Least-Cost Cobb-Douglas Production

Production at Minimum Cost

A production target can often be met using many combinations of inputs. If labor and capital are the inputs, for example, a firm might substitute one for the other while still producing the required output. When the input prices differ, however, those combinations generally have different costs. The least-cost production problem asks which feasible combination meets the target at the lowest total input cost.

For a two-input Cobb-Douglas technology, this problem has a useful structure: the cost-minimizing ratio of inputs depends on the input prices and production exponents, while the target output determines the overall scale of the inputs. The solution also produces a formula for minimum cost as a function of the required output.

Model and Assumptions

Let \(x\) and \(y\) be the quantities of two inputs, with respective positive prices \(w\) and \(r\) per unit. Suppose output is described by the Cobb-Douglas production function \(f(x,y)=A x^\alpha y^\beta\), where \(A>0\) and \(\alpha,\beta>0\). The required output is \(q>0\). The firm seeks to minimize the total cost \(wx+ry\), subject to producing at least \(q\).

The assumptions \(q>0\), \(A>0\), \(\alpha,\beta>0\), \(w>0\), and \(r>0\) matter. Positive output cannot be produced if either input is zero, and positive prices make unnecessary input expenditure costly. Define \(s=\alpha+\beta\), which is positive under the assumptions.

Why the Output Constraint Binds

At a least-cost solution, output is exactly \(q\), not more. To see why, suppose a feasible pair \((x,y)\) produced output \(Q=A x^\alpha y^\beta>q\). Scale both inputs by \(t=(q/Q)^{1/s}\). Since \(0<t<1\), both inputs fall, so their cost falls. The new output is

Thus the original pair could not have been least-cost. The constraint is therefore an equality at an optimum.

Deriving the Optimal Input Ratio

For positive inputs, the Lagrange equations for minimizing cost subject to \(A x^\alpha y^\beta=q\) are

Dividing the first equation by the second eliminates the multiplier and gives the cost-minimizing input ratio. Equivalently, the ratio of input prices equals the ratio of the corresponding marginal products.

This ratio expresses a basic economic trade-off. A higher price \(w\) for input \(x\) makes the least-cost mix use less \(x\) relative to \(y\); a higher exponent \(\alpha\) raises the relative productive contribution of \(x\) and raises its optimal share.

To turn the ratio into input quantities, let \(k=\alpha r/(\beta w)\), so \(x=ky\). Substitution into the binding production constraint gives \(A k^\alpha y^s=q\). Solving for the positive inputs yields the conditional demands.

The minimum cost can be written directly using these demands. A particularly useful equivalent form follows from the expenditure shares implied by the ratio. Since \(x/y=\alpha r/(\beta w)\), we have \(wx/ry=\alpha/\beta\). Consequently, the proportions of total expenditure spent on \(x\) and \(y\) are \(\alpha/s\) and \(\beta/s\), respectively.

Substituting \(x=\alpha C/(s w)\) and \(y=\beta C/(s r)\) into the production equality gives the minimum-cost function:

Why This Is a Global Minimum

The Lagrange equations identify a stationary candidate, but by themselves they do not establish that the candidate is the global cost minimum. The following argument proves the minimum-cost formula and shows that the optimal expenditure shares are uniquely determined.

Take any positive input pair and let its total expenditure be \(C=wx+ry\). Define \(t=wx/C\), the fraction of expenditure spent on input \(x\). Then \(0<t<1\), and the remaining fraction \(1-t\) is spent on input \(y\). It follows that \(x=Ct/w\) and \(y=C(1-t)/r\), so output can be written in terms of total expenditure and its allocation:

For a fixed total expenditure \(C\), maximizing output is therefore equivalent to maximizing \(g(t)=t^\alpha(1-t)^\beta\) on \(0<t<1\). Its logarithmic derivative is \(\alpha/t-\beta/(1-t)\). This derivative vanishes exactly when \(t=\alpha/s\). Moreover, the second derivative of \(\log g(t)\) is \(-\alpha/t^2-\beta/(1-t)^2<0\), and \(g(t)\) tends to zero at both endpoints. Thus this is the unique maximum, with value \((\alpha/s)^\alpha(\beta/s)^\beta\).

Every input pair with expenditure \(C\) consequently satisfies

To produce at least \(q\), the right side must be at least \(q\). Solving this inequality for \(C>0\) gives \(C\ge C^*(q;w,r)\). Equality is attainable by choosing the expenditure share \(t=\alpha/s\) and total expenditure \(C=C^*(q;w,r)\). The resulting inputs are \(x=\alpha C/(s w)\) and \(y=\beta C/(s r)\); they produce exactly \(q\) and are precisely the conditional demands above. Since the maximizing share is unique and the required total cost is fixed at equality, this also proves uniqueness of the cost-minimizing input pair.

Worked Examples

Common Mistake: Assuming Equal Exponents Mean Equal Inputs

Equal production exponents do not, on their own, imply equal cost-minimizing input quantities. The input ratio also depends on prices. In the first example, one might incorrectly set \(x=y\) because both exponents are \(1/2\). Combining that incorrect ratio with the binding output condition gives \(2x=100\), so \(x=y=50\). This pair does produce 100 units, but its cost is \(4(50)+1(50)=250\), which is greater than 200.

The correction is to use the price-adjusted ratio \(x/y=\alpha r/(\beta w)=1/4\), not \(x/y=1\). Equal exponents imply equal expenditure shares at the optimum, not equal input quantities when the prices differ. In this example both inputs account for 100 of the total cost, even though the quantities are 25 and 100.

Interpreting the Result

The exponents determine how total spending is divided: the cost shares are \(\alpha/s\) for input \(x\) and \(\beta/s\) for input \(y\). The prices determine the quantities needed to achieve those shares. As the target \(q\) increases, both conditional demands scale by \(q^{1/s}\), and minimum cost has that same output scaling. These formulas give a direct way to compare feasible production plans without solving a new optimization problem from scratch for each target.

The assumptions remain important when applying the formulas. They require positive input prices, positive production exponents, a positive technology coefficient, and a positive target. If an input price is zero or an exponent is zero, the interior ratio derivation does not apply as stated. In the stated model, however, the output constraint binds, the conditional demands are positive, and the global argument confirms that the derived plan is the unique least-cost plan.