Why Zero-Consumption Choices Need Their Own Test
An interior tangency is useful only when the consumer chooses positive quantities of both goods. If the best affordable bundle uses none of one good, the usual marginal-utility-per-dollar equality need not hold. A reliable solution therefore checks the zero-consumption possibilities as well as any feasible interior candidate.
This tutorial develops that comparison for a consumer choosing between two goods. The method is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus. It uses the earlier material on utility and consumer choice, but focuses on how to test the edges of the budget set without assuming in advance that the optimum is interior.
Reduce the Problem to the Budget Line
Suppose utility is strictly increasing in each good: holding the other quantity fixed, more of either good raises utility. Then a bundle that leaves some income unspent cannot be optimal, because the consumer could buy more of a good and attain higher utility. The budget binds, so the relevant line is \(p_xx+p_yy=m\). Solving for \(y\) gives
This parameterization turns the two-good problem into a one-variable problem. The endpoints of the interval are exactly the zero-consumption bundles: \(x=0\) gives \((0,m/p_y)\), and \(x=m/p_x\) gives \((m/p_x,0)\). Between them, both goods are consumed.
If utility \(U(x,y)\) is differentiable at an interior point of the budget line, define utility along that line by substituting the budget equation:
Any interior maximum of \(V\) must satisfy \(V'(x)=0\). By the chain rule,
where \(MU_x\) and \(MU_y\) are the marginal utilities evaluated at the candidate bundle. The equality is an interior condition, not an endpoint test. At an endpoint, the consumer cannot move in both directions along the budget line, so the derivative need not equal zero. Moreover, a derivative may not exist there even when utility itself is well defined.
A Candidate-Comparison Procedure
For a smooth, strictly increasing utility function on the two-good budget set, the practical procedure is to find feasible interior stationary points and compare their utility with utility at both budget intercepts. If there is no feasible interior stationary point, the intercepts still need to be compared. This procedure follows from one-variable optimization: a maximum on a closed interval can occur at an endpoint or at an interior point where the derivative is zero.
Use \(y=(m-p_xx)/p_y\) and \(0\leq x\leq m/p_x\). This identifies the feasible range and both zero-consumption bundles.
Differentiate utility along the budget line and solve \(V'(x)=0\), retaining only roots strictly inside the interval. Check that the utility and derivative conditions used are valid there.
Calculate utility at \((0,m/p_y)\) and \((m/p_x,0)\). Do not substitute an endpoint into an interior first-order condition as if it were an interior point.
Choose the candidate with the highest utility. If utility is not differentiable at an interior point, or if there are kinks or other irregularities, include those candidates in the comparison rather than relying only on \(V'(x)=0\).
The procedure identifies candidates; it does not justify ignoring candidates or assuming that every stationary point is a maximum. For example, if there are several feasible stationary points, each must be evaluated. In the worked examples below, the derivative or its sign is also checked so the comparison is complete for the stated utility function.
Worked Example: An Interior Candidate Beats Both Zero-Consumption Bundles
A consumer has utility \(U(x,y)=2\sqrt{x}+\sqrt{y}\), income \(m=12\), and prices \(p_x=2\), \(p_y=1\). Find the maximizing bundle. Quantities must be nonnegative, and utility is strictly increasing in both goods, so the budget binds.
Method and budget line. Parameterize the budget by \(y=12-2x\), where \(0\leq x\leq6\). Utility along the line is \(V(x)=2\sqrt{x}+\sqrt{12-2x}\).
Interior candidate. For \(0<x<6\),
Setting the derivative equal to zero gives \(\sqrt{x}=\sqrt{12-2x}\), so \(x=12-2x\), \(3x=12\), and \(x=4\). The corresponding quantity is \(y=12-2(4)=4\). The candidate is feasible and interior.
Endpoint comparison. At \(x=0\), the bundle is \((0,12)\), with utility \(2\sqrt{0}+\sqrt{12}=2\sqrt{3}\), approximately \(3.464\). At \(x=6\), the bundle is \((6,0)\), with utility \(2\sqrt{6}+\sqrt{0}\), approximately \(4.899\). At the interior candidate, utility is \(2\sqrt{4}+\sqrt{4}=4+2=6\).
Check. The derivative is positive when \(x<4\) and negative when \(x>4\), since \(12-2x>x\) is equivalent to \(x<4\). Thus utility rises up to \(x=4\) and falls after it. The maximizing bundle is \((4,4)\), and its utility \(6\) exceeds both endpoint utilities. The method avoids treating the tangency as sufficient without checking the endpoints.
Worked Example: A Zero-Consumption Bundle Is Best
A consumer has utility \(U(x,y)=\ln(1+x)+2\ln(1+y)\), income \(m=10\), and prices \(p_x=10\), \(p_y=1\). Find the maximizing bundle. Both goods have positive marginal utility, so spending the full budget is optimal.
Method and budget line. Write \(y=10-10x\), with \(0\leq x\leq1\). Then \(V(x)=\ln(1+x)+2\ln(11-10x)\). Its derivative is
At \(x=0\), the derivative from the right is \(1-20/11=-9/11<0\). Further, throughout the interval,
Thus \(V'(x)\) is decreasing. Since it is already negative at \(x=0\), it remains negative for \(0\leq x<1\). Utility falls as \(x\) increases, so there is no feasible interior stationary point and the best bundle is the left endpoint \((0,10)\).
Check against the other endpoint. Utility at \((0,10)\) is \(\ln(1)+2\ln(11)=2\ln(11)\), approximately \(4.796\). At \((1,0)\), utility is \(\ln(2)+2\ln(1)=\ln(2)\), approximately \(0.693\). The derivative check shows that no interior bundle does better than \((0,10)\). The method avoids assuming that positive marginal utility automatically makes an interior choice optimal.
Worked Example: A Square-Root Utility with a Boundary Derivative Issue
A consumer has utility \(U(x,y)=\sqrt{x}+2\sqrt{y}\), income \(m=4\), and prices \(p_x=1\), \(p_y=4\). Find the maximizing bundle and compare it with the zero-consumption choices. The utility function is increasing, and the budget binds.
Method and budget line. Use \(y=(4-x)/4=1-x/4\), with \(0\leq x\leq4\). For an interior point,
Set this equal to zero. Then \(1/(2\sqrt{x})=1/(4\sqrt{y})\), so \(\sqrt{x}=2\sqrt{y}\), or \(x=4y\). Combining this with \(x+4y=4\) gives \(8y=4\), hence \(y=0.5\) and \(x=2\).
Compare candidate utilities. At \((2,0.5)\), utility is \(\sqrt{2}+2\sqrt{0.5}=2\sqrt{2}\), approximately \(2.828\). At \((0,1)\), utility is \(2\); at \((4,0)\), utility is also \(2\). To check the interior candidate, note that
The derivative is strictly decreasing, and it changes from positive near \(x=0\) to negative near \(x=4\). Therefore the stationary point is the unique interior maximum, and its utility exceeds both endpoint values. The square-root derivative is not finite at \(x=0\); the endpoint is correctly tested using utility itself, not by requiring a derivative there.
Common Mistake: Using the Wrong Price Ratio
A frequent error is to differentiate utility along the budget line but use the wrong rate at which one quantity changes when the other increases. From \(p_xx+p_yy=m\), differentiating gives \(dy/dx=-p_x/p_y\), not \(-p_y/p_x\). Consequently, the correct interior condition is \(MU_x/p_x=MU_y/p_y\).
For example, in the first worked example the budget line is \(y=12-2x\), so \(dy/dx=-2\). The correct derivative is \(V'(x)=1/\sqrt{x}-1/\sqrt{y}\), giving \(x=y\) and the feasible candidate \((4,4)\). If one incorrectly uses \(dy/dx=-1/2\), one obtains \(V'(x)=1/\sqrt{x}-1/(4\sqrt{y})\) and the false condition \(x=16y\). Combined with the budget, that gives \(x=48/17\), \(y=3/17\), not the correct stationary point. The error changes the candidate because it reverses the relative-price adjustment.
A second trap is to solve an interior equation and stop. The condition \(V'(x)=0\) does not test either endpoint, and it may have no feasible solution. Always check \((0,m/p_y)\) and \((m/p_x,0)\) directly. Also verify that a root lies inside the feasible interval; algebraic solutions outside it are not affordable bundles.
What the Boundary Comparison Tells You
A zero-consumption optimum is an economically meaningful result: given income, prices, and preferences, the consumer gains more utility by allocating the budget to one good rather than retaining positive amounts of both. It is not a failure of the optimization method. The candidate comparison makes this possibility visible even when the problem initially appears to call for an interior tangency.
The method depends on its stated setup. If utility is not strictly increasing, the budget need not bind, so the full budget line alone may not contain the optimum; the entire feasible set must then be considered. If utility has a kink, a discontinuity, or an interior point where it is not differentiable, that point may be a candidate even though the derivative equation does not identify it. In all cases, the central discipline is the same: state the feasible set, identify its relevant boundaries, and compare the utility of all candidates justified by the problem.
Check Your Understanding
For each problem, use the budget line to identify any interior stationary candidate, then compare it with the zero-consumption bundles. Quantities are nonnegative, and each stated utility function is increasing.
- A consumer has \(U(x,y)=\sqrt{x}+\sqrt{y}\), income \(8\), and prices \(p_x=p_y=1\). What bundle maximizes utility? (Answer: \((4,4)\), with utility \(4\); each endpoint gives \(\sqrt{8}\).)
- A consumer has \(U(x,y)=\ln(1+x)+\ln(1+y)\), income \(5\), and prices \(p_x=2\), \(p_y=1\). Find the maximizing bundle and its utility. (Answer: \((1,3)\), with utility \(\ln(8)\); the endpoint utilities are \(\ln(6)\) and \(\ln(3.5)\).)
- A consumer has \(U(x,y)=\ln(1+x)+2\ln(1+y)\), income \(8\), and prices \(p_x=8\), \(p_y=1\). Does the optimum consume zero of either good? (Answer: \((0,8)\); utility is \(2\ln(9)\), and utility decreases as \(x\) increases from zero.)
- A consumer has \(U(x,y)=\sqrt{x}+\sqrt{y}\), income \(8\), and prices \(p_x=1\), \(p_y=4\). Find the interior candidate and determine whether it beats both endpoints. (Answer: \((6.4,0.4)\), with utility \(\sqrt{10}\), which exceeds the endpoint utilities \(\sqrt{8}\) and \(\sqrt{2}\).)
- A consumer has \(U(x,y)=\ln(1+x)+\ln(1+y)\), income \(8\), and prices \(p_x=1\), \(p_y=3\). Find the maximizing bundle and compare its utility with the budget intercepts. (Answer: \((5,1)\), with utility \(\ln(12)\); the endpoint utilities are \(\ln(9)\) and \(\ln(11/3)\), both lower.)