Solving Consumer Choice with Piecewise Utility

Why Piecewise Utility Needs a Careful Search

A consumer's utility formula may change when consumption crosses a threshold. The threshold might represent a different valuation of additional units, a service tier, or a change in how two goods are combined. Within each region, familiar optimization tools may apply, but a regional optimum is not automatically the best feasible choice overall. The consumer could do better at the threshold itself or at an endpoint of the budget line.

The reliable approach is to divide the feasible choices into the regions where each formula applies, optimize within each region, and then compare the utility of every feasible candidate. This tutorial develops that procedure for two goods. It is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.

Definition: A piecewise utility function assigns utility using different formulas in different parts of the consumption set. A point where the applicable formula changes is a kink if the utility function is continuous there but its slope may change. A kink need not maximize utility; it must be checked as a candidate.

Set Up the Feasible Choice

Let $x$ and $y$ denote quantities of two goods, with positive prices $p_x$ and $p_y$, and income $m$. The consumer's budget set is

p_xx+p_yy\leq m,\qquad x\geq 0,\quad y\geq 0.

If utility is strictly increasing in $y$ throughout the feasible set, any bundle that leaves income unspent can be improved by buying more $y$. The budget therefore binds, and $y=(m-p_xx)/p_y$. Feasible values of $x$ run from $0$ to $m/p_x$. Substitution turns the problem into maximizing utility along the budget line.

For a piecewise utility function, this substitution also turns each region into a one-variable optimization problem. If the formula changes at $x=k$, for example, the budget line can be divided into the portions $x\leq k$ and $x\geq k$, provided $k$ is feasible. Optimize on each portion, and include $x=k$ among the candidates. The feasible endpoints $x=0$ and $x=m/p_x$ must also be considered.

First-order condition: For a differentiable utility function and a smooth interior optimum on a binding budget line, differentiating utility after substituting $y=(m-p_xx)/p_y$ gives $$ U_x-\frac{p_x}{p_y}U_y=0, $$ or $U_x=(p_x/p_y)U_y$. This condition alone does not establish a maximum; candidates must still be feasible and compared. The ratio form $U_x/U_y=p_x/p_y$ is valid only when $U_y\ne 0$.

Why the Regional First-Order Condition Holds

Within one smooth region, define $V(x)=U(x,(m-p_xx)/p_y)$, the utility attainable at quantity $x$ when the budget binds. Increasing $x$ by a small amount requires reducing $y$ to pay for it. In particular, $dy/dx=-p_x/p_y$. By the chain rule,

V'(x)=U_x+U_y\frac{dy}{dx} =U_x-\frac{p_x}{p_y}U_y.

At a differentiable interior maximum within that region, $V'(x)=0$, which gives the stated first-order condition. This derivation applies only where the utility formula is differentiable and the choice is interior to the region and budget interval. At a kink, the derivative may not exist, so evaluate the utility there directly. At a budget endpoint, there may be no feasible movement in both directions, so the interior condition need not hold.

A Candidate-by-Candidate Procedure

Write the feasible budget line.
If utility increases with $y$, substitute $y=(m-p_xx)/p_y$ and restrict $x$ to $0\leq x\leq m/p_x$.

Mark each feasible region.
Record the formula that applies on each interval and every threshold that lies within the feasible range.

Optimize within each region.
Find stationary points that lie inside the region. Also retain the region's endpoints; a stationary point outside its interval is not a candidate there.

Evaluate and compare.
Calculate utility at each feasible stationary point, kink, and budget endpoint. The candidate with the greatest utility is a global optimum if the candidate list covers the whole feasible interval.

This procedure does not require utility to be concave. A stationary point might be a minimum, and regional optima can have different utility levels. Checking and comparing the candidates is what identifies the global maximum. In the examples below, utility increases with $y$, so the budget binds. If that monotonicity condition fails in another problem, do not assume the budget binds; candidates away from the budget line may need to be considered.

Worked Example: A Kink Is the Best Bundle

A consumer has income $m=14$, faces prices $p_x=2$ and $p_y=1$, and has utility $U(x,y)=f(x)+y$, where

f(x)= \begin{cases} 6x-\frac{1}{2}x^2, & 0\leq x\leq 4,\\ 16+2(x-4)-\frac{1}{4}(x-4)^2, & x\geq 4. \end{cases}

Find the utility-maximizing bundle. The method is to substitute the binding budget, optimize on each side of the kink, and compare the kink and budget endpoints with any regional stationary points.

Step 1: Establish feasibility. Utility increases with $y$, since its marginal utility is $1$. Thus the budget binds: $y=14-2x$, and $0\leq x\leq 7$. The kink at $x=4$ is feasible.

Step 2: Optimize on $0\leq x\leq 4$. On this interval,

U(x,14-2x)=6x-\frac{1}{2}x^2+14-2x =14+4x-\frac{1}{2}x^2.

Its derivative is $4-x$, which is positive for $0\leq x<4$. Utility rises across this region, so its largest value on the interval occurs at $x=4$, not at an interior stationary point.

Step 3: Optimize on $4\leq x\leq 7$. Write $t=x-4$, so $0\leq t\leq 3$. After substitution,

U=16+2t-\frac{1}{4}t^2+14-2(4+t) =22-\frac{1}{4}t^2.

For $t>0$, this is less than $22$, so the largest value on this region is at $t=0$, the kink $x=4$. Equivalently, the derivative with respect to $t$ is $-t/2$, which is negative for $t>0$.

Step 4: Check the candidates and conclude. At $x=0$, $y=14$, and $U=14$. At $x=4$, $y=6$, and $U=f(4)+6=16+6=22$. At $x=7$, $y=0$, and $f(7)=16+2(3)-\frac14(3^2)=16+6-2.25=19.75$. The kink bundle $(4,6)$ gives the greatest utility, $22$, so it is optimal. The common mistake this method avoids is selecting a smooth stationary point while failing to check a kink where the best attainable utility can occur.

Worked Example: A Smooth Optimum Inside One Region

A hypothetical consumer has income $m=10$, prices $p_x=2$ and $p_y=1$, and utility $U(x,y)=f(x)+y$, with

f(x)= \begin{cases} 5x-\frac{1}{2}x^2, & 0\leq x\leq 2,\\ 8+3(x-2)-\frac{1}{2}(x-2)^2, & x\geq 2. \end{cases}

Find the maximizing bundle and verify it against the kink and both budget endpoints.

Step 1: Substitute the budget. Since utility increases with $y$, $y=10-2x$, and $0\leq x\leq 5$. The kink at $x=2$ is feasible.

Step 2: Check the first region. For $0\leq x\leq 2$,

U=5x-\frac{1}{2}x^2+10-2x =10+3x-\frac{1}{2}x^2.

The derivative $3-x$ is positive throughout this interval, so utility increases up to the kink. The best point in this region is $x=2$.

Step 3: Find the optimum in the second region. Let $t=x-2$, where $0\leq t\leq 3$. Then

U=8+3t-\frac{1}{2}t^2+10-2(2+t) =14+t-\frac{1}{2}t^2.

The derivative is $1-t$, which is zero at $t=1$. The second derivative is $-1$, so utility is strictly concave on this region and this stationary point is its regional maximum. It gives $x=3$ and $y=10-2(3)=4$. At this bundle, $f(3)=8+3-\frac12=10.5$, so $U=10.5+4=14.5$.

Step 4: Compare all candidates. At $x=0$, $y=10$, and $U=10$. At the kink $x=2$, $y=6$, and $U=8+6=14$. At the other budget endpoint $x=5$, $y=0$, and $f(5)=8+9-\frac12(9)=12.5$, giving $U=12.5$. The interior regional maximum gives $14.5$, greater than all three alternatives, so the optimum is $(3,4)$. The method avoids the trap of treating the kink as automatically optimal merely because it is where the formula changes.

Worked Example: A Regional Stationary Point Is Not the Global Optimum

A consumer has $m=12$, $p_x=2$, $p_y=1$, and $U(x,y)=f(x)+y$, where

f(x)= \begin{cases} 5x-\frac{1}{2}x^2, & 0\leq x\leq 3,\\ 10.5+(x-3)+\frac{1}{2}(x-3)^2, & x\geq 3. \end{cases}

Find the global optimum. In particular, do not assume that every regional stationary point is a maximum.

Step 1: Find the feasible interval. Utility increases with $y$, so $y=12-2x$, and $0\leq x\leq 6$. The kink at $x=3$ is feasible.

Step 2: Check the first region. For $0\leq x\leq 3$, utility along the budget is

U=5x-\frac{1}{2}x^2+12-2x =12+3x-\frac{1}{2}x^2.

Its derivative $3-x$ is positive for $0\leq x<3$. Therefore the best value in this region is at the kink.

Step 3: Check the second region. Set $t=x-3$, with $0\leq t\leq 3$. Then

U=10.5+t+\frac{1}{2}t^2+12-2(3+t) =16.5-t+\frac{1}{2}t^2.

The derivative is $-1+t$, so the stationary point is $t=1$, or $x=4$. The second derivative is $1>0$, making this point a regional minimum, not a maximum. The maximum on the closed interval $0\leq t\leq 3$ must therefore occur at an endpoint.

Step 4: Compare the full candidate set. At $x=0$, $y=12$, and $U=12$. At the kink $x=3$, $y=6$, and $U=10.5+6=16.5$. At the second-region stationary point $x=4$, $y=4$, and $f(4)=10.5+1+0.5=12$, so $U=16$. At $x=6$, $y=0$, and $f(6)=10.5+3+4.5=18$, so $U=18$. The global optimum is $(6,0)$, not the kink or the stationary point. The key mistake avoided is assuming that a derivative equal to zero identifies a maximum; the candidate still needs a second-order or other classification and a global comparison.

Common Errors and a Final Check

Piecewise optimization is a finite comparison problem only after the candidate set has been built correctly. If utility is continuous and differentiable within each region, a regional maximum on a closed interval can occur at an interior stationary point or at an endpoint of that interval. The endpoints include kinks and, where applicable, the ends of the feasible budget line. If the formula is not differentiable at a threshold, evaluate utility there directly rather than forcing a derivative condition.

Using a regional formula outside its interval. A stationary point counts only if it lies in the region where the formula used to calculate it applies.
Checking only smooth stationary points. Kinks and budget endpoints can be optimal even though the differentiable first-order condition does not apply there.
Calling every stationary point a maximum. A positive second derivative can identify a local minimum. Compare all feasible candidates.
Dividing by a marginal utility without checking it. The budget-line first-order condition is $U_x=(p_x/p_y)U_y$. The ratio form requires $U_y\ne 0$.
Assuming the budget binds automatically. It binds in the examples because utility strictly increases with $y$. Without that condition, the consumer may not benefit from spending all income.

Key takeaway: Partition the feasible set using the utility formula's regions, optimize within each region, include every kink and feasible boundary, and compare utility values. A global solution is the best feasible candidate, not merely the most convenient stationary point.

Check Your Understanding

For each problem, identify the regional candidates and compare their utility values; do not rely on a stationary point alone.

A consumer has $m=8$, $p_x=2$, $p_y=1$, and $U=f(x)+y$, where $f(x)=4x-\frac12x^2$ for $0\leq x\leq2$, and $f(x)=6+2(x-2)-\frac14(x-2)^2$ for $x\geq2$. Find the optimal bundle and utility. (Answer: $(x,y)=(2,4)$, $U=10$.)
A consumer has $m=8$, $p_x=2$, $p_y=1$, and $U=f(x)+y$, where $f(x)=4x-\frac12x^2$ for $0\leq x\leq1$, and $f(x)=3.5+3(x-1)-\frac12(x-1)^2$ for $x\geq1$. Find the optimal bundle and check the kink and budget endpoints. (Answer: $(x,y)=(2,4)$, $U=10$.)
A consumer has $m=5$, $p_x=p_y=1$, and $U=f(x)+y$, where $f(x)=4x-\frac12x^2$ for $0\leq x\leq2$, and $f(x)=6+(x-2)+\frac12(x-2)^2$ for $x\geq2$. Find the global optimum, including the second-region stationary point if there is one. (Answer: $(x,y)=(5,0)$, $U=13.5$.)
At a smooth interior optimum on a binding budget line, what first-order condition follows by differentiating utility after substituting $y=(m-p_xx)/p_y$? When may it be written as a ratio? (Answer: $U_x=(p_x/p_y)U_y$; write $U_x/U_y=p_x/p_y$ only if $U_y\ne0$.)
A consumer has $m=12$, $p_x=3$, $p_y=1$, and $U=f(x)+y$, where $f(x)=5x$ for $0\leq x\leq2$, and $f(x)=10+(x-2)$ for $x\geq2$. Find the optimal bundle and compare it with both budget endpoints. (Answer: $(x,y)=(2,6)$, $U=16$.)

What You'll Learn