Why the Relationship Matters
A firm may ask how much output it obtains per unit of an input, and whether that average is improving as it uses more of the input. The key comparison is between average product, output per unit of input, and marginal product, the additional output associated with a small increase in input. Their relationship reveals whether average product is rising or falling.
This tutorial considers a production function with one variable input, such as labor, while other inputs and production conditions are held fixed. The input level is denoted by \(L\), and total output is \(Q(L)\). The input level is positive. For the derivative results, assume \(Q\) is differentiable at the input levels considered.
Average product and marginal product answer different questions. Average product summarizes output across all \(L\) units currently used. Marginal product describes the change in output at the current input level. As a result, a high marginal product does not automatically mean that average product is high; what matters for average product is whether marginal product is above or below the existing average.
The Exact Relationship
Start from the definition \(AP(L)=Q(L)/L\). Since \(Q(L)=L\cdot AP(L)\), differentiate both sides with respect to \(L\). The product rule gives the central identity.
Rearranging shows directly how the gap between marginal and average product determines the slope of average product.
Because \(L>0\), the denominator does not change the sign of the right-hand side. Thus, when marginal product is above average product, average product is rising; when marginal product is below average product, average product is falling. When the two are equal, average product has zero derivative.
From \(AP(L)=Q(L)/L\), it follows that \(Q(L)=L\cdot AP(L)\) for \(L>0\).
The product rule gives \(Q'(L)=AP(L)+L\,dAP(L)/dL\). By definition, \(Q'(L)=MP(L)\).
Subtract \(AP(L)\) and divide by \(L>0\). The resulting identity says the sign of the average-product slope is the sign of \(MP(L)-AP(L)\).
At an interior local maximum of a differentiable average-product function, the first-order condition from calculus is \(dAP(L)/dL=0\). Substituting that condition into the identity gives \(MP(L)=AP(L)\). This is why the marginal-product and average-product curves meet at a smooth interior maximum of average product.
The condition is interior: the input level must lie inside the range being considered, rather than at an endpoint. It also relies on differentiability. If the best average product occurs at a boundary, or if average product has a corner where its derivative does not exist, the derivative condition need not apply. Even at an interior differentiable point, equality only establishes a stationary point. To confirm a maximum, check that average product rises before the point and falls after it, or use an appropriate second-derivative test.
Worked Examples
Worked Example: Find and Verify the Average-Product Maximum
A workshop’s short-run production function, with other conditions fixed, is \(Q(L)=8L+6L^2-L^3\) for \(0<L\leq 6\). Find the input level that maximizes average product, then verify the marginal-product relationship there. Use the derivative method, and check that the stationary point is a maximum.
Step 1: Calculate average product. Divide total output by the input:
Step 2: Find the stationary point. Differentiate average product and set its derivative equal to zero:
Step 3: Verify it is a maximum. The second derivative is \(d^2AP(L)/dL^2=-2<0\). Average product is therefore strictly concave on the stated range, and \(L=3\) is its maximum there.
Step 4: Compute output and average product at that input.
Step 5: Compute marginal product independently. Differentiate total output, then evaluate at \(L=3\):
The results agree: \(MP(3)=AP(3)=17\). As a further check on the interpretation, at \(L=2\), \(AP(2)=16\) and \(MP(2)=20\), so average product is rising there. At \(L=4\), \(AP(4)=16\) and \(MP(4)=8\), so average product is falling there. The method avoids assuming that equality alone establishes a maximum; the second derivative and the values on either side confirm it.
Worked Example: Read the Relationship From a Discrete Schedule
A hypothetical production schedule records output in batches for an input that must be assigned in whole units. Calculate average and marginal product at each level, identify the highest average product, and explain why the discrete schedule does not require exact equality between marginal and average product at the peak.
| Input units \(n\) | Total output \(Q_n\) |
|---|---|
| 0 | 0 |
| 1 | 12 |
| 2 | 30 |
| 3 | 51 |
| 4 | 64 |
| 5 | 70 |
Step 1: Calculate average product. For a positive whole-number input, divide total output by the number of units:
The greatest listed average product is 17 batches per input unit, at \(n=3\).
Step 2: Calculate discrete marginal product. For the move from \(n-1\) units to \(n\) units, define the additional output as \(MP_n=Q_n-Q_{n-1}\):
The marginal product of the third unit is 21, greater than the preceding average product of 15, and the average rises from 15 to 17. The fourth unit adds 13, less than the current average of 17, and average product falls from 17 to 16.
Step 3: Check the discrete identity. Since \(Q_n=Q_{n-1}+MP_n\), dividing by \(n\) and using \(Q_{n-1}=(n-1)AP_{n-1}\) gives:
For the move to three units, the left side is \(17-15=2\), and the right side is \((21-15)/3=2\). For the move to four units, the left side is \(16-17=-1\), and the right side is \((13-17)/4=-1\). The checks agree. The method avoids treating a discrete input as if it were a continuously adjustable variable: the observed maximum is at an integer input, and no equality at that integer is required by this finite-difference calculation.
Worked Example: Confirm Equality With a Nonlinear Production Function
For a hypothetical production process, let \(Q(L)=5L+12L^{3/2}-3L^2\) for \(L>0\). Find the input level where average product is maximized, calculate total, average, and marginal product there, and verify the maximum. This example uses the derivative method and avoids differentiating total output when the target is the maximum of average product.
Step 1: Form average product.
Step 2: Find and classify the stationary point.
The second derivative is \(-3L^{-3/2}\), which is negative for every \(L>0\). Thus average product is strictly concave, and this stationary point is a maximum.
Step 3: Evaluate total and average product.
Step 4: Evaluate marginal product.
Therefore \(MP(4)=AP(4)=17\), as the theorem predicts. The numerical check also confirms the units: both products are 17 units of output per unit of input.
Common Mistakes and Exam Traps
A frequent error is to calculate average product at an input level by dividing marginal product by the input. In the first worked example, this incorrect calculation would give \(MP(3)/3=17/3\), rather than the correct \(AP(3)=Q(3)/3=51/3=17\). Average product uses total output in its numerator; marginal product uses the derivative of total output.
Another trap is to treat \(MP=AP\) as proof of a maximum. It proves that the derivative of average product is zero at a positive input, but a stationary point might be a minimum. For instance, if \(Q(L)=10L+L(L-2)^2\), then \(AP(L)=10+(L-2)^2\). At \(L=2\), both average and marginal product are 10, but \(AP\) has a minimum there because its second derivative is positive. Check the direction of change or curvature before calling a stationary point a maximum.
- Do not confuse the two derivatives. Marginal product is \(Q'(L)\); the slope of average product is \(dAP(L)/dL\). They are connected by the identity, but they are not the same quantity.
- Do not omit the positive-input condition. Average product \(Q(L)/L\) is undefined at \(L=0\), and the rearranged identity divides by \(L\).
- Do not force a continuous condition onto a discrete schedule. With whole input units, compare each added unit’s output with the average before that addition; equality at the integer maximum is not guaranteed.
Check Your Understanding
Work through each problem by distinguishing total, average, and marginal product.
- A production function is \(Q(L)=6L+4L^2-L^3\). Calculate average and marginal product at \(L=2\), and use the second derivative of average product to classify the stationary point there. (Answer: \(AP(2)=10\), \(MP(2)=10\), and \(L=2\) is a maximum because \(AP''(L)=-2\).)
- For \(Q(L)=9L+8L^2-2L^3\), find the input that maximizes average product, calculate the maximum average product, and verify marginal product at that input. (Answer: \(L=2\), maximum \(AP=17\), and \(MP(2)=17\).)
- A discrete schedule has output \(Q_0=0,Q_1=14,Q_2=34,Q_3=57,Q_4=72\). Calculate average product at inputs 1 through 4 and the marginal product of the fourth unit. Identify the highest listed average product and state whether the fourth unit raises it. (Answer: average products are \(14,17,19,18\); the fourth unit has \(MP_4=15\), so the maximum listed average product is 19 at \(n=3\), and the fourth unit lowers it.)
- Let \(Q(L)=7L+12L^{3/2}-2L^2\). Find the input that maximizes average product and verify the equality with marginal product at that input. (Answer: \(L=9\); \(AP(9)=25\), \(MP(9)=25\), and \(AP''(L)=-3L^{-3/2}<0\), so it is a maximum.)
- Suppose \(Q(L)=10L+L(L-2)^2\). At \(L=2\), calculate average and marginal product and decide whether their equality identifies a maximum. (Answer: \(AP(2)=MP(2)=10\), but this is a minimum because \(AP(L)=10+(L-2)^2\) has \(AP''(L)=2>0\).)