What Diminishing Marginal Utility Means
When a consumer has more of a good, an additional unit may add less utility than the unit before it. This is the idea of diminishing marginal utility. The comparison concerns changes in one good at a time: to assess the marginal utility of good \(x\), hold the quantities of other goods fixed. Without that condition, a change in utility could reflect changes in several goods and would not isolate the marginal utility of \(x\).
In the earlier material on utility functions, marginal utility was defined as a partial derivative. For a differentiable utility function \(U(x,y)\), the marginal utility of \(x\) is \(MU_x=U_x(x,y)\), the change in utility associated locally with a small increase in \(x\), with \(y\) held fixed. Diminishing marginal utility means this marginal utility decreases as \(x\) increases over the range being considered.
The condition is local to the specified range and fixed quantity of the other good. It does not, by itself, say that marginal utility is positive: a declining marginal utility can eventually reach zero or become negative. Nor does diminishing marginal utility for \(x\) alone describe how a consumer trades \(x\) against \(y\). These distinctions matter when interpreting a formula or table.
Checking the Condition With Derivatives
The second derivative provides a direct test when \(U\) is twice differentiable. Since \(MU_x=U_x\), the rate at which marginal utility changes as \(x\) changes is the derivative of \(U_x\) with respect to \(x\), namely \(U_{xx}\). A negative value means that \(MU_x\) is falling locally. If it is negative throughout an interval, the mean value theorem implies that \(MU_x\) is strictly lower at any larger quantity in that interval.
This test assumes the relevant derivatives exist on the range under study. A negative second derivative at just one point establishes a local direction of change there; it does not establish diminishing marginal utility over a larger interval. For a utility table, derivatives may not be available, so compare the utility gained per unit over successive changes in \(x\), again holding other goods fixed.
Worked Example: Reading Diminishing Marginal Utility From a Table
A hypothetical consumer’s utility at a fixed quantity \(y=4\) is recorded below. Assess whether the table supports diminishing marginal utility of \(x\) over the listed quantities. The entries are utility units, and each step in \(x\) is one unit.
| \(x\) | \(U(x,4)\) |
|---|---|
| 0 | 0 |
| 1 | 12 |
| 2 | 21 |
| 3 | 27 |
Method. Calculate the utility gain for each one-unit increase. Because the step size is one, each gain is also the average utility gain per unit over that step.
From \(x=0\) to \(x=1\), the gain is \(12-0=12\). From \(x=1\) to \(x=2\), the gain is \(21-12=9\). From \(x=2\) to \(x=3\), the gain is \(27-21=6\). Thus the successive one-unit gains are \(12,9,6\), which decline.
Conclusion and check. The table supports diminishing marginal utility over these observed one-unit changes, with \(y=4\) fixed. The arithmetic checks: the three gains sum to \(12+9+6=27\), equal to \(U(3,4)-U(0,4)=27-0\). This finite table supports a conclusion about the listed steps; it does not prove that marginal utility declines at every unobserved quantity between them.
Common mistake avoided. Comparing utility levels alone is not the test. Utility rises from 0 to 27 here; diminishing marginal utility is shown by the smaller successive gains, not by a decline in total utility.
Why a Negative Second Derivative Works
The derivative test follows directly from how marginal utility changes. Fix \(y\), and suppose \(U\) is twice differentiable in \(x\) on an interval. If \(U_{xx}<0\) throughout the interval, then \(U_x\) has a negative derivative there. Therefore, for any two quantities \(x_2>x_1\) in that interval, \(U_x(x_2,y)<U_x(x_1,y)\). This is precisely a declining marginal utility of \(x\).
The same conclusion can be expressed in terms of total utility changes. By the fundamental theorem of calculus, the gain from increasing \(x\) from \(a\) to \(b\), while holding \(y\) fixed, is the integral of marginal utility over that range. If marginal utility declines throughout the range, equal-sized successive intervals have lower average marginal utility, and thus smaller utility gains.
This equality explains why derivative and table methods are related, but not identical. A derivative describes a local rate of change. A table gives total changes over finite intervals; dividing each change by the interval width gives an average rate over that interval. With equal interval widths, smaller successive gains correspond to smaller average rates. For unequal widths, compare gains per unit rather than raw gains.
Worked Example: Testing Quadratic Utility and Computing a Finite Gain
Suppose a hypothetical consumer’s utility is \(U(x,y)=24x-3x^2+2xy\), for nonnegative quantities \(x\) and \(y\). At the fixed quantity \(y=2\), determine whether marginal utility of \(x\) diminishes, calculate marginal utility at \(x=1\) and \(x=2\), and find the exact utility gain from increasing \(x\) from 1 to 2.
Method. Differentiate with respect to \(x\), treating \(y\) as fixed. Then use the second derivative for the diminishing-marginal-utility test and evaluate the utility function at both endpoints for the finite gain.
The first partial derivative is \(MU_x=U_x=24-6x+2y\). The second partial derivative is \(U_{xx}=-6\). Since \(-6<0\) for every \(x\) and \(y\), marginal utility of \(x\) strictly declines as \(x\) increases, including when \(y=2\).
At \(y=2\), \(MU_x=24-6x+4=28-6x\). Thus \(MU_x(1,2)=28-6=22\), and \(MU_x(2,2)=28-12=16\). The marginal utility is lower at \(x=2\), as the second-derivative test predicts.
Now evaluate total utility at each endpoint: \(U(2,2)=24(2)-3(2^2)+2(2)(2)=48-12+8=44\). Also, \(U(1,2)=24(1)-3(1^2)+2(1)(2)=24-3+4=25\). The utility gain is \(44-25=19\) utility units. As a check, the gain is between 1 and 2 units of \(x\), so its average marginal utility is 19, between the endpoint marginal utilities 22 and 16.
Conclusion and common mistake avoided. The second derivative verifies diminishing marginal utility, and the exact finite gain is 19. Do not omit the cross term \(2xy\): at \((2,2)\) it contributes \(2(2)(2)=8\), making utility 44, not 40.
Interpreting the Test Carefully
Diminishing marginal utility is stated for a particular utility representation and a particular direction of change. In an ordinal model, a strictly increasing transformation of utility preserves preference rankings, but it need not preserve the curvature of the utility function. For a twice-differentiable transformation \(\widetilde U=f(U)\), the chain rule gives
Even when \(f'(U)>0\), the first term can affect the sign of \(\widetilde U_{xx}\). Thus diminishing marginal utility is a property of the specified utility function, not automatically a representation-independent fact about preferences. When a problem supplies a utility function, assess the stated function; do not silently replace it with another representation and assume its marginal-utility curvature is unchanged.
Also keep the other good fixed. If both \(x\) and \(y\) change, a change in \(U_x\) may come from the dependence of marginal utility on \(y\), as well as from the change in \(x\). For example, with \(U(x,y)=24x-3x^2+2xy\), marginal utility is \(24-6x+2y\). Changing \(y\) changes \(MU_x\), but the diminishing-in-\(x\) test is still \(U_{xx}=-6\), evaluated while \(y\) is held fixed.
Worked Example: Comparing Average Gains Over Unequal Intervals
A hypothetical utility function is \(U(x,y)=30\ln(x+1)+5y\), defined for \(x\geq 0\) and \(y\geq 0\). Hold \(y=2\) fixed. Find the marginal utility function, verify whether it diminishes, and compare the average utility gain per unit over \(x=0\) to \(x=1\) with that over \(x=1\) to \(x=3\). Report final decimal values to three places.
Method. Differentiate for marginal utility and its rate of change. For the finite intervals, compute the utility gain and divide by the interval width, since the two intervals are not the same size.
Differentiating with respect to \(x\) gives \(MU_x=30/(x+1)\). Differentiating again gives \(U_{xx}=-30/(x+1)^2\), which is negative for every \(x\geq0\). Therefore marginal utility strictly diminishes throughout the specified domain.
At \(y=2\), utility is \(U(x,2)=30\ln(x+1)+10\). From \(x=0\) to \(x=1\), the gain is \(U(1,2)-U(0,2)=30\ln 2\approx20.794\). The interval has width 1, so the average gain per unit is \(20.794/1=20.794\).
From \(x=1\) to \(x=3\), the gain is \(U(3,2)-U(1,2)=30(\ln4-\ln2)=30\ln2\approx20.794\). This interval has width 2, so the average gain per unit is \(20.794/2\approx10.397\). The average gain per unit is smaller on the later interval.
Check and common mistake avoided. The gains themselves are equal, but their interval widths differ. Comparing raw gains would miss the decline in average gain per unit; dividing by the respective widths gives 20.794 and 10.397. Values are rounded only at the final step, to three decimal places.
Common Errors and a Reliable Routine
The main exam challenge is often not differentiation but identifying exactly what is being compared. A disciplined method keeps the quantity, direction, and interval clear.
State which quantities besides \(x\) are held constant. This isolates a change in the marginal utility of \(x\).
For a differentiable function, calculate \(U_x\) and \(U_{xx}\). For a table, calculate successive utility gains and divide by the quantity changes when interval widths differ.
A negative second derivative must hold throughout the interval to establish strict decline throughout it. A finite table supports a conclusion only about its listed changes.
- Confusing total utility with marginal utility. Total utility can rise even while marginal utility falls. Compare derivatives or successive per-unit gains.
- Changing more than one good. A bundle-to-bundle utility difference does not isolate \(MU_x\) if \(y\) changes at the same time.
- Using the first derivative as the diminishing test. \(U_x>0\) says utility is increasing in \(x\) locally; the sign of \(U_{xx}\) determines whether \(MU_x\) is falling.
- Comparing unequal intervals by raw gains. Divide each utility change by its quantity change before comparing average gains per unit.
Check Your Understanding
For each problem, keep the specified quantities fixed and distinguish marginal utility from total utility change.
- At fixed \(y=1\), a utility table gives \(U(0,1)=0\), \(U(1,1)=9\), \(U(2,1)=16\), and \(U(3,1)=21\). Calculate the successive one-unit gains and state whether they diminish. (Answer: Gains are 9, 7, and 5 utility units; yes, they diminish.)
- Let \(U(x,y)=15x-2x^2+3y\). Find \(MU_x\), \(U_{xx}\), and the utility gain from \(x=1\) to \(x=2\) at \(y=1\). (Answer: \(MU_x=15-4x\), \(U_{xx}=-4\), and the gain is \(U(2,1)-U(1,1)=25-16=9\) utility units.)
- Let \(U(x,y)=8x-x^2+2xy\), with \(y=3\). Find \(MU_x\) at \(x=2\) and \(x=3\), calculate the utility gain from \(x=2\) to \(x=3\), and determine whether marginal utility diminishes. (Answer: \(MU_x=14-2x\), giving 10 at \(x=2\) and 8 at \(x=3\); the gain is \(U(3,3)-U(2,3)=33-24=9\); yes, since \(U_{xx}=-2\).)
- A utility table at fixed \(y\) records \(U(0,y)=0\), \(U(2,y)=14\), and \(U(5,y)=29\). Compare the average utility gain per unit on the two intervals. Does the table support declining average gains? (Answer: The first average is \(14/2=7\) per unit; the second is \((29-14)/(5-2)=5\) per unit; yes, the average gain per unit declines.)
- For \(x>0\), compare \(U(x)=x\) with the increasing transformation \(\widetilde U(x)=\sqrt{x}\). State the marginal utility in each representation and whether each has diminishing marginal utility. (Answer: \(U'(x)=1\), so marginal utility is constant; \(\widetilde U'(x)=1/(2\sqrt{x})\) and \(\widetilde U''(x)=-1/(4x^{3/2})<0\), so the transformed representation has diminishing marginal utility.)