Long-Run Cost-Minimization Problem Workshop

Setting Up a Long-Run Cost-Minimization Problem

A firm that can vary all its inputs in the long run chooses the least-cost input bundle capable of producing a specified output. The task is to connect three pieces of information: the production technology, the output target, and the prices of the inputs. A correct solution identifies the minimizing bundle, calculates its cost, and checks that the bundle actually produces the target.

Let $K$ denote capital and $L$ denote labor. Let $r>0$ be the price per unit of capital and $w>0$ the price per unit of labor. For target output $q>0$, the firm minimizes $rK+wL$ subject to $f(K,L)\geq q$, where $f$ is the production function. Inputs are nonnegative, prices are taken as given, and the firm is free to adjust both inputs. Because prices are positive, a cost-minimizing bundle will not use more input than needed to meet the target when output can be reduced by reducing inputs.

Definition: The conditional factor demands are the cost-minimizing quantities $K^c(r,w,q)$ and $L^c(r,w,q)$ needed to produce target output $q$ at input prices $r$ and $w$. The long-run cost of that output is $C(r,w,q)=rK^c+wL^c$.

An isocost line contains all input bundles with the same total cost: $rK+wL=C$. With $L$ on the horizontal axis and $K$ on the vertical axis, its slope is $-w/r$. An isoquant contains input bundles producing the same output. The long-run cost-minimization problem is to find the lowest isocost that reaches the target isoquant. For a smooth interior solution, the curves are tangent. But a tangency is not guaranteed: perfect substitutes can yield a corner, and fixed-proportions production can yield a kink. The technology and the feasible input bundles determine which case applies.

Interior Solutions for Cobb-Douglas Technology

For a Cobb-Douglas technology $q=A K^\alpha L^\beta$, take $A>0$, $\alpha>0$, and $\beta>0$. At an interior optimum with a differentiable isoquant, the marginal rate of technical substitution equals the input-price ratio. The marginal products are $MP_K=\alpha q/K$ and $MP_L=\beta q/L$. Since $MP_K/MP_L$ is the amount of labor that can be saved by a small increase in capital while holding output fixed, tangency to the isocost requires $MP_K/MP_L=r/w$. Thus:

\frac{\alpha}{\beta}\frac{L}{K}=\frac{r}{w} \qquad\Longrightarrow\qquad \frac{K}{L}=\frac{\alpha w}{\beta r}.

The ratio gives the cost-minimizing input mix, but it does not by itself give the quantities. Substitute $K=tL$, where $t=\alpha w/(\beta r)$, into the production constraint. This determines $L$, and then $K$ follows from the ratio. Because positive-exponent Cobb-Douglas isoquants are strictly convex in the positive-input region, this interior tangency is the unique cost-minimizing mix under the stated positive-price assumptions.

Result: For $q=A K^\alpha L^\beta$, an interior cost-minimizing solution satisfies $K/L=\alpha w/(\beta r)$. Substituting that ratio into $A K^\alpha L^\beta=q$ gives the conditional factor demands. The result assumes positive target output, positive input prices, positive exponents, and an interior solution; it is not a formula for a corner or a fixed-proportions kink.

Worked Example: Finding the Mix and Classifying Scale

A workshop produces $q=6$ units according to $q=K^{1/3}L^{1/3}$. Capital costs $r=\$2$ per unit and labor costs $w=\$4$ per unit. Find the cost-minimizing inputs and cost, verify the output, and classify returns to scale. Method: use the Cobb-Douglas tangency condition, then substitute into the output and cost equations. This avoids confusing the input ratio with its reciprocal.

The exponents are $\alpha=\beta=1/3$, so the optimal ratio is $K/L=(\alpha w)/(\beta r)=((1/3)4)/((1/3)2)=2$. Therefore $K=2L$.

Substitute into the production target: $6=(2L)^{1/3}L^{1/3}=(2L^2)^{1/3}$. Cubing both sides gives $216=2L^2$, so $L^2=108$, and hence $L=6\sqrt{3}$. The nonnegative solution is required because labor cannot be negative. Then $K=2L=12\sqrt{3}$.

Check output: $KL=(12\sqrt{3})(6\sqrt{3})=216$, so $K^{1/3}L^{1/3}=(KL)^{1/3}=216^{1/3}=6$. Cost is $2(12\sqrt{3})+4(6\sqrt{3})=24\sqrt{3}+24\sqrt{3}=48\sqrt{3}$, or $\$83.14$ to the nearest cent.

For scale, multiplying both inputs by $t$ multiplies output by $t^{1/3+1/3}=t^{2/3}$. Since $2/3<1$, this technology has decreasing returns to scale. The cost-minimizing inputs and the returns-to-scale classification answer different questions; the latter depends on how output changes when all inputs scale together.

From the Optimal Ratio to Conditional Factor Demands

The ratio condition becomes a reusable method when written as a sequence. First calculate $t=\alpha w/(\beta r)$; next set $K=tL$; then solve the production constraint for $L$; finally calculate $K$ and total cost. Keeping these steps separate makes it easier to detect an algebra error before reporting the answer.

Find the input ratio.
For an interior Cobb-Douglas solution, set $K/L=\alpha w/(\beta r)$. Check that the ratio responds sensibly: a higher labor price $w$ raises optimal capital relative to labor, while a higher capital price $r$ lowers it.

Impose the output target.
Replace $K$ by $tL$ in $A K^\alpha L^\beta=q$. Solve for the nonnegative input level and use the ratio to find the other input.

Verify both constraints and the objective.
Substitute the candidate bundle into the production function to confirm it yields $q$, then calculate $rK+wL$. If the technology can have a corner or kink, compare the relevant alternatives rather than relying on tangency alone.

This procedure also explains why conditional demands depend on prices and the target, not on a particular chosen output price. The firm is asked to produce a specified quantity at minimum cost; the revenue from selling that quantity does not change which input bundle is cheapest for the target.

Worked Example: Conditional Demands with Constant Returns to Scale

A producer uses $q=K^{2/3}L^{1/3}$, faces capital price $r=\$2$ and labor price $w=\$4$, and must produce $q=8$. Find the cost-minimizing bundle, its cost, and the returns-to-scale classification. Method: find the tangency ratio and solve the production constraint. This avoids rounding the ratio or guessing input quantities before enforcing the output requirement.

Here $\alpha=2/3$ and $\beta=1/3$. Thus $K/L=((2/3)4)/((1/3)2)=4$, or $K=4L$. Substitute into production: $8=(4L)^{2/3}L^{1/3}=4^{2/3}L=2^{4/3}L$. Therefore $L=8/2^{4/3}=2^{5/3}$, and $K=4(2^{5/3})=2^{11/3}$.

Check output: $(2^{11/3})^{2/3}(2^{5/3})^{1/3}=2^{22/9}2^{5/9}=2^{27/9}=2^3=8$. Cost is $2(2^{11/3})+4(2^{5/3})=8(2^{5/3})+4(2^{5/3})=12(2^{5/3})$, which is approximately $\$38.10$ to the nearest cent.

The exponents sum to $2/3+1/3=1$. Scaling both inputs by $t$ therefore scales output by $t$, so the technology has constant returns to scale. The computed input quantities are approximately $K=12.6992$ and $L=3.1748$; their ratio is $4$, as required.

Corner and Kink Solutions

An interior first-order condition is useful only when an interior solution exists. With perfect substitutes, the firm compares output per dollar for each input and uses the input with the higher output per dollar; if the two are equal, there may be multiple minimizing bundles. With fixed proportions, extra units of just one input do not raise output once the other input is the bottleneck. The cost-minimizing bundle is then at the kink where the required proportions are met. Both cases are applications of the same principle: find the least-cost bundle that reaches the target isoquant.

Worked Example: A Fixed-Proportions Kink

A small assembly operation has production function $q=\min(K,2L)$. Capital costs $r=\$2$ per unit and labor costs $w=\$4$ per unit. Find the least-cost bundle for $q=10$, check its cost, and classify scale. Method: enforce both production requirements at the kink. This avoids buying excess units of one input that cannot compensate for a shortage of the other.

To produce at least 10 units, both $K\geq10$ and $2L\geq10$ must hold. Thus $L\geq5$. Since both input prices are positive, any amount above these minimum requirements raises cost without being needed for the target. The minimizing bundle is $K=10$, $L=5$, where $K=2L$ and the two arguments of the minimum are equal.

Check output: $\min(10,2(5))=\min(10,10)=10$. Cost is $2(10)+4(5)=\$40$. For any $t>0$, $\min(tK,2tL)=t\min(K,2L)$, so this technology has constant returns to scale.

Checks That Make a Solution Reliable

A complete cost-minimization answer should survive three checks. First, the input ratio should satisfy the appropriate optimality condition, or the candidate should be justified as a corner or kink. Second, substituting the proposed bundle into the production function should meet the target exactly. Third, the reported cost should equal the sum of each input quantity multiplied by its own price. These checks are especially important when the calculation contains fractional exponents.

A frequent exam trap is to classify scale from a numerical input bundle instead of the production function. Returns to scale ask what happens to output when all inputs are multiplied by the same factor; for Cobb-Douglas production, the exponent sum determines the answer. Another trap is to reverse the factor ratio. The correct Cobb-Douglas ratio is $K/L=\alpha w/(\beta r)$, not $\beta r/(\alpha w)$. A reversed ratio generally fails the tangency condition even if the proposed bundle happens to satisfy the output target.

The method is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus: identify the relevant technology, choose the proper optimality condition, solve for the required inputs, and interpret the result rather than stopping at a formula.

Key takeaway: For a smooth Cobb-Douglas technology, use the price-weighted marginal-product condition to find the input ratio, then use the production target to find the quantities. For corners and kinks, apply the technology’s structure directly. Always verify output and cost, and classify scale from the production function.

Check Your Understanding

For each problem, identify the relevant method, calculate the requested bundle or cost, and check the technology’s scale classification where requested.

A greenhouse produces $q=K^{1/2}L^{1/2}$. Capital costs $r=\$4$, labor costs $w=\$1$, and the target is $q=9$. Find the cost-minimizing inputs and cost, and classify returns to scale. (Answer: $K=4.5$, $L=18$, cost $=\$36$; constant returns to scale.)
A repair service produces $q=K^{1/3}L^{2/3}$, with $r=\$3$, $w=\$1$, and target $q=6$. Find the cost-minimizing input ratio, inputs, and cost, and classify scale. (Answer: $K/L=1/6$, $K=6^{1/3}$, $L=6^{4/3}$, cost $=9\cdot6^{1/3}\approx\$16.35$; constant returns to scale.)
A printing process has $q=\min(3K,L)$. Capital costs $r=\$5$, labor costs $w=\$2$, and output must be $15$. Find the least-cost inputs and cost, then classify scale. (Answer: $K=5$, $L=15$, cost $=\$55$; constant returns to scale.)
A craft producer uses $q=K^{0.4}L^{0.4}$, with both input prices equal to $\$1$, and target $q=4$. Find the minimizing bundle and cost, and classify scale. (Answer: $K=L=4\sqrt{2}$, cost $=8\sqrt{2}\approx\$11.31$; decreasing returns to scale.)
A technology is $q=K+3L$. Capital costs $r=\$2$, labor costs $w=\$9$, and the target is $q=12$. Find the minimizing input choice and cost, and classify scale. (Answer: use $K=12$, $L=0$, cost $=\$24$; constant returns to scale.)

What You'll Learn