Cobb-Douglas Cost-Minimization Problems

Why Cobb-Douglas Cost Minimization Matters

A firm that can adjust both capital and labor in the long run must choose an input combination that produces its required output at the lowest possible cost. For a Cobb-Douglas production function, the technology and input prices together determine the cost-minimizing ratio of capital to labor. The output target then determines the scale of both inputs.

The method brings together marginal products, the tangency condition, and the output constraint. The central caution is to keep the input-price ratio and the marginal-product ratio in the correct order: an inverted ratio can produce a plausible-looking input mix that is not cost minimizing.

Definition: A two-input Cobb-Douglas production function has the form $Q=A K^\alpha L^\beta$, where $Q$ is output, $K$ is capital, $L$ is labor, $A>0$ is a productivity parameter, and $\alpha,\beta>0$ are output elasticities. Given positive input prices $r$ for capital and $w$ for labor, a cost-minimization problem chooses $K,L\geq 0$ to minimize $C=rK+wL$, subject to producing at least the required output $\bar Q>0$.

Assume inputs are divisible and the firm can adjust both inputs; input prices are fixed and positive; and the production function is the stated Cobb-Douglas function. Because output is increasing in each input, a least-cost choice for a positive target will produce exactly $\bar Q$: producing more would require weakly more inputs and would cost more. With both exponents positive, the optimum for a positive target is interior.

From Marginal Products to the Input Ratio

The marginal product of an input is the additional output produced by a small increase in that input, holding the other input fixed. For $Q=A K^\alpha L^\beta$, differentiating gives:

MP_K=\frac{\partial Q}{\partial K}=\alpha A K^{\alpha-1}L^\beta=\frac{\alpha Q}{K}, \qquad MP_L=\frac{\partial Q}{\partial L}=\beta A K^\alpha L^{\beta-1}=\frac{\beta Q}{L}.

At an interior cost-minimizing tangency, the marginal rate of technical substitution equals the input-price ratio: $MP_K/MP_L=r/w$. Substituting the Cobb-Douglas marginal products and cancelling the common positive output $Q$ gives the input ratio:

\frac{MP_K}{MP_L} = \frac{\alpha L}{\beta K} = \frac{r}{w} \quad\Longrightarrow\quad \frac{K}{L}=\frac{\alpha w}{\beta r}.

This ratio is a useful check before solving for input levels. For example, if $r$ rises while the other parameters stay fixed, the ratio $K/L$ falls: capital is relatively more expensive, so the least-cost bundle uses less capital relative to labor. The ratio alone is not the answer, however. The output constraint is needed to determine the quantities.

Key result: For $Q=A K^\alpha L^\beta$ with positive parameters and positive input prices, an interior cost-minimizing bundle for target $\bar Q>0$ satisfies $K/L=\alpha w/(\beta r)$ and $A K^\alpha L^\beta=\bar Q$. Both conditions are needed to identify the bundle.

Solving the Ratio and Output Constraint

Let $t=\beta r/(\alpha w)$, so that the cost-minimizing ratio can be written $L=tK$. Substitution into the output constraint gives a direct solution for capital, followed by labor:

\bar Q=A K^\alpha(tK)^\beta=A t^\beta K^{\alpha+\beta}, \qquad K=\left(\frac{\bar Q}{A t^\beta}\right)^{1/(\alpha+\beta)}, \qquad L=tK.

The sum $\alpha+\beta$ is not required to equal one. The method applies to any positive exponents under the stated assumptions. Once the quantities are known, minimum expenditure is $C_{\min}=rK+wL$. It is important to multiply each input quantity by its own price and add the two expenditures.

Calculate marginal products.
For Cobb-Douglas production, use $MP_K=\alpha Q/K$ and $MP_L=\beta Q/L$.

Apply the tangency condition.
Set $MP_K/MP_L=r/w$ to obtain $K/L=\alpha w/(\beta r)$.

Use the output constraint.
Substitute the input ratio into $A K^\alpha L^\beta=\bar Q$ and solve for both inputs.

Compute and check cost.
Calculate $rK+wL$, then verify that the proposed bundle produces exactly $\bar Q$ and satisfies the tangency ratio.

Why the Tangency Gives the Minimum

The tangency condition has a direct economic interpretation. If $MP_K/MP_L>r/w$, a small amount of expenditure shifted from labor to capital would produce more output for the same cost; the original bundle could not be least cost. If $MP_K/MP_L<r/w$, shifting expenditure toward labor would be more productive. At the least-cost interior bundle, neither shift reduces the cost of achieving the target, so the two ratios are equal.

The solution is not merely a local candidate. To see why, impose the output constraint and express labor as a function of capital:

L(K)=\left(\frac{\bar Q}{A K^\alpha}\right)^{1/\beta} =\left(\frac{\bar Q}{A}\right)^{1/\beta}K^{-\alpha/\beta}.

The corresponding expenditure is $C(K)=rK+w(\bar Q/A)^{1/\beta}K^{-\alpha/\beta}$. Its first derivative is

C'(K)=r-\frac{\alpha}{\beta}w\left(\frac{\bar Q}{A}\right)^{1/\beta}K^{-\alpha/\beta-1}.

Setting $C'(K)=0$ yields $rK=(\alpha/\beta)wL$, which is equivalent to $K/L=\alpha w/(\beta r)$. Further, $C(K)$ tends to infinity as $K$ approaches zero, because the required labor tends to infinity, and it also tends to infinity as $K$ grows without bound, because the capital expenditure does. The second derivative is positive for $K>0$:

C''(K)=\frac{\alpha}{\beta}\left(\frac{\alpha}{\beta}+1\right) w\left(\frac{\bar Q}{A}\right)^{1/\beta}K^{-\alpha/\beta-2}>0.

Therefore the stationary point is the unique global minimum along the output constraint. This justification depends on positive exponents and prices and on the Cobb-Douglas technology; it is not a general rule for every possible production function.

Worked Example: A Square-Root Production Function

A hypothetical producer has $Q=\sqrt{KL}$, pays $r=\$4$ per unit of capital and $w=\$1$ per unit of labor, and must produce $\bar Q=24$. Find the cost-minimizing inputs and minimum expenditure. Use the tangency ratio and then the output constraint; the common mistake to avoid is reversing $K/L$ and $L/K$.

Here $A=1$, $\alpha=\beta=0.5$. The marginal products are $MP_K=0.5Q/K$ and $MP_L=0.5Q/L$. The tangency condition gives

$K/L=\alpha w/(\beta r)=(0.5)(1)/((0.5)(4))=1/4$, so $L=4K$.

The output constraint is $24=\sqrt{K(4K)}=\sqrt{4K^2}=2K$, since $K>0$. Thus $K=12$ and $L=4(12)=48$.

Capital expenditure is $rK=4(12)=\$48$. Labor expenditure is $wL=1(48)=\$48$. Therefore $C_{\min}=48+48=\$96$.

Check. Output is $\sqrt{12(48)}=\sqrt{576}=24$, and $K/L=12/48=1/4$, as required. The bundle meets both conditions, and the minimum expenditure is $\$96$.

Worked Example: Checking the Full Expenditure Calculation

A hypothetical workshop produces $Q=\sqrt{KL}$, with a target $\bar Q=115.2$. Capital costs $r=\$4$ per unit and labor costs $w=\$1$ per unit. Find the optimal inputs and expenditure. The method is again tangency followed by substitution; the trap is to stop after finding the input ratio and omit the cost calculation.

With $\alpha=\beta=0.5$, the optimal ratio is $K/L=(0.5)(1)/((0.5)(4))=1/4$. Hence $L=4K$. Squaring the output constraint gives $KL=(115.2)^2=13{,}271.04$. Substituting $L=4K$ gives $4K^2=13{,}271.04$, so $K^2=3{,}317.76$ and $K=57.6$, using the positive root. Therefore $L=4(57.6)=230.4$.

Capital expenditure is $4(57.6)=\$230.40$. Labor expenditure is $1(230.4)=\$230.40$. The minimum total expenditure is $\$230.40+\$230.40=\$460.80$.

Check. The product of the inputs is $57.6(230.4)=13{,}271.04$, so output is $\sqrt{13{,}271.04}=115.2$. Also $57.6/230.4=1/4$, and the two expenditure components sum to $\$460.80$.

Worked Example: Unequal Exponents and Different Input Prices

A hypothetical firm has production function $Q=K^{1/3}L^{2/3}$, with $r=\$2$, $w=\$8$, and target output $\bar Q=12\sqrt[3]{2}$. Find the least-cost input bundle and expenditure. Use the marginal-product tangency condition; the common mistake to avoid is treating unequal exponents as though the optimal input ratio were one.

Here $A=1$, $\alpha=1/3$, and $\beta=2/3$. The ratio is

$K/L=(\alpha w)/(\beta r)=((1/3)(8))/((2/3)(2))=(8/3)/(4/3)=2$, so $K=2L$.

Substitute this into the output constraint: $12\sqrt[3]{2}=(2L)^{1/3}L^{2/3}=\sqrt[3]{2}L$. Dividing both sides by $\sqrt[3]{2}$ gives $L=12$, and hence $K=24$.

Capital expenditure is $2(24)=\$48$, while labor expenditure is $8(12)=\$96$. Thus $C_{\min}=\$48+\$96=\$144$.

Check. Production is $24^{1/3}12^{2/3}=\sqrt[3]{2}\,(12)=12\sqrt[3]{2}$, and $K/L=24/12=2$, as required. The expenditure calculation assigns each price to the correct input.

Common Mistakes and Practical Checks

Cost minimization is often lost through a small algebra or interpretation error rather than a difficult calculation. Before accepting an answer, check the input ratio, the output constraint, and the expenditure independently.

Inverting the price ratio. The condition is $MP_K/MP_L=r/w$, which gives $K/L=\alpha w/(\beta r)$. Reversing this result changes the input mix.
Using the ratio as if it were an input quantity. A result such as $K/L=1/4$ means $L=4K$; it does not determine $K$ until the output target is used.
Forgetting the productivity parameter. If $A\ne1$, it remains in the output equation and affects the required scale of inputs.
Assuming constant returns to scale. The exponents need not sum to one for this fixed-output cost-minimization method. Do not silently replace $\alpha+\beta$ by one.
Reporting the input ratio but not total cost. Minimum expenditure is $rK+wL$, not simply $K+L$ and not one input’s expenditure alone.

Key takeaway: For a positive-output Cobb-Douglas technology with positive input prices, first use $K/L=\alpha w/(\beta r)$ to find the least-cost mix. Then use $A K^\alpha L^\beta=\bar Q$ to find the input levels, calculate $rK+wL$, and verify both the target output and the ratio.

Check Your Understanding

For each problem, find the requested input bundle and expenditure where applicable. Answers are provided for checking.

A producer has $Q=\sqrt{KL}$, with $r=\$2$, $w=\$8$, and target $\bar Q=30$. Find the least-cost inputs and minimum expenditure. (Answer: $K=60,\ L=15,\ C_{\min}=\$240$)
A producer has $Q=3\sqrt{KL}$, with $r=\$6$, $w=\$2$, and target $\bar Q=72$. Find the least-cost inputs and expenditure. (Answer: $K=8\sqrt{3},\ L=24\sqrt{3},\ C_{\min}=\$96\sqrt{3}$, approximately $\$166.28$)
A firm has $Q=K^{1/4}L^{3/4}$, $r=\$3$, and $w=\$1$. Its target is the output produced by $K=1,L=9$. Use the tangency rule to determine whether this bundle is cost minimizing, and find its expenditure. (Answer: Yes; $K/L=1/9$, output $3\sqrt{3}$, and $C_{\min}=\$12$)
A workshop has $Q=2K^{0.4}L^{0.6}$, $r=\$3$, $w=\$2$, and target $\bar Q=2(8^{0.4})(18^{0.6})$. Find the cost-minimizing inputs and expenditure. (Answer: $K=8,\ L=18,\ C_{\min}=\$60$)
For $Q=K^{1/3}L^{2/3}$, let $r=\$2$, $w=\$8$, and $\bar Q=12\sqrt[3]{2}$. A proposed bundle is $K=12,L=24$. Check whether it meets the target and the tangency condition; if not, give the least-cost bundle and expenditure. (Answer: The proposed bundle has the wrong ratio, $K/L=1/2$; the least-cost bundle is $K=24,L=12$, with $C_{\min}=\$144$)

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