Quadratic Utility and Diminishing Marginal Utility

Why Quadratic Utility Is Useful

A quadratic utility function is a simple way to represent a good whose additional units contribute progressively less utility. Its marginal utility is a straight line: the first units may have a large marginal contribution, while later units have a smaller one. Because that line can eventually reach zero and become negative, it is important to check the quantity range over which the function represents preferences that favor more of the good.

This tutorial treats utility as a function of a feasible quantity, or of quantities of two goods. It focuses on the function’s marginal utilities, second derivatives, and monotonicity on the stated domain. These calculations are aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.

One-Good Quadratic Utility

Definition: A one-good quadratic utility function can be written as \(U(x)=\alpha+\beta x-\gamma x^2\), where \(x\) is the quantity consumed and \(\alpha,\beta,\gamma\) are constants. When \(\gamma>0\), the function is strictly concave and its marginal utility decreases as \(x\) increases. The feasible range of \(x\) must be stated when assessing whether the function is increasing.

The constant \(\alpha\) shifts utility levels but does not affect marginal utility. Differentiating with respect to \(x\) gives the marginal utility, and differentiating again gives its rate of change:

U'(x)=\beta-2\gamma x,\qquad U''(x)=-2\gamma.

Thus, if \(\gamma>0\), then \(U''(x)<0\) at every quantity. This is diminishing marginal utility: the marginal utility itself falls as quantity rises. It does not mean that marginal utility must be positive. At sufficiently high quantities, the marginal utility may be negative, so utility falls when quantity increases.

If \(\beta>0\) and \(\gamma>0\), marginal utility equals zero at \(x^*=\beta/(2\gamma)\). It is positive for \(x<x^*\), zero at \(x=x^*\), and negative for \(x>x^*\). This is the quantity at which the quadratic function reaches its maximum. If the feasible domain extends past \(x^*\), the function is not increasing throughout that domain.

Why the Marginal-Utility Formula Works

The derivative describes the change in utility per small change in quantity. For a quadratic function, the utility change from increasing quantity by a finite amount \(h\) can also be calculated exactly. Expanding the function gives:

\begin{aligned} U(x+h)-U(x) &=\alpha+\beta(x+h)-\gamma(x+h)^2-(\alpha+\beta x-\gamma x^2)\\ &=h(\beta-2\gamma x)-\gamma h^2\\ &=hU'(x)-\gamma h^2. \end{aligned}

For a small positive \(h\), the first term is the current marginal utility multiplied by the change in quantity, while the negative second term accounts for the fact that marginal utility declines during the change. The same expression shows why marginal utility at the starting quantity alone is not generally the exact utility change over a finite interval.

Marginal Utility and Strict Increase on a Feasible Interval

Suppose the feasible quantity is \(0\leq x\leq K\), with \(K>0\), and \(\gamma>0\). Since \(U'(x)=\beta-2\gamma x\) falls as \(x\) rises, its smallest value on this interval occurs at \(K\). Therefore, \(U'(x)>0\) everywhere on the interval exactly when \(\beta-2\gamma K>0\).

That condition is stronger than strict increase of the function. If \(\beta-2\gamma K=0\), marginal utility is positive for every \(x<K\) and zero only at \(K\); the function is still strictly increasing on the whole closed interval. In fact, for this quadratic with \(\gamma>0\), it is strictly increasing on \([0,K]\) whenever \(\beta-2\gamma K\geq 0\). If \(\beta-2\gamma K<0\), marginal utility is negative near \(K\), so the function decreases over part of the interval and is not strictly increasing throughout it.

Key distinction: \(U'(x)>0\) at every point is a sufficient condition for strict increase, but it is not necessary. On \([0,K]\), the quadratic can be strictly increasing even when \(U'(K)=0\), provided \(U'(x)>0\) before that endpoint.

Worked Examples

Worked Example: Finding the Diminishing-Marginal-Utility Region

A planner uses the hypothetical utility index \(U(q)=18+12q-1.5q^2\) for a quantity \(q\) measured in units. Calculate marginal utility and its second derivative, find where the function is increasing, and evaluate marginal utility at \(q=2\) and \(q=3\).

Method. Differentiate once to find marginal utility and again to assess whether it diminishes:

U'(q)=12-3q,\qquad U''(q)=-3.

Since \(U''(q)=-3<0\), marginal utility decreases at every quantity. To find where the utility function is increasing, solve \(U'(q)>0\):

12-3q>0\quad\Longleftrightarrow\quad q<4.

Marginal utility reaches zero at \(q=4\), and is negative for \(q>4\). At \(q=2\), \(U'(2)=12-3(2)=6\). At \(q=3\), \(U'(3)=12-3(3)=3\). The marginal utility has fallen from 6 to 3 as quantity rises from 2 to 3, consistent with \(U''(q)=-3\).

Check. The zero of the marginal-utility formula is \(q=4\), and \(U'(q)\) is positive below 4 and negative above 4. Common mistake avoided: diminishing marginal utility does not imply that utility is increasing at every quantity; here utility decreases after \(q=4\).

Worked Example: Strict Increase When Marginal Utility Is Zero at the Endpoint

A model assigns utility \(U(x)=7+10x-2x^2\) on the feasible interval \(0\leq x\leq 2.5\). Find the marginal utility, determine whether utility is strictly increasing over the entire interval, and check whether it remains increasing if the interval is extended to \(x=3\).

Method. Differentiate and evaluate at the upper endpoint:

U'(x)=10-4x,\qquad U''(x)=-4,\qquad U'(2.5)=10-4(2.5)=0.

For every \(x<2.5\), \(10-4x>0\). Therefore utility is strictly increasing up to the endpoint. The zero derivative at \(x=2.5\) does not prevent strict increase on the closed interval.

The finite-change expression confirms this. For any \(0\leq x_1<x_2\leq 2.5\), the exact difference is:

\begin{aligned} U(x_2)-U(x_1) &=(x_2-x_1)\bigl(10-2(x_1+x_2)\bigr). \end{aligned}

Because \(x_1<2.5\) and \(x_2\leq 2.5\), \(x_1+x_2<5\). Hence \(10-2(x_1+x_2)>0\), while \(x_2-x_1>0\). The utility difference is positive, proving strict increase for any two distinct quantities in the interval.

If the interval is extended to \(x=3\), the function is not increasing throughout: \(U(2.5)=7+25-12.5=19.5\), whereas \(U(3)=7+30-18=19\). Check. Utility rises to \(x=2.5\) and then falls. Common mistake avoided: requiring \(U'(x)>0\) at the endpoint would incorrectly reject strict increase on the original closed interval.

Worked Example: Separable Quadratic Utility in Two Goods

Consider the hypothetical utility function \(U(x,y)=5+8x-x^2+6y-1.5y^2\), with feasible quantities \(0\leq x\leq4\) and \(0\leq y\leq2\). Find each marginal utility and second own partial derivative. Then calculate both marginal utilities at \((2,1)\) and the utility change when both quantities rise by \(0.5\).

Method. Differentiate with respect to one good while holding the other fixed:

U_x=8-2x,\qquad U_{xx}=-2,\qquad U_y=6-3y,\qquad U_{yy}=-3.

Both own second partial derivatives are negative, so marginal utility diminishes in each good as its own quantity increases. On the stated rectangle, \(U_x\geq0\) and \(U_y\geq0\); each is zero only at its respective upper endpoint. Thus increasing either good alone does not reduce utility anywhere on its feasible range.

At \((2,1)\), the marginal utilities are \(U_x(2,1)=8-2(2)=4\) and \(U_y(2,1)=6-3(1)=3\). For an increase of \(0.5\) in \(x\), the exact utility change is \(0.5(4)-(0.5)^2=2-0.25=1.75\). For an increase of \(0.5\) in \(y\), it is \(0.5(3)-1.5(0.5)^2=1.5-0.375=1.125\). The total change is \(1.75+1.125=2.875\), since the function is additively separable.

Check. Direct evaluation gives \(U(2,1)=5+16-4+6-1.5=21.5\) and \(U(2.5,1.5)=5+20-6.25+9-3.375=24.375\). The difference is \(24.375-21.5=2.875\), as calculated. Common mistake avoided: multiplying each starting marginal utility by the full change gives only an approximation; the quadratic terms must be included for an exact finite change.

Interpreting the Model and Avoiding Common Traps

A quadratic with a negative squared term is useful because it makes diminishing marginal utility especially transparent: the marginal utility is linear, and its slope is the constant negative second derivative. But the same feature means that marginal utility eventually becomes negative if quantity is allowed to grow without bound. A quadratic of this form therefore cannot represent a preference for more of the good over every nonnegative quantity. Check the feasible domain before interpreting the function as increasing.

Do not confuse diminishing with positive. \(U''(x)<0\) means marginal utility decreases; it does not guarantee \(U'(x)>0\).
Check the full feasible interval. A positive marginal utility at one quantity does not show that utility increases throughout a larger domain.
Handle endpoint equality correctly. On \([0,K]\), \(U'(K)=0\) can coexist with strict increase throughout the interval if marginal utility is positive before \(K\).
Use the correct derivative. \(U'(x)\) is marginal utility; \(U''(x)\) describes how marginal utility changes.
Do not treat a finite change as exactly marginal utility times quantity change. For a quadratic, the additional squared-change term is part of the exact calculation.

Key takeaway: For \(U(x)=\alpha+\beta x-\gamma x^2\) with \(\gamma>0\), marginal utility is \(\beta-2\gamma x\) and declines at the constant rate \(2\gamma\). On a bounded feasible interval, check its endpoint value to determine whether utility remains increasing; equality to zero at the upper endpoint still permits strict increase on the interval.

Check Your Understanding

For each problem, calculate the requested quantities and check the stated feasible range.

Let \(U(x)=4+9x-1.5x^2\). Find \(U'(x)\), \(U''(x)\), and \(U'(2)\). State whether marginal utility diminishes. (Answer: \(U'(x)=9-3x\), \(U''(x)=-3\), \(U'(2)=3\); yes, marginal utility diminishes.)
Let \(U(x)=2+14x-2x^2\) on \([0,3.5]\). Is it strictly increasing on the whole interval, and is marginal utility positive at the upper endpoint? (Answer: Yes, it is strictly increasing; \(U'(3.5)=0\), while \(U'(x)>0\) for \(x<3.5\).)
Let \(U(x)=10+6x-0.5x^2\) on \([0,7]\). Find the quantity where marginal utility is zero, and identify where the function increases and decreases within the feasible interval. (Answer: \(U'(x)=6-x\), so marginal utility is zero at \(x=6\); utility increases for \(0\leq x<6\), reaches its maximum at 6, and decreases for \(6<x\leq7\).)
For \(U(x,y)=3+10x-2x^2+8y-y^2\), find both marginal utilities and own second partial derivatives. Evaluate the marginal utilities at \((1,2)\), and determine whether each good alone raises utility over the rectangle \(0\leq x\leq2.5,\ 0\leq y\leq4\). (Answer: \(U_x=10-4x,\ U_{xx}=-4,\ U_y=8-2y,\ U_{yy}=-2\); at \((1,2)\) the marginal utilities are 6 and 4; each good alone raises utility throughout its interval, with marginal utility zero only at its upper endpoint.)
Let \(U(x)=12+8x-x^2\) on \([0,4]\). Calculate the exact utility change from \(x=2\) to \(x=3\), and decide whether the function is strictly increasing on the whole feasible interval. (Answer: \(U(3)-U(2)=27-24=3\); it is strictly increasing on \([0,4]\), with \(U'(4)=0\) and \(U'(x)>0\) for \(x<4\).)

What You'll Learn