Comparative Statics of the Optimal Input Mix

Why Comparative Statics Matters

A cost-minimizing firm chooses a combination of inputs to produce a required output at the lowest possible cost. Once the initial choice is understood, comparative statics asks how that choice changes when the output requirement or an input price changes. The answer is useful for interpreting substitution: a firm facing a higher price for one input may economize on it, but the response to a higher output requirement depends on how its technology changes the value of different input combinations.

The analysis here uses two inputs, labor \(L\) and capital \(K\), with respective prices \(w>0\) and \(r>0\). The production function \(q=f(L,K)\) gives output from an input bundle. This treatment is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.

Definition: For a required output \(q\), a cost-minimizing input bundle is a feasible pair \((L,K)\) that produces at least \(q\) and minimizes total input cost \(C=wL+rK\). At an interior optimum, both inputs are positive. The interior tangency condition equates the marginal rate of technical substitution, \(f_L/f_K\), to the input-price ratio \(w/r\), provided the production function is differentiable and both marginal products are positive.

An interior tangency is a condition to check, not a guarantee that the optimum is interior. The firm may instead choose a boundary bundle or encounter a kink. Smoothness, positive marginal products, and convex isoquants do not by themselves rule out such outcomes. Verify that a proposed tangency is feasible and compare it with relevant boundaries when the problem permits them.

What the Tangency Condition Says

The isocost line for a given expenditure \(C\) is \(wL+rK=C\). Its slope, when capital \(K\) is plotted vertically against labor \(L\), is \(-w/r\). Along a differentiable isoquant, the slope is \(-f_L/f_K\). At an interior tangency, the slopes match:

\frac{f_L}{f_K}=\frac{w}{r}.

This condition says that the marginal product gained per dollar is equal for the two inputs: \(f_L/w=f_K/r\). If one input delivered more marginal product per dollar, a small reallocation toward that input could reduce cost while maintaining output. The equality is necessary for an interior optimum; by itself, it does not establish that a solution is globally optimal or that a boundary choice cannot do better.

For comparative statics, it helps to describe a smooth isoquant locally as \(K=g(L,q)\): the amount of capital needed for output \(q\) when labor is \(L\). Along this isoquant, cost is \(wL+rg(L,q)\). If the relevant portion of the isoquant is strictly convex in this representation, then \(g_{LL}>0\), where the subscripts denote partial derivatives. The tangency condition becomes \(w/r+g_L=0\).

How a Relative-Price Change Affects the Mix

Suppose output \(q\) is fixed and the optimal solution is interior on a smooth, strictly convex portion of the isoquant. Write \(t=w/r\). The tangency condition is \(t+g_L(L,q)=0\). Differentiating with respect to \(t\), holding \(q\) fixed, gives:

1+g_{LL}\frac{dL}{dt}=0 \qquad\Longrightarrow\qquad \frac{dL}{dt}=-\frac{1}{g_{LL}}<0.

Because \(K=g(L,q)\) along the isoquant, the change in capital is

\frac{dK}{dt}=g_L\frac{dL}{dt} =(-t)\left(-\frac{1}{g_{LL}}\right) =\frac{t}{g_{LL}}>0.

Thus, under these stated interior and curvature conditions, an increase in the wage relative to the rental rate reduces labor and increases capital along the fixed-output isoquant. The direction follows from the convex isoquant: at the new tangency, the firm moves toward the relatively cheaper input. This local result does not imply that every price change has the same effect when the optimum is at a boundary or the technology has a kink.

Worked Example: A Higher Wage at Fixed Output

A firm uses \(f(L,K)=\sqrt{LK}\) and must produce \(q=12\). Initially \(w=4\) and \(r=4\). The wage then rises to \(w=16\), while the rental rate and required output stay fixed. Find the input bundle before and after the change, and verify the output and cost.

Method. On the required-output isoquant, \(LK=12^2=144\), so \(K=144/L\). Cost as a function of labor is \(C=wL+144r/L\). Its derivative is \(dC/dL=w-144r/L^2\). Setting this equal to zero gives \(L=12\sqrt{r/w}\) and then \(K=12\sqrt{w/r}\). The second derivative, \(288r/L^3\), is positive for \(L>0\), so this stationary point minimizes cost on the positive isoquant.

Before the wage change: \(w=r=4\), so \(L=12\sqrt{4/4}=12\) and \(K=12\sqrt{4/4}=12\). Output is \(\sqrt{12\cdot12}=\sqrt{144}=12\). Cost is \(4(12)+4(12)=48+48=96\).

After the wage change: \(w=16\) and \(r=4\), so \(L=12\sqrt{4/16}=12(1/2)=6\) and \(K=12\sqrt{16/4}=12(2)=24\). Output is \(\sqrt{6\cdot24}=\sqrt{144}=12\). Cost is \(16(6)+4(24)=96+96=192\).

The firm uses less labor and more capital at the higher relative wage, while preserving output. Common mistake avoided: changing the wage but leaving the old input bundle in place without checking the new tangency. The new marginal-product ratio is \(K/L=24/6=4\), equal to \(w/r=16/4=4\).

Output Changes Do Not Have One Universal Mix Effect

At fixed prices, increasing required output does not, by itself, tell us whether the firm should use more labor relative to capital, more capital relative to labor, or keep the same proportions. The technology determines how the shape and position of successive isoquants affect the tangency. In the local representation \(K=g(L,q)\), the tangency condition is \(w/r+g_L=0\). Differentiating with respect to \(q\), at fixed prices, yields

g_{LL}\frac{dL}{dq}+g_{Lq}=0 \qquad\Longrightarrow\qquad \frac{dL}{dq}=-\frac{g_{Lq}}{g_{LL}}.

Even when \(g_{LL}>0\), convexity does not set the sign of \(g_{Lq}\). Therefore, it does not determine the direction of the labor response to higher output. Capital’s response also depends on the technology: \(dK/dq=g_q+g_L(dL/dq)\). A specific production function, or additional restrictions on how its isoquants change with output, is needed to make a definite prediction.

Worked Example: More Output Changes Labor but Not Capital

A firm has production function \(f(L,K)=L+\sqrt K\). Its input prices are \(w=2\) and \(r=1\). Compare the cost-minimizing bundles for required outputs \(q=2\) and \(q=3\). Assume inputs are nonnegative.

Method. For an interior bundle, the output constraint gives \(\sqrt K=q-L\), so \(K=(q-L)^2\), with \(0<L<q\). Cost along the isoquant is \(C(L)=2L+(q-L)^2\). Its first derivative is \(2-2(q-L)\), and its second derivative is \(2>0\). The interior candidate satisfies \(2-2(q-L)=0\), or \(L=q-1\); then \(K=1\). For both outputs in this problem, this candidate has \(0<L<q\), so it is feasible.

At \(q=2\): \(L=2-1=1\), \(K=1\). Output is \(1+\sqrt1=2\), and cost is \(2(1)+1(1)=3\). The cost function is strictly convex, so this feasible stationary point is the minimum.

At \(q=3\): \(L=3-1=2\), \(K=1\). Output is \(2+\sqrt1=3\), and cost is \(2(2)+1(1)=5\). Again the strictly convex cost function makes this feasible stationary point the minimum.

When required output rises from 2 to 3, labor rises from 1 to 2 while capital stays at 1. Common mistake avoided: assuming that every input must increase in the same proportion when output rises. The bundle must meet the production constraint at minimum cost, and the technology determines how its composition changes.

Checking the Scope of an Interior Result

The derivative results above are conditional: they describe a smooth interior optimum on a suitably convex part of an isoquant. They are not a shortcut for skipping feasibility or boundary checks. A tangency may fail to exist, may lie outside the feasible range, or may not be the lowest-cost choice when boundary bundles are allowed. The earlier material on perfect substitutes, fixed proportions, and boundary solutions illustrates why kinks and corners require separate methods.

For example, let \(f(L,K)=L+\ln(1+K)\), require \(q=1\), and set \(w=1\), \(r=2\). Along the isoquant, \(L=1-\ln(1+K)\), with \(0\leq K\leq e-1\). Cost is

C(K)=1-\ln(1+K)+2K, \qquad C'(K)=-\frac{1}{1+K}+2>0.

Since \(1/(1+K)\leq1\) for \(K\geq0\), the derivative is at least 1 throughout the feasible interval. Cost is minimized at \(K=0\), \(L=1\), a boundary choice. Both marginal products are positive, and the isoquant can be written \(K=e^{1-L}-1\), which is strictly convex. This example shows why those features alone do not establish an interior optimum.

When a problem asks for comparative statics, first identify what is held fixed. A change in \(w/r\) at fixed output concerns substitution along one isoquant. A change in required output at fixed prices compares choices on different isoquants. Mixing these two comparisons can lead to a plausible calculation that answers the wrong question.

State the fixed conditions.
Record whether output, one input price, or both prices are held constant.

Check the solution type.
Determine whether an interior smooth solution is appropriate; inspect boundaries or kinks when they are feasible.

Apply the tangency condition when valid.
Set \(f_L/f_K=w/r\), or minimize cost along the required-output isoquant.

Compare choices and verify.
Substitute the input bundle into the production function and cost expression, and check that the stated change matches the question.

Common Mistakes and Takeaways

Treating an interior condition as proof of an interior optimum. The tangency condition applies only when an interior optimum exists; check feasible boundaries and kinks.
Claiming output changes always preserve input proportions. That conclusion needs additional structure in the technology. For a general smooth convex technology, the output effect on the mix is not determined by convexity alone.
Confusing the price level with the relative price. The tangency depends on \(w/r\). If both input prices change by the same proportion, this ratio is unchanged, though the cost level changes.
Failing to verify the resulting bundle. Check both that it produces the required output and that its cost is computed using the prices for the scenario being analyzed.

Key takeaway: At a regular interior optimum, the marginal rate of technical substitution equals the input-price ratio. Under strict convexity along the isoquant, a higher relative price of labor reduces labor and increases capital at fixed output. The effect of a higher output requirement on the mix is technology-dependent, and smoothness and convexity do not eliminate the need to check for boundaries.

Check Your Understanding

For each problem, distinguish the fixed conditions from the changing one and verify the bundle against the production constraint.

A firm uses \(f(L,K)=\sqrt{LK}\), must produce \(q=8\), and faces \(w=2\), \(r=8\). Find its interior cost-minimizing bundle and its cost. (Answer: \(L=16\), \(K=4\), cost \(64\).)
For \(f(L,K)=\sqrt{LK}\) and fixed output \(q=10\), compare the optimal bundles when \(r=1\) and \(w\) rises from 1 to 4. (Answer: initially \((L,K)=(10,10)\); after the wage increase, \((L,K)=(5,20)\).)
A firm has \(f(L,K)=L+\sqrt K\), \(w=2\), and \(r=1\). Find the interior cost-minimizing bundles and costs for \(q=4\) and \(q=5\). (Answer: at \(q=4\), \((L,K)=(3,1)\) and cost \(7\); at \(q=5\), \((L,K)=(4,1)\) and cost \(9\).)
At fixed output, an isoquant is locally \(K=g(L,q)\), with \(g_{LL}=4\). If \(w/r=3\), find \(dL/dt\) and \(dK/dt\) as the relative price \(t=w/r\) increases, assuming the interior convex conditions hold. (Answer: \(dL/dt=-1/4\) and \(dK/dt=3/4\).)
For \(f(L,K)=L+\ln(1+K)\), \(q=1\), \(w=1\), and \(r=2\), the feasible isoquant has \(0\leq K\leq e-1\). Determine whether the cost minimum is interior or at a boundary and give the bundle. (Answer: boundary minimum at \(K=0\), \(L=1\); cost is \(1\).)

What You'll Learn