From Diminishing Marginal Product to Rising Marginal Cost

Why Productivity and Marginal Cost Move in Opposite Directions

In the short run, a firm may have some inputs fixed while adjusting a variable input such as labor. If each additional unit of that input produces less additional output than the one before it, the firm must use more input to obtain each further unit of output. When the input’s price is constant, that reduced productivity translates into a higher marginal cost.

The relationship is useful because it connects two ways of describing production. Marginal product measures the additional output associated with an additional unit of input. Marginal cost measures the additional cost associated with an additional unit of output. Under specific conditions, one is the inverse of the other after accounting for the input’s price.

Definition: Suppose a firm produces output $Q=f(L)$, where $L$ is a variable input and other inputs are fixed. The marginal product of $L$ is $MP_L=dQ/dL$. If each unit of $L$ costs $w$, where $w$ is constant, then variable cost is $VC=wL$. For a positive marginal product, marginal cost is $MC=dTC/dQ=w/MP_L$.

The formula describes a local, or instantaneous, change. Its standard short-run assumptions are that the production function is differentiable at the input level being considered; the input price $w$ is constant; other inputs are held fixed; and $MP_L>0$. The firm’s fixed cost may be positive, but it does not change as output changes in this calculation.

Deriving the Relationship

The derivation follows the chain of how an input change affects cost and output. With one variable input, an increase in $L$ changes variable cost at the rate $dVC/dL=w$, and changes output at the rate $dQ/dL=MP_L$. To find the cost change per unit of output, divide the first rate by the second.

TC=FC+wL \qquad \frac{dTC}{dL}=w \qquad \frac{dQ}{dL}=MP_L

Since $MP_L>0$, output can be used as the local measure of change in place of the input. Applying the chain rule gives the result:

MC=\frac{dTC}{dQ} =\frac{dTC/dL}{dQ/dL} =\frac{w}{MP_L}

This explains why the formula is an inverse relationship: at a constant input price, a smaller positive marginal product means more input cost is needed for each marginal unit of output. If marginal product is positive and diminishing as $L$ rises, $w/MP_L$ rises as $L$ rises. Thus marginal cost rises over that range, holding the other assumptions fixed. This conclusion is not a claim about every production range: it requires declining, positive marginal product.

Worked Example: Marginal Cost From a Production Function

A hypothetical workshop’s short-run production function is $Q=12\sqrt{L}$, where $L$ is labor-hours per shift and $Q$ is units produced per shift. Other inputs are fixed. Each labor-hour costs $w=\$18$, and fixed cost is $\$240$ per shift. Find the marginal product and marginal cost at $L=9$. Use the derivative method, and state the common mistake the calculation avoids.

Method and assumptions. The function is differentiable for $L>0$. The labor-hour price is constant, and fixed cost does not vary with output. We calculate a local marginal cost, not the cost of a finite output increase.

Step 1: Differentiate output with respect to labor.

MP_L=\frac{dQ}{dL} =\frac{d}{dL}(12\sqrt{L}) =\frac{6}{\sqrt{L}}

At $L=9$, $\sqrt{9}=3$, so $MP_L=6/3=2$ units per labor-hour.

Step 2: Apply the constant-price formula.

MC=\frac{w}{MP_L} =\frac{\$18\text{ per labor-hour}}{2\text{ units per labor-hour}} =\$9\text{ per unit}

As a check, $VC=18L$, so $dTC/dL=18$; dividing this by $dQ/dL=2$ also gives $dTC/dQ=18/2=9$ dollars per unit. The production level at $L=9$ is $Q=12(3)=36$ units per shift, which is consistent with evaluating the marginal quantities at that input level.

Common mistake avoided: Multiplying wage by marginal product would give $18(2)=36$, but that does not measure cost per unit of output. The correct operation is to divide the input price by output per input unit.

Discrete Changes: An Interval Version

Production schedules often show output at whole input levels rather than a differentiable function. For a finite change from $L_1$ to $L_2$, let $\Delta L=L_2-L_1$ and $\Delta Q=Q_2-Q_1$. If the input price is constant, the variable-cost change is $w\Delta L$. For $\Delta Q>0$, the interval marginal cost is the cost increase divided by the output increase.

MC_{\text{interval}} =\frac{\Delta VC}{\Delta Q} =\frac{w\Delta L}{\Delta Q} =\frac{w}{\Delta Q/\Delta L}

Here, $\Delta Q/\Delta L$ is the average additional output per additional input unit over that particular interval. The expression is an exact finite-change calculation under the stated assumptions. It is not automatically the instantaneous marginal cost at either endpoint. If the schedule uses successive one-unit input increases, each interval can be calculated separately.

Worked Example: Marginal Cost Across Successive Input Intervals

A hypothetical packaging line has the following short-run output per shift. Labor is the only variable input, each worker-shift costs $\$24$, and fixed costs are unchanged across the listed levels. Calculate the interval marginal cost for each step from two through six workers. Use the discrete-change method, and identify the mistake of using total cost rather than the cost change.

Workers per shift $L$	Output $Q$, units per shift
2	30
3	45
4	57
5	66
6	72

Method and assumptions. Each interval adds one worker-shift, so $\Delta L=1$ and $\Delta VC=\$24$ each time. We divide that incremental cost by the corresponding output increase.

From two to three workers, $\Delta Q=45-30=15$ units. Therefore, $MC_{\text{interval}}=24/15=\$1.60$ per unit.

From three to four workers, $\Delta Q=57-45=12$ units. Therefore, $MC_{\text{interval}}=24/12=\$2.00$ per unit.

From four to five workers, $\Delta Q=66-57=9$ units. Therefore, $MC_{\text{interval}}=24/9=\$2.666\ldots$, or $\$2.67$ per unit rounded to the nearest cent.

From five to six workers, $\Delta Q=72-66=6$ units. Therefore, $MC_{\text{interval}}=24/6=\$4.00$ per unit. The successive output gains are $15,12,9,6$, while each added worker costs $\$24$; the calculated interval marginal costs therefore rise as the additional output falls.

Check and common mistake avoided. The total variable cost at six workers is $6(24)=\$144$, but $\$144$ is not the cost of moving from five to six workers. That interval adds only one worker, so its cost increase is $\$24$, and the correct interval calculation is $24/6=\$4.00$ per unit.

When Diminishing Marginal Product Raises Marginal Cost

The inverse formula also makes it possible to compare marginal cost at different input levels without first calculating total cost. The key is to verify that marginal product is still positive. A fall in marginal product from one positive value to a smaller positive value raises $w/MP_L$ when $w$ is constant.

Worked Example: Comparing Marginal Cost at Two Labor Levels

A hypothetical firm has short-run production $Q=24L-L^2$, where $L$ is labor-hours per shift and $Q$ is units per shift. Its other inputs are fixed, and each labor-hour costs $w=\$20$. Compare marginal product and marginal cost at $L=4$ and $L=8$. Use derivatives, and avoid treating marginal cost as wage times marginal product.

Method and assumptions. The function is differentiable, and the price per labor-hour is constant. Its marginal product is positive at both specified labor levels, so the formula applies at both points.

Step 1: Differentiate production.

MP_L=\frac{dQ}{dL}=24-2L

At $L=4$, $MP_L=24-2(4)=16$ units per labor-hour. At $L=8$, $MP_L=24-2(8)=8$ units per labor-hour. Marginal product is positive at both points and is lower at the higher labor level.

Step 2: Calculate marginal cost at each point.

MC\big|_{L=4}=\frac{20}{16}=\$1.25\text{ per unit} \qquad MC\big|_{L=8}=\frac{20}{8}=\$2.50\text{ per unit}

The marginal cost doubles as marginal product falls from 16 to 8, because the numerator $w$ is unchanged. As a check, variable cost is $VC=20L$, so $dTC/dL=20$. Dividing by the respective marginal products gives $20/16=1.25$ and $20/8=2.50$. Output at the two levels is $Q(4)=96-16=80$ and $Q(8)=192-64=128$ units per shift.

Common mistake avoided: At $L=4$, multiplying $20$ by $16$ gives $320$, not marginal cost per output unit. The input price must be divided by output per input unit: $20/16=\$1.25$ per unit.

Limits, Interpretation, and a Common Trap

The formula $MC=w/MP_L$ is valid as a finite numerical ratio only when marginal product is positive. If $MP_L=0$, the instantaneous marginal output at that point is zero and $w/MP_L$ is undefined. This says nothing by itself about the output change from a finite increase in labor: a derivative describes the local rate at a point, while a finite change must be assessed from $\Delta Q$.

For example, in the mathematical production function $Q=L^2$, $MP_L=2L$, which is zero at $L=0$. Yet increasing labor from zero to one gives $\Delta Q=1^2-0^2=1$. The zero derivative at the starting point does not mean that this finite labor increase produces no output. At that point the ratio $w/MP_L$ remains undefined; use finite changes to evaluate the stated finite interval.

A related trap is to assume that every decline in marginal product implies rising marginal cost, without checking the relevant conditions. The inverse relationship requires a constant input price and a positive marginal product. If marginal product is zero, the ratio is undefined; if it is negative, the formula does not describe ordinary marginal cost over a range where more input reduces output. If the input price changes, the constant-$w$ derivation also no longer applies as written.

Key takeaway: With one variable input, fixed other inputs, a constant input price $w$, and positive marginal product, $MC=w/MP_L$. Diminishing positive marginal product raises marginal cost because each additional unit of output requires more of the priced input.

This production-cost relationship is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus. For calculations, first identify whether the question provides a differentiable production function or finite production changes. Use derivatives for local marginal quantities and $\Delta VC/\Delta Q$ for finite intervals.

Check Your Understanding

For each problem, identify whether the data call for a local derivative or a finite-change calculation.

A variable input costs $\$15$ per unit, and its positive marginal product is 5 units of output per input unit. Find marginal cost. (Answer: $\$3.00$ per unit of output.)
A hypothetical production function is $Q=10\sqrt{L}$, with a constant input price of $\$12$ per labor-hour. Find marginal cost at $L=25$. (Answer: $\$12.00$ per unit of output.)
A firm adds two worker-shifts at $\$18$ each, and output rises from 80 to 92 units. Find interval marginal cost. (Answer: $\$3.00$ per unit of output.)
A hypothetical production function is $Q=30L-L^2$, and the variable input costs $\$24$ per unit. Find marginal product and marginal cost at $L=9$. (Answer: $MP_L=12$ units per input unit; $MC=\$2.00$ per output unit.)
With a constant input price of $\$28$, marginal product falls from 14 to 7 units per input unit. Calculate marginal cost at each point and state its percentage increase. (Answer: marginal cost rises from $\$2.00$ to $\$4.00$ per unit, a 100% increase.)
For $Q=L^2$, marginal product is zero at $L=0$. What is $w/MP_L$ there, and how much does output change when labor rises from zero to one? (Answer: the ratio is undefined; the finite output change is $1$ unit.)

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