Solving a Cost Minimization Problem Using Isoquants and Isocosts with a Cobb-Douglas Production Function
Continuing my recent streak of micro 101/201 walkthroughs, here we’ll walk through a classic cost minimization problem using a Cobb-Douglas production function with capital and labour.
We’ll see how firms can minimize costs for a given level of output, and graph this visually through isoquants and isocosts (the firm version of indifference curves and budget constraints, which we consumers get!).
We’ll be deriving the marginal rate of technical substitution (MRTS), apply a tangency optimality condition, and finally solving for an equilibrium condition of labour and capital.
So, this article will certainly be useful if you’re a student in econ or 201 — or for anyone learning for fun. Leave any questions in the comments!
For this problem, we start with a firm production function featuring capital (K) and labour (L) as production inputs:
- Q(K,L) = K*L
Notation-wise, note that “Q(K,L)” denotes Q as being a function of the inputs K and L, while K*L is the form of that function—5 units of capital and 5 units of labour as inputs, for example, yields Q = 25.
Note that this production function exhibits constant returns to scale (CRS) because doubling both inputs doubles output (you can test this by adding a lambda to the inputs and seeing if it equals lambda times the output).
We’re also given standard costs associated with capital and labour, such being wage (w) and interest rate (r). You can think of the interest rate generally as the price of capital. These are linear costs associated with the production inputs, producing a cost function as follows:
- C = rK + wL
So, the cost to a given firm to produce Q units through K units of capital and L units of labour is K multiplied by the interest rate r and L multiplied by the wage rate w. For example, a firm employing 5 people at $100 per day to work on 4 machines has a cost function 4r + (100)(5) = 4r + 500.
To graph this problem, we’ll set up a graph with L on the x-axis and K on the y-axis. To find the slope of the relationship between these inputs, we can adjust the cost function for KL
- C = rK + wL
- C — wL = rK
- K = C/r — (w/r)L
This is an isocost, which shows all combinations of inputs at a fixed cost (all combinations of K and L that result in the same total cost C). There are an infinite number of isocosts, each for a different cost (5, 50, 500, 500.000001 and so on). You’ll note isocosts are similar to indifference curves, whereas an infinite number of indifference curves exist for different levels of utility. C/r is the Y-intercept, and -w/r is the slope. Let’s graph our K/L relationship:
(note that a bar over a variable represents a fixed quantity)
Hey! Quick note from Jon. If you’re a student in econ 101 or 201, you’ll find these articles useful (exact problem walkthroughs like the ones you’ll see on exams). Back to the problem!
So, each of these lines represents a different cost level. Next, we’ll plot an isoquant, which is similar to an isocost, except a quantity (q-bar) is fixed instead of cost. Isoquants look like indifference curves.
There are also infinite isoquants representing different quantity levels. To find an equilibrium, we’ll find the closest isocost to the origin (farthest “left”) touching an isoquant which intercepts the equilibrium values of K and L.
Let’s start by calculating the marginal rate of technical substitution (MRTS), which is the slope of the isoquant. It’s calculated as the marginal product of labour (the x-axis good) divided by the marginal product of capital (the y-axis good). These marginal products are found by taking the partial derivative of the production function Q with respect to the production input at hand (K or L), as follows:
MRTS = MPL/MPK = (∂Q/∂L)/(∂Q/∂K) = (∂(K*L)/∂L)/(∂(K*L)/∂K) = K/L
Since we know the isoquant is downward sloping, it’s slope is therefore -K/L. Now that we know the equations for the isocost and isoquant, let’s add them to the graph:
Next, we’ll find the cost-minimizing equilibrium bundle for the firm. We’ve already done most of the hard work—all we need to do is set the slopes of the isoquant and isocost equal to one another, as this point is where our firm gets the same additional output per dollar spent on labor and capital. To go deeper on that:
- MRTS = w/r
- MPL/MPK = w/r
- MPL/w = MPK/r
So, at the point where the isocost is tangential to the isoquant, firms get the same marginal value per dollar spent whether they invest in labour or capital. That must mean they’re at an equilibrium. That said, let’s set those values equal to each other:
- -K/L = -w/r
- K/L = w/r
If you’re given this problem in a course (let’s say Microeconomics 101 or 201), you’ll likely be given simply values for w, r, and Q to solve out the cost-minimizing allocation (basically, the question of “Given a required output level Q, what’s the cheapest way to produce it?”).
In our problem, here’s the values we’ll move ahead with, which produces the question what’s the cheapest way to produce an output of 100?:
- w = 1
- r = 1
- Q = 100
Let’s return to our optimality condition and plug in w and r:
- K/L = w/r
- K/L = 1/1
- K/L = 1
- K = L(1)
- K = L
So, given w and r equal 1, K and L are equivelant. If our values were w,r = 1,2, K would equal 2L, and so on. Knowing this, let’s plug in values for our production function to find what are called conditional demands:
- Q = KL
- Q = KK = LL
- Q = K² = L²
- K = L = square root of Q = 10
Now, as a last step, let’s plug our conditional demands into the cost function (as we already know the values for w and r):
- C = rK + wL
- C = (1)(sqrQ) + (1)(sqrL)
- C = (1)(sqr100) + (1)(sqr100)
- C = (1)(10) + (1)(10)
- C = 10 + 10
- C = 20
In summary, the cost-minimizing levels of inputs required to produce an output of 100 are 10 units of capital and 10 units of labour, which produce a total cost of 20.
Hope you found this helpful, and leave any questions in the comments! Check out my other articles on economics here (useful for studying, or just learning).
