand barrels of root beer:
- At [latex]Q=2,000[/latex]
- [latex]L(2,000)=\frac{2,000}{20}=100[/latex] and [latex]K(2,000)=\frac{2,000}{5}=400[/latex].
- At [latex]Q=3,000[/latex]
- [latex]L(3,000)=\frac{3,000}{20}=150[/latex] and [latex]K(3,000)=\frac{3,000}{5}=600[/latex].
We can graph this firm’s expansion path ( figure 7.4.1 ) from the input demands when [latex]Q[/latex] equals one thousand, two thousand, and three thousand. We can also immediately derive the long-run total cost curve from these factor demands by putting them into the long-run cost function, [latex]C(!)=wL+rK[/latex]:
[latex]C(Q)=wL+rK=wL(Q)+rK(Q)=w\frac{Q}{20}+r\frac{q}{5}[/latex]
At input prices [latex]w[/latex] = $20 and [latex]r[/latex] = $5, the function becomes [latex]C(Q)=20\frac{Q}{20}+5\frac{Q}{5}=2Q[/latex].
The long-run total cost curve shows us the specific total cost for each output amount when the firm is minimizing input costs.
Graphically, the expansion path and associated long-run total cost curve look like figure 7.4.1 .
Figure 7.4.1 illustrates how the solution to the cost minimization problem translates into factor demands and long-run total cost. We can solve for the factor demands and the total cost function more generally by replacing our specific input prices with w and r in the following way. The solution to the cost minimization problem is characterized by the MRTS equaling the input price ratio: the [latex]MRTS=\frac{K}{L}[/latex] and the [latex]\text{input price ratio}=\frac{w}{r}[/latex], so [latex]\frac{K}{L}=\frac{w}{r}[/latex], or [latex]rK=wL[/latex], or [latex]L=\frac{rK}{w}[/latex]. We can plug this into the production function to get
[latex]Q=10L^{\frac{1}{2}}K^{\frac{1}{2}}=10(\frac{rK}{w})^{\frac{1}{2}}\,K^{\frac{1}{2}}=10(\frac{r}{w})^{\frac{1}{2}}\,K[/latex].
Solving for the input demand for capital yields
[latex]K^*=\frac{Q}{10}(\frac{w}{r})^{\frac{1}{2}}[/latex].
Since [latex]L=\frac{rK}{w}[/latex], we can find the input demand for labor:
[latex]L^*=\frac{Q}{10}(\frac{r}{w})^{\frac{1}{2}}[/latex]
Now we have input-demand functions that are functions of both output, [latex]Q[/latex], and the input prices, [latex]w[/latex] and [latex]r[/latex]. Note that when [latex]Q[/latex] rises, the inputs of both capital and labor rise as well. Also note that when the price of labor, [latex]w[/latex], rises relative to the price of capital, [latex]r[/latex], or when [latex]\frac{w}{r}[/latex] rises, the use of the capital input rises and the use of the labor input falls. And when the price of capital rises relative to the price of labor, the use of labor rises, and the use of capital falls. So from these functions, we can see the firm’s optimal adjustment to changing input costs in the form of substituting the relatively cheaper input for the relatively more expensive input.
Perfect Complement and Perfect Substitute Production Functions
Perfect complements and perfect substitutes in production are not uncommon. Suppose our pencil-making firm needs exactly one operator (labor) to operate one pencil-making machine. A second worker per machine adds nothing to output, and a second machine per worker also adds nothing to output. In this case, the pencil-making firm would have a perfect complement production function. Alternatively, suppose our root beer producer could either use two workers (labor) to measure and mix up the ingredients or employ one machine to do the same job. Either combination yields the same output. In this case, the root beer producer would have a perfect substitute production function.
Similar to the consumer choice problem, for production functions where inputs are perfect complements or substitutes, the condition that MRTS equals the price ratio will no longer hold. To see this, consider figure 7.4.2 .
In panel (a), a perfect complement isoquant just intersects the isocost line at the corner of the isoquant. (Take a moment and confirm to yourself that any other combination of labor and capital on the isoquant would be more expensive.) However, at the corner of the isoquant, the slope is undefined, so there is no MRTS. For perfect complements, using inputs in any combination other than the optimal ratio is not cost minimizing. So we can immediately express the optimal ratio as a condition of cost. If the production function is of the perfect complement type, [latex]Q=min[\alpha L,\beta K][/latex], the optimal input ratio is [latex]\alpha L=\beta K[/latex]. And since output is equal to the minimum of the two arguments of the function, that means [latex]Q=\alpha L=\beta K[/latex]. So the optimal amount of inputs for any output level q is [latex]L^*=\frac{q}{a’}[/latex] and [latex]K^*=\frac{q}{\beta}[/latex].
Panels (b) and (c) of figure 7.4.2 show the optimal solution to the long-run cost minimization problem when the production function is a perfect substitute type. The solution is on one corner or the other of the isocost line, depending on the marginal productivities of the inputs and their costs. In (b), the MRTS or the slope of the isoquant is lower (less steep) than the slope of the isocost line or the ratio of the input prices. Since this is the case, it is much less costly to employ only capital to produce [latex]Q[/latex]. In (c), the MRTS or the slope of the isoquant is higher (steeper) than the slope of the isocost line or the ratio of the input prices. Since this is the case, it is much less costly to employ only labor to produce [latex]Q[/latex].
Recall that a perfect substitute production function is of the additive type:
[latex]Q=\alpha L+\beta K[/latex]
The marginal product of labor is [latex]\alpha[/latex], and the marginal product of capital is [latex]\beta[/latex].
Since the MRTS is the ratio of the marginal products, the MRTS is [latex]\frac{\alpha}{\beta}[/latex], which is also the slope of the isoquant.
The ratio of input prices is [latex]\frac{w}{r}[/latex]. This price ratio is the slope of the isocost.
From the graphs we can see that if [latex]\frac{\alpha }{\beta } \lt \frac{w}{r}[/latex], or the isoquant is less steep than the isocost, only capital is used, thus we know that no labor will be employed, or [latex]L^*= 0[/latex], and output must equal [latex]\beta K[/latex], or [latex]Q=\beta K[/latex]. Solving this for [latex]K[/latex] gives us [latex]K^*=\frac{q}{\beta}[/latex]. Alternatively, if [latex]\frac{\alpha }{\beta } \gt \frac{w}{r}[/latex], only labor is used, so [latex]K^*=0[/latex], [latex]Q=\alpha L[/latex], and [latex]L^*=\frac{q}{\alpha}[/latex].
On May 15, 2014, the city of Seattle, Washington, passed an ordinance that established a minimum wage of $15 an hour, almost $5 more than the statewide minimum wage and more than double the federal minimum of $7.25 an hour.
A minimum wage increase brings up many issues about its impact, particularly for a city surrounded by suburbs that allow much lower rates of pay. One question we can answer with our current tools is how Seattle-based businesses affected by the increased minimum wage are likely to react to the higher cost of labor.
Most businesses that employ labor have many other inputs as well, some of which can be substituted for labor. Consider a janitorial firm that sells floor-cleaning services to office buildings, restaurants, and industrial plants. The janitorial firm can choose to clean floors using a small amount of capital and a large amount of labor: they can employ many cleaners and equip them with a simple mop.
Or they could choose to employ more capital in the form of a modern floor-cleaning machine and employ fewer cleaners.
Our theory of cost minimization can help us understand and predict the consequences of making the labor input for cleaning the floors 50 percent more expensive. Figure 7.5.1 shows a typical firm’s long-run cost minimization problem. It is reasonable to consider the long run in this case because it would not take the firm very long to lease or purchase and have delivered a floor-cleaning machine. It is also reasonable to assume that floor-cleaning machines and workers are substitutes but not perfect ones—meaning that machines can be used to replace some labor hours, but some machine operators are still needed. The opposite is also true: the restaurant can replace machines with labor, but labor needs some capital (a simple mop) to clean a floor.
In figure 7.5.1 , the isoquant, [latex]\overline{Q}[/latex], represents the fixed amount of floor the firm needs to clean each day and the different combinations of capital and labor it can use to achieve that output target. When the cost of labor, [latex]w[/latex], increases and the cost of capital, [latex]r[/latex], stays the same, the isocost line gets steeper as w/r increases. We can see in the figure that when this happens, the firm will naturally shift away from using the relatively more expensive input and toward the relatively cheaper input. The restaurant will decrease the amount of labor it employs and increase the amount of capital it uses.
From this specific firm, we can generalize that a dramatic permanent increase in the minimum wage will cause affected firms to employ fewer hours of labor and that employment overall will fall in the affected area. The magnitude of employment change caused by such a policy depends on the production technology of all affected firms—that is, how easy it is for them to substitute more capital. All we can predict with our model currently is the fact that such shifts away from labor will likely occur. Whether the cost of this decrease in employment is outweighed by the benefit of such a policy is beyond the scope of the current analysis, but our model of cost minimization has provided useful insight into the decisions firms will make in reaction to the increase in minimum wage.
- Do you support a national minimum wage increase? Why or why not?
- Do you think the benefits of a minimum wage increase outweigh the costs? Explain your answer.
- What do you predict would happen if, instead of a minimum wage, a tax on the purchase or rental of capital equipment was imposed?
Review: Topics and learning outcomes
Learn: key topics.
A cost that does not change as output changes.
A cost that changes as output changes.
An expenditure that is not recoverable, i.e., the cost of the paint a business owner uses to paint the leased storefront of his coffee shop.
The loss of value of a durable good or asset over time.
A good that has a long usable life.
A function that describes the optimal factor input level for every possible level of output, i.e.,
A graph of every possible combination of inputs that yields the same cost of production.
The MRTS is also the slope of the isoquant
Long-run cost minimization problem
The slope of the isoquant is the MRTS, and the slope of the is
[latex]-w/r[/latex]
[latex]MRTS=-\frac{w}{r}[/latex]
[latex]\frac{MP_L}{w}=\frac{MP_K}{r}[/latex]
This formula has many different calculus derived conclusions that should be reviewed.
[latex]\frac{\partial \wedge}{\partial L}=0\Rightarrow w=\lambda \frac{\partial f(L,K)}{\partial L}[/latex]
[latex]\frac{\partial \wedge }{\partial K}=0\Rightarrow r=\lambda \frac{\partial f(L,K)}{\partial K}[/latex] {~?~ST: end equation}
[latex]\frac{MP_L}{MP_K}=\frac{w}{r}[/latex]
[latex]\frac{\partial \wedge }{\partial \lambda }=0\Rightarrow Q=f(L,K)[/latex]
[latex]Q=f(L,K)[/latex]
Intermediate Microeconomics by Patrick M. Emerson is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.
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When you’re dealing with money, you want a maximum value if you’re receiving cash. But if you’re on a tight budget and have to watch those pennies, then you’re concerned with minimizing your expenses. The following is a minimization problem dealing with saving money on supplements.
You’re on a special diet and know that your daily requirement of five nutrients is 60 milligrams of vitamin C, 1,000 milligrams of calcium, 18 milligrams of iron, 20 milligrams of niacin, and 360 milligrams of magnesium. You have two supplements to choose from: Vega Vita and Happy Health. Vega Vita costs 20 cents per tablet, and Happy Health costs 30 cents per tablet. Vega Vita contains 20 milligrams of vitamin C, 500 milligrams of calcium, 9 milligrams of iron, 2 milligrams of niacin, and 60 milligrams of magnesium. Happy Health contains 30 milligrams of vitamin C, 250 milligrams of calcium, 2 milligrams of iron, 10 milligrams of niacin, and 90 milligrams of magnesium. How many of each tablet should you take each day to meet your minimum requirements while spending the least amount of money?
A good way to organize this is to make a chart or table listing the requirements, costs, and amount of nutrients in each tablet.
With all the information organized into the table, it’s time to solve for the number of tablets that will minimize your cost using linear programming.
- Choose variables to represent the quantities involved. The quantities here are the number of tablets. Let a tablet of Vega Vita be represented by v and a tablet of Happy Health be represented by h .
- Write an expression for the objective function using the variables. The goal is to spend the smallest amount of money necessary (so you want the minimum). Vega Vita costs 20 cents per tablet, and Happy Health costs 30 cents per tablet. Minimize: $0.20 v + $0.30 h
- Intercept on vertical axis: (0, 9)
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Solving cost minimization problems
Situation: Goal: minimize TC = PL . L + PK . K Constraint: produce amount Qo = Q(L,K)
Key relationships: (1) Tangency Condition (tc): MPL/MPK = ( D Q/ D L)/( D Q/ D K) = PL/PK (2) Output Constraint: Qo = Q(L,K)
How can we solve minimization problem using simplex method
Table of Contents
How can we solve linear programming problem using simplex method?
- Standard form.
- Introducing slack variables.
- Creating the tableau.
- Pivot variables.
- Creating a new tableau.
- Checking for optimality.
- Identify optimal values.
What is simplex method maximization?
How do you find the maximum using the simplex method, what is minimization and maximization in linear programming, what type of problem is solved by simplex method, how does the simplex method work, how do you solve a two phase simplex method, what is simplex method with example, why do we use simplex method, what are the advantages of simplex method.
- Given n decision variables, usually converges in O(n) operations with O(n) pivots.
- Takes advantage of geometry of problem: visits vertices of feasible set and checks each visited vertex for optimality. (In primal simplex, the reduced cost can be used for this check.)
- Good for small problems.
What is minimization and maximization?
What is minimization in linear programming, how do i use simplex algorithm, how do you solve maximization and minimization problem, how will you convert maximization and minimization in assignment problem, how are minimization and maximization techniques different, how can we minimize feasible region, what is profit maximization and cost minimization, what are minimization tactics, you may also like, how to spell dollars, how to clean and stain a deck, how can you tell a black rock, how to cook perfect pancakes, why is it important to write an excuse letter, how long does it take to receive mail, how to make a line graph, how to install solar screens, how do you store and reheat pancakes, what to do if your partner accuses you of cheating, what is the importance of curriculum to a teacher, how to update car registration, how to wash brooks shoes, how do you make a turkey, where to buy swords, how to screenshot a certain area on windows, how to cut breakfast potatoes, how to know sugar level is high or low, what is the study of cytology, how to spell tomorrow.
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Solving the following cost minimization problem using Kuhn-Tucker conditions
I am currently getting my Masters in Economics. I did not get any exposure to optimization with inequality constraints in my undergrad. I would like to ensure that I am doing this problem correctly. Note: I understand that Khun-Tucker is not necessary for solving this following problem. That being said, it is important that I master this technique, so I want all the practice I can get:
Problem: A firm has two plants with cost functions $c_{1}(y_{1})=4\sqrt{y_{1}}$ and $c_{2}(y_{2})=2\sqrt{y_{2}}$ . What is the cost of producing $y=y_{1}+y_{2}$ ? Assume that $y_{1},y_{2} \geq 0$
I set up the Lagrange in the following way:
$$\mathcal{L}=4\sqrt{y_{1}}+2\sqrt{y_{2}}-\lambda(y_{1}+y_{2}-y)-\mu_{1}y_{1}-\mu_{2}y_{2}$$
The FOC: $$\mathcal{L}_{y_{1}}=\frac{2}{\sqrt{y_{1}}}-\lambda-\mu_{1}=0$$ $$\mathcal{L}_{y_{2}}=\frac{1}{\sqrt{y_{2}}}-\lambda-\mu_{2}=0$$ $$\mathcal{L}_{\lambda}=0 \Longrightarrow y_{1}+y_{2}=y$$ $$\mu_{1}y_{1}=0$$ $$\mu_{2}y_{2}=0$$
My complementary slackness conditions are: $$\mu_{1} \geq 0; \ \mu_{1}=0 \ if \ y_{1}>0$$ $$\mu_{2} \geq 0; \ \mu_{2}=0 \ if \ y_{2}>0$$
I first check for an interior solution. That is, I look to see if there is a solution where $y_{1},y_{2}>0$ . This means that $\mu_{1}=\mu_{2}=0$ . This means that $$\frac{2}{\sqrt{y_{1}}}=\lambda \ \ and \ \ \frac{1}{\sqrt{y_{2}}}=\lambda$$ I can take the ratio to find that $$\frac{2 \sqrt{y_{2}}}{\sqrt{y_{1}}}=1$$ Using the fact that $y_{1}+y_{2}=y$ , we can see that we have the following solution $$y_{1}=\frac{4}{5}y \ \ and \ \ y_{2}=\frac{1}{5}y$$
I now check the second order conditions. We have $C(y)=4\sqrt{y_{1}}+2\sqrt{y_{2}}$ which means $$C_{1}=\frac{2}{\sqrt{y_{1}}}, \ \ C_{2}=\frac{1}{\sqrt{y_{2}}}$$ $$\Longrightarrow C_{11}=-\frac{1}{y_{1}^{3/2}}, \ \ C_{22}=-\frac{1}{2y_{2}^{3/2}}, \ \ C_{12}=C_{21}=0$$ Plugging in the optimal values into the hessian give us:
$$H=\begin{bmatrix} -\frac{5\sqrt{5}}{8y^{3/2}} & 0 \\ 0 & -\frac{5\sqrt{5}}{2y^{3/2}} \end{bmatrix}$$
Now this is where things get messy for me. I know that if the Hessian is semi positive definite, then we have a minimum. If we have that the hessian is semi negative definite, then we have a maximum. This Hessian doesn't appear to be either. I believe this indicates to us that we have a saddle point.
That is, using principal minors, we have:
$$\mid H \mid ^{(1)}<0, \ \ and \ \ \mid H \mid ^{(2)}=|H|>0$$
That is, $$|H|^{(1)}=det(-\frac{5\sqrt{5}}{8y^{3/2}})=-\frac{5\sqrt{5}}{8y^{3/2}}<0$$
$$|H|^{(2)}=|H|=det(\begin{bmatrix} -\frac{5\sqrt{5}}{8y^{3/2}} & 0 \\ 0 & -\frac{5\sqrt{5}}{2y^{3/2}} \end{bmatrix})=(-\frac{5\sqrt{5}}{8y^{3/2}})(-\frac{5\sqrt{5}}{2y^{3/2}})-0=\frac{125}{16y^{3}}>0$$
That being said, I'm pretty sure we are supposed to find that the interior solution is a maximum. I believe I have done something wrong, but I am not sure what.
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$$H=\begin{bmatrix} -\frac{5\sqrt{5}}{8y^{3/2}} & 0 \\ 0 & -\frac{5\sqrt{5}}{2y^{3/2}} \end{bmatrix}$$ Now this is where things get messy for me. I know that if the Hessian is semi positive definite, then we have a minimum. If we have that the hessian is semi negative definite, then we have a maximum. This Hessian doesn't appear to be either. I believe this indicates to us that we have a saddle point.
The Hessian above has two eigenvalues (denoted by $\lambda$ ): $$ \lambda_1 = -\frac{5\sqrt{5}}{8y^{3/2}} \ \text{ and } \ \lambda_2 = -\frac{5\sqrt{5}}{2y^{3/2}} $$ The expression $y^{3/2}$ is defined only when $y \geq 0$ . This implies that both eigenvalues are negative or undefined. Thus the Hessian does not appear to be indefinite.
- $\begingroup$ Thank you, this was a rookie mistake. This next question is more of a math question, why am I finding that |H|^(1)<0 and |H|^(2)=|H|>0 (ie. since the upper left entry is negative and the lower right entry is negative). Wouldn't the determinant be positive? $\endgroup$ – Mistah White Oct 20, 2021 at 16:08
- $\begingroup$ What exactly are you denoting by $|H|^{(1)}<0$? $\endgroup$ – Giskard Oct 20, 2021 at 16:10
- $\begingroup$ @MistahWhite Perhaps you should edit your calculations into the body of the question, as otherwise it may be nigh-on-impossible to determine why you are reaching erroneus results. $\endgroup$ – Giskard Oct 20, 2021 at 16:11
- $\begingroup$ Thank you for your suggestion. I just edited the calculations into the question. $\endgroup$ – Mistah White Oct 20, 2021 at 16:18
- $\begingroup$ @MistahWhite I think you may have misread Sylvester's criterion ? Or why is it that you think this contradicts negative definiteness? $\endgroup$ – Giskard Oct 20, 2021 at 16:21
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Linear programming cost minimization example
Example: Assume that a pharmaceutical firm is to produce exactly 40 gallons of mixture in which the basic ingredients, x and y, cost $8 per gallon and $15 per gallon, respectively, No more than 12 gallons of x can be used, and at least 10 gallons of y must be used. The firm wants to minimize cost.
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Example: Assume that a pharmaceutical firm is to produce exactly 40 gallons of mixture in which the basic ingredients, x and y, cost $8 per gallon and $15 per gallon, respectively, No more than 12 gallons of x can be used, and at least 10 gallons of y must be used. The firm wants to minimize cost.
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Lagrangian: How to Solve Cost Minimization Problem - YouTube hey guys! so today we'll talk about the lagrangian cost minimization problem and in this picture Macroeconomics: Supply and...
The goal of the firm's cost minimization problem is to produce a given quantity at the lowest possible cost: that is, find the point along an isoquant which is along the lowest possible isocost line. The key thing here is that we're treating the amount of output as fixed; that is, some target amount q q.
In this episode I describe the Cost Minimization Problem in case of two inputs and show how we solve for optimal cost functions.Important Note for Navigating...
The short-run cost minimization problem is straightforward: since the only adjustable input is labor, the solution to the problem is to employ just enough labor to produce a given level of output. Figure 7.2.1 illustrates the solution to the short-run cost minimization problem.
Solving the cost-minimization problem Conceptually: find best (lowest) isocost given the required isoquant. Graphically... Draw the "target" output level (isoquant) q 0. Ask: what is the cheapest combination of L,K that makes q 0 possible. Solving the cost-minimization problem What if prices of factors change? What if target output changes?
The following is a minimization problem dealing with saving money on supplements. You're on a special diet and know that your daily requirement of five nutrients is 60 milligrams of vitamin C, 1,000 milligrams of calcium, 18 milligrams of iron, 20 milligrams of niacin, and 360 milligrams of magnesium.
Solving cost minimization problems Situation: Goal: minimize TC = PL. L + PK. K Constraint: produce amount Qo = Q (L,K) Key relationships: (1) Tangency Condition (tc): MPL/MPK = ( D Q/ D L)/ ( D Q/ D K) = PL/PK (2) Output Constraint: Qo = Q (L,K) To do: Try the following example: Given: Q = L 1/2 K 1/2 PL = $4, PK = $1
A fully worked example going through how to find cost-minimizing combinations of inputs with three classic production functions: linear, Leontief, and Cobb-Douglas. Regression Exercise Solution...
For those of you interested in math and knowing what the Lagrange method of solving optimization problems is, the cost minimization problem of the rm stated above can be solved also using this method. What we do is write the Lagrangean: ( x 1;x 2; ) = w 1x 1 +w 2x 2 (f(x 1;x 2) y) where is the Lagrange multiplier on the constraint.
To every minimization problem there corresponds a dual problem. The solution of the dual problem is used to find the solution of the original problem. The dual problem is a maximization problem, which we learned to solve in the last section. We first solve the dual problem by the simplex method.
This video uses a lagrangian to minimize the cost of producing a given level of output.
Solution: The given maximization problem is converted into minimization problem by subtracting from the highest sales value (i.e., 41) with all elements of the given table. Reduce the matrix column-wise and draw minimum number of lines to cover all the zeros in the matrix, as shown in Table.
The ratio pi=MUi measures the cost of increasing utility by one util, or the \cost{per{bang". At the optimum the agent equates the cost{per{bang of the two goods. Intuitively, if good 1 has a higher cost{per{bang than good 2, then the agent should spend less on good 1 and more on good 2. In doing so, she could attain the same utility at a lower ...
% Minmization optimization while (sum (abs (beta_d-beta_u))>0.1) initial_guess = randn (2,2); OLS = @ (B,input_vars)sum (abs (myfun (B,input_vars)-beta_u).^2); % ordinary least squares cost function opts = optimoptions (@fminunc, 'MaxIterations', 10000, 'MaxFunctionEvaluations', 50000, 'Display', 'Iter', 'FiniteDifferenceStepSize', 1e-3);
I can take the ratio to find that 2 y 2 y 1 = 1 Using the fact that y 1 + y 2 = y, we can see that we have the following solution y 1 = 4 5 y a n d y 2 = 1 5 y I now check the second order conditions. We have C ( y) = 4 y 1 + 2 y 2 which means C 1 = 2 y 1, C 2 = 1 y 2 C 11 = − 1 y 1 3 / 2, C 22 = − 1 2 y 2 3 / 2, C 12 = C 21 = 0
Describes how to set up and solve cost minimization problems. AboutPressCopyrightContact usCreatorsAdvertiseDevelopersTermsPrivacyPolicy & SafetyHow YouTube worksTest new features © 2022...
Cost Minimization Given factors of production and , with rental prices and , find cheapest way to produce a given level of output : such that: wK K L wL y ()wLL wK K L K +, min y = F(L,K) Cost Function Solution to minimization problem is cost function: wLL ()wL,wK , y wK K wL,wK, y * + * c()wL,wK, y = Finding the Cost Function Cost of using and
Linear programming cost minimization example - This video shows how to solve a linear programming problem in excel to minimize cost. ... Solve math problems If you're struggling to clear up a math equation, try breaking it down into smaller, more manageable pieces. This will help you better understand the problem and how to solve it.
In this video we learn how to solve cost minimization problem of a firm with production function given by f(l,k) = min{2l+k, l+2k}. And we also solve profit...
Solving a minimization problem with linear programming. This video is provided by the Learning Assistance Center of Howard Community College ... Linear Programming (Profit Maximization, Cost Minimization Linear Programming (Cost Minimization) Part 1 of 3. Watch later. Share. Copy link. Info. Shopping. Tap to unmute. 970 Consultants
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Solving cost minimization problems. A fully worked example going through how to find cost-minimizing combinations of inputs with three classic production functions: linear ` minimum calculator. Free Minimum Calculator - find the Minimum of a data set step-by-step. Determine math question ...
Cost Minimization with a Lagrangian
Minimization linear programming problems are solved in much the same way as the maximization problems. For the standard minimization linear program, the constraints are of the form a x + b y ≥ c, as opposed to the form a x + b y ≤ c for the standard maximization problem.
Solve a Minimization Problem Using Linear Programming. ... Subsequently, we show how linear programming can be used to solve complex constrained profit maximization and cost minimization problems, and we estimate the. 24/7 Customer Help If you need help, we're here for you 24/7. ...