Deriving Marginal Products In Cobb-Douglas Functions

Why Derive Marginal Products?

A production function describes the maximum output a firm can produce from specified quantities of inputs, given its technology. A Cobb-Douglas production function is a particular form that makes it possible to connect output to labor and capital using powers. Once its marginal products are derived, the firm can assess how output responds locally to a change in either input while holding the other input fixed.

The earlier material on marginal product as a derivative defines a marginal product as the partial derivative of output with respect to one input. The key word is partial: when finding labor’s marginal product, capital is held constant; when finding capital’s marginal product, labor is held constant. The resulting formulas also help compare an input’s marginal output per unit with its proportional contribution to production.

Definition: A two-input Cobb-Douglas production function is \(Q=A L^\alpha K^\beta\), where \(Q\) is output, \(L\) is labor, \(K\) is capital, \(A>0\) is a technology parameter, and \(\alpha,\beta\) are fixed exponents. For positive \(L\) and \(K\), labor’s marginal product is \(MP_L=\partial Q/\partial L\), and capital’s marginal product is \(MP_K=\partial Q/\partial K\). Each measures the local change in output from a one-unit increase in that input, holding the other input fixed.

For the standard case considered here, \(L>0\), \(K>0\), and \(\alpha,\beta>0\). These conditions make the marginal-product formulas well-defined and positive. The formulas describe local rates of change: for a finite input increase, the output change is generally not exactly the marginal product multiplied by the increase, because the marginal product can change along the way.

Deriving the Two Marginal Products

To find labor’s marginal product, treat \(A\), \(K^\beta\), and the exponent \(\alpha\) as constants while differentiating with respect to \(L\). The power rule gives the derivative of \(L^\alpha\) as \(\alpha L^{\alpha-1}\). For capital, hold \(A\), \(L^\alpha\), and \(\beta\) constant and apply the same rule to \(K^\beta\).

\begin{aligned} MP_L &= \frac{\partial}{\partial L}\left(A L^\alpha K^\beta\right) = A K^\beta \left(\alpha L^{\alpha-1}\right) = \alpha A L^{\alpha-1}K^\beta = \alpha\frac{Q}{L},\\[4pt] MP_K &= \frac{\partial}{\partial K}\left(A L^\alpha K^\beta\right) = A L^\alpha \left(\beta K^{\beta-1}\right) = \beta A L^\alpha K^{\beta-1} = \beta\frac{Q}{K}. \end{aligned}

The last equality in each line follows by dividing the original production function by the input being differentiated. For example, \(Q/L=A L^{\alpha-1}K^\beta\), so multiplying this quantity by \(\alpha\) gives \(MP_L\). These equivalent forms are often convenient when \(Q\), \(L\), and \(K\) are already known.

The formulas also reveal a direct distinction between marginal output per unit and proportional input contribution. Multiplying each marginal product by its input gives \(MP_L L=\alpha Q\) and \(MP_K K=\beta Q\). Equivalently, for small proportional input changes, the output response is approximately the sum of the exponent-weighted input changes:

\frac{dQ}{Q} = \alpha\frac{dL}{L} + \beta\frac{dK}{K}.

This relationship follows by taking the total differential of \(Q=A L^\alpha K^\beta\), then dividing by \(Q\). It is a local approximation for changes in both inputs, not an exact formula for arbitrary finite changes. The exponents therefore describe proportional sensitivity, while the marginal products describe output change per additional unit at the current input levels.

Worked Example: Deriving and Evaluating Both Marginal Products

A workshop’s output is \(Q=12L^{1/2}K^{1/2}\). At \(L=16\) units of labor and \(K=9\) units of capital, derive both marginal-product formulas, evaluate them, and compare their meanings. Use output units per input unit.

Method. Differentiate with respect to one input at a time, then substitute the specified input bundle. This avoids treating both inputs as changing in the same partial derivative.

Step 1: Find output. Since \(16^{1/2}=4\) and \(9^{1/2}=3\),

$$ Q=12(4)(3)=144. $$

Step 2: Differentiate with respect to labor. The capital term is held fixed:

MP_L=12\left(\frac{1}{2}\right)L^{-1/2}K^{1/2} =6L^{-1/2}K^{1/2}.

At \(L=16\), \(L^{-1/2}=1/4\); at \(K=9\), \(K^{1/2}=3\). Therefore \(MP_L=6(1/4)(3)=18/4=4.5\) output units per additional unit of labor, locally.

Step 3: Differentiate with respect to capital. This time labor is held fixed:

MP_K=12L^{1/2}\left(\frac{1}{2}\right)K^{-1/2} =6L^{1/2}K^{-1/2}.

At the specified bundle, \(L^{1/2}=4\) and \(K^{-1/2}=1/3\), so \(MP_K=6(4)(1/3)=8\) output units per additional unit of capital, locally.

Step 4: Compare and check. At this bundle, an extra unit of capital has the larger marginal product: \(8>4.5\). Using the equivalent formulas gives \(MP_L=(1/2)(144/16)=4.5\) and \(MP_K=(1/2)(144/9)=8\), confirming both calculations. This comparison is per input unit; it does not say that capital’s proportional contribution is larger. In fact, \(MP_L L=4.5(16)=72\) and \(MP_K K=8(9)=72\).

Common mistake avoided: differentiating \(L^{1/2}K^{1/2}\) as though both inputs change when finding one marginal product. A partial derivative changes only the named input.

Interpreting the Exponents and Comparing Contributions

At any positive input bundle, \(MP_L=\alpha Q/L\) and \(MP_K=\beta Q/K\). Thus, knowing output and both input quantities is enough to evaluate the marginal products without repeating the differentiation. The ratio of the two marginal products is also useful:

\frac{MP_L}{MP_K} = \frac{\alpha}{\beta}\frac{K}{L}.

This ratio compares the output change per extra unit of labor with the output change per extra unit of capital at the current bundle. Its value depends on both the exponents and the input quantities. It does not compare the inputs in equal physical units: a labor unit and a capital unit may represent quite different amounts of resources.

By contrast, \(\alpha\) and \(\beta\) give the proportional sensitivity of output to each input. For example, a small one-percent increase in labor, with capital unchanged, corresponds approximately to an \(\alpha\)-percent increase in output. A small one-percent increase in capital, with labor unchanged, corresponds approximately to a \(\beta\)-percent increase. These statements are local and follow from the proportional-change formula above.

Worked Example: Separating Marginal Product from Proportional Sensitivity

A production process follows \(Q=4L^{1/3}K^{2/3}\). Evaluate output, both marginal products, and the proportional output effects of small five-percent increases in labor and capital at \(L=8\), \(K=27\). Assume each input change is considered separately.

Method. First evaluate the production function, then use \(MP_L=\alpha Q/L\) and \(MP_K=\beta Q/K\). Use the exponents for proportional sensitivity rather than comparing the raw marginal products for that question.

Step 1: Evaluate output. Since \(8^{1/3}=2\) and \(27^{2/3}=9\),

$$ Q=4(2)(9)=72. $$

Step 2: Calculate marginal products. Labor’s marginal product is

MP_L=\frac{1}{3}\frac{72}{8}=3.

Capital’s marginal product is

MP_K=\frac{2}{3}\frac{72}{27} =\frac{48}{81} =\frac{16}{27} \approx 0.5926.

Thus labor has the larger output change per additional input unit at this bundle: \(3>16/27\). The units used for labor and capital differ, so this is a per-unit comparison, not a claim that labor is more important in proportional terms.

Step 3: Compare proportional effects. For a small five-percent increase in labor alone, the approximate output increase is \((1/3)(5\%)=1.6667\%\). For a small five-percent increase in capital alone, it is approximately \((2/3)(5\%)=3.3333\%\). Capital therefore has the greater proportional sensitivity even though its marginal product per input unit is smaller here.

Step 4: Check the input-weighted marginal products. \(MP_L L=3(8)=24=(1/3)(72)\). Also, \(MP_K K=(16/27)(27)=16=(2/3)(72)\). Both identities agree with the exponents and the computed output.

Common mistake avoided: concluding that labor has the greater proportional effect just because \(MP_L>MP_K\). The marginal products have different input units; the exponents are the relevant comparison for equal percentage changes.

Comparing Input Bundles at the Same Output

Marginal products are evaluated at a particular input bundle, not assigned once and for all to a production function. The same technology can produce the same output with different combinations of labor and capital, and the marginal products can differ across those bundles. The identities \(MP_L=\alpha Q/L\) and \(MP_K=\beta Q/K\) make this dependence explicit: at a given output, a larger quantity of an input corresponds to a smaller marginal product of that input when its exponent is positive.

Worked Example: Comparing Marginal Products Across Two Bundles

A small fabrication process has production function \(Q=5L^{1/2}K^{1/2}\). Compare the marginal products at bundle A, \(L=9\), \(K=16\), and bundle B, \(L=16\), \(K=9\). Calculate output first and then compare each input’s marginal product across the bundles.

Method. Evaluate the production function separately at each bundle, then use the output-over-input marginal-product formulas. This prevents assuming that equal output implies equal marginal products.

Step 1: Find output at bundle A. \(9^{1/2}=3\) and \(16^{1/2}=4\), so \(Q_A=5(3)(4)=60\).

Step 2: Find bundle A marginal products.

MP_{L,A}=\frac{1}{2}\frac{60}{9}=\frac{10}{3}, \qquad MP_{K,A}=\frac{1}{2}\frac{60}{16}=\frac{15}{8}.

Step 3: Find output at bundle B. The inputs are reversed, so \(Q_B=5(4)(3)=60\), the same output.

Step 4: Find bundle B marginal products.

MP_{L,B}=\frac{1}{2}\frac{60}{16}=\frac{15}{8}, \qquad MP_{K,B}=\frac{1}{2}\frac{60}{9}=\frac{10}{3}.

At bundle A, labor’s marginal product is \(10/3\) and capital’s is \(15/8\). At bundle B those values reverse. Each bundle produces 60 units, but the input quantities differ, so the marginal product of each input differs. As a check, bundle A has \(MP_{L,A}L=30\) and \(MP_{K,A}K=30\), while bundle B has \(MP_{L,B}L=30\) and \(MP_{K,B}K=30\).

Common mistake avoided: transferring a marginal-product value from one input bundle to another merely because output is unchanged. The formulas depend on both output and the quantity of the input being varied.

Common Errors and Practical Checks

Before interpreting a computed result, verify the differentiation and the point at which it was evaluated. A useful check is to compare the direct derivative with the shortcut formula \(MP_L=\alpha Q/L\) or \(MP_K=\beta Q/K\). If they disagree, a power, input value, or exponent has likely been mishandled.

Reducing the exponent incorrectly. Differentiating \(L^\alpha\) gives \(\alpha L^{\alpha-1}\), not \(\alpha L^\alpha\). The exponent falls by one.
Forgetting which input is fixed. When differentiating with respect to labor, \(K^\beta\) is a constant multiplier. When differentiating with respect to capital, \(L^\alpha\) is the constant multiplier.
Confusing units with percentages. A marginal product measures output per additional input unit. An exponent measures proportional sensitivity to that input.
Treating a local rate as a finite change. The marginal product at the starting bundle is a local rate. It generally does not give the exact output change from a large input increase.
Ignoring the evaluation point. State the values of both labor and capital used to evaluate a marginal product. The result is specific to that bundle.

Key takeaway: For \(Q=A L^\alpha K^\beta\), differentiate with respect to one input while holding the other fixed. Then use \(MP_L=\alpha Q/L\) and \(MP_K=\beta Q/K\) to evaluate output per additional input unit. Use the exponents—not a raw comparison of marginal products—to compare proportional sensitivities.

Check Your Understanding

For each problem, distinguish a per-unit marginal product from a proportional input effect where relevant.

A process has \(Q=6L^{1/2}K^{1/2}\). At \(L=4\), \(K=25\), calculate output and both marginal products. (Answer: \(Q=60\), \(MP_L=7.5\), \(MP_K=1.2\).)
A production function is \(Q=3L^{1/3}K^{2/3}\), with \(L=27\) and \(K=8\). Find output and both marginal products. (Answer: \(Q=36\), \(MP_L=4/3\), \(MP_K=3\).)
For \(Q=2L^{1/3}K^{2/3}\), evaluate output and both marginal products at \(L=64\), \(K=8\). Which input has the larger marginal product per unit at this bundle? (Answer: \(Q=32\), \(MP_L=1/6\), \(MP_K=8/3\); capital has the larger marginal product per unit.)
A technology follows \(Q=4L^{1/2}K^{1/2}\). Compare the marginal products at \((L,K)=(4,25)\) and \((25,4)\), calculating output at each bundle. (Answer: Both bundles produce \(Q=40\). At \((4,25)\), \(MP_L=5\), \(MP_K=0.8\); at \((25,4)\), \(MP_L=0.8\), \(MP_K=5\).)
For \(Q=7L^{0.3}K^{0.7}\), estimate the percentage output change from a small ten-percent increase in labor alone, then from a small five-percent increase in capital alone. (Answer: Approximately \(3\%\) for the labor increase and \(3.5\%\) for the capital increase.)

What You'll Learn