When a Tangency Is a Global Maximum

Why a Tangency Needs More Than Matching Slopes

For two goods, an interior consumer optimum is often located by matching the slope of an indifference curve to the slope of the budget line. That condition identifies a candidate bundle, but it does not by itself show that the consumer does at least as well there as at every other affordable bundle. A stationary point can be a minimum, and a tangent can face the wrong way relative to the feasible region.

The key question is therefore not only whether the slopes match, but also whether the utility surface bends in a way that supports a maximum and whether the gradient points toward higher utility on the correct side of the budget line. Concave utility provides a direct sufficient condition. Convex preferences provide a related geometric explanation when the supporting boundary is correctly oriented.

Definition: A utility function \(U(x,y)\) is concave on a convex domain if, for any two bundles \(z_1,z_2\) in that domain and any \(t\in[0,1]\), \(U(tz_1+(1-t)z_2)\geq tU(z_1)+(1-t)U(z_2)\). A preference relation is convex if each upper contour set—the bundles at least as preferred as a given bundle—is convex. Concavity of a utility representation implies convex upper contour sets, though convex preferences do not require that every utility representation be concave.

Take positive prices \(p_x,p_y\) and income \(m\). The budget set consists of bundles satisfying \(p_xx+p_yy\leq m\), with \(x,y\geq0\). At an interior, budget-exhausting candidate \((x^*,y^*)\), the usual smooth tangency condition is \(U_x(x^*,y^*)/U_y(x^*,y^*)=p_x/p_y\), provided both marginal utilities are positive. Equivalently, the gradient of utility is a scalar multiple of the price vector.

\nabla U(x^*,y^*)=\lambda(p_x,p_y),\qquad p_xx^*+p_yy^*=m,\qquad \lambda\geq 0.

The nonnegative sign on \(\lambda\) matters. The price vector points toward bundles that use more expenditure; when \(\lambda\geq0\), the utility gradient points in that same direction. Thus the locally preferred side of the tangent is away from the affordable side of the budget line. If \(\lambda<0\), the gradient points toward lower expenditure, and the tangency geometry can instead indicate that affordable bundles are preferred.

Why Concavity Makes the Candidate Globally Optimal

The result below uses a differentiable concave utility function, a convex budget set, an interior candidate that exhausts the budget, and the first-order condition with \(\lambda\geq0\). These are sufficient conditions for global optimality; they are not a claim that every consumer problem has an interior solution. Boundary choices, nondifferentiable utility, or a nonbinding budget require an appropriately modified analysis.

Theorem: Suppose \(U\) is differentiable and concave on the feasible domain. If an affordable bundle \((x^*,y^*)\) satisfies \(p_xx^*+p_yy^*=m\) and \(\nabla U(x^*,y^*)=\lambda(p_x,p_y)\) for some \(\lambda\geq0\), then \((x^*,y^*)\) is a global utility maximum over the budget set. If \(U\) is strictly concave, there can be no other maximizing bundle.

To see why, concavity gives a supporting-plane inequality: at any feasible bundle \(z=(x,y)\), utility cannot exceed the value at the candidate plus the candidate’s gradient dotted with the displacement. The budget constraint makes that gradient term nonpositive.

U(z)\leq U(z^*)+\nabla U(z^*)\cdot(z-z^*) =U(z^*)+\lambda\bigl[p_x(x-x^*)+p_y(y-y^*)\bigr].

Since \(z\) is affordable, \(p_xx+p_yy\leq m\), while the candidate spends exactly \(m\). Therefore \(p_x(x-x^*)+p_y(y-y^*)\leq0\). With \(\lambda\geq0\), the added term is nonpositive, so \(U(z)\leq U(z^*)\). This comparison applies to every feasible bundle, not merely to nearby bundles. That is why the result establishes a global maximum.

Check feasibility and binding.
Confirm that the candidate has nonnegative quantities and spends the full budget.

Check the gradient condition.
Verify that the gradient is a multiple of the price vector and that the multiplier is nonnegative.

Check curvature.
Establish concavity of utility on the relevant feasible domain, not just a slope match at one point.

Conclude globally.
Apply the supporting-plane inequality to any affordable alternative bundle.

The same conclusion has a geometric interpretation. Concave utility implies convex upper contour sets: mixtures of bundles that each provide at least a specified utility level also provide at least that level. At a smooth boundary point of such a set, a supporting line keeps the upper contour set on its preferred side. For the tangent to establish optimality, however, its preferred side must point away from feasible bundles. The condition \(\nabla U=\lambda(p_x,p_y)\) with \(\lambda\geq0\) supplies this orientation: moving to the preferred side requires increasing expenditure, which is infeasible from a budget-exhausting candidate.

Convexity alone should not be used as a shortcut that ignores orientation. A geometric argument needs a supporting boundary and must identify which side is preferred and which side is affordable. The gradient-and-concavity proof above makes those requirements explicit.

Worked Example: A Concave Utility Tangency

A consumer chooses nonnegative quantities of two goods with utility \(U(x,y)=\sqrt{xy}\), prices \(p_x=2\) and \(p_y=1\), and income \(m=24\). Find the interior tangency and verify that it is a global maximum. Method: solve the marginal-rate-of-substitution condition, then use concavity and the global-optimality theorem. This avoids treating a slope match by itself as proof.

For positive \(x,y\), the marginal utilities are \(U_x=\frac12\sqrt{y/x}\) and \(U_y=\frac12\sqrt{x/y}\). Hence \(U_x/U_y=y/x\). The tangency condition sets this ratio equal to the price ratio: \(y/x=2/1=2\), so \(y=2x\).

Substitute into the binding budget \(2x+y=24\): \(2x+2x=24\), so \(4x=24\) and \(x^*=6\). Then \(y^*=2(6)=12\). The budget check is \(2(6)+12=12+12=24\).

At this bundle, utility is \(U(6,12)=\sqrt{72}=6\sqrt{2}\), approximately \(8.485\). The function \(\sqrt{xy}\) is concave for nonnegative \(x,y\). At the candidate, \(U_x=\frac12\sqrt{2}\) and \(U_y=\frac{1}{2\sqrt{2}}\); their ratio is \(2\), as required. Both are positive, so the gradient equals a positive multiple of the price vector. The theorem therefore establishes global optimality. The common mistake avoided here is to stop after finding \(y=2x\) without checking budget feasibility and curvature.

Worked Example: Concave Utility with Unequal Weights

Suppose \(U(x,y)=\ln x+2\ln y\) for positive quantities, with prices \(p_x=3\), \(p_y=1\), and income \(m=30\). Find the interior candidate and justify its global status. Method: use marginal utilities to find the tangency, then check the concavity condition. This guards against assuming that the form of a first-order equation alone proves a maximum.

The marginal utilities are \(U_x=1/x\) and \(U_y=2/y\). Thus \(U_x/U_y=y/(2x)\). Equating this to \(p_x/p_y=3\) gives \(y/(2x)=3\), or \(y=6x\).

Insert this relation into \(3x+y=30\): \(3x+6x=30\), so \(9x=30\) and \(x^*=10/3\). Then \(y^*=6(10/3)=20\). The spending check is \(3(10/3)+20=10+20=30\).

At the candidate, \(U_x=1/(10/3)=3/10\) and \(U_y=2/20=1/10\); their ratio is \(3\). The utility function is strictly concave on the positive quadrant because its second derivatives are \(U_{xx}=-1/x^2<0\), \(U_{yy}=-2/y^2<0\), and \(U_{xy}=0\). Its Hessian is therefore negative definite there. The candidate satisfies the correctly oriented tangency condition, so it is a global maximum; strict concavity also rules out a second maximizing bundle. The common mistake avoided is to confuse a valid positive-multiplier tangency with an unverified stationary point.

Two Ways a Tangency Can Mislead

A tangency can fail as a global-maximization argument for two different reasons. Utility may not be concave, so a stationary point can be a minimum. Or the gradient may point in the wrong direction, even when the utility function is concave. These are separate checks: curvature does not repair a negative multiplier, and a positive multiplier does not by itself establish the needed curvature.

Worked Example: A First-Order Tangency That Is a Minimum

Let \(U(x,y)=(x-5)^2+y\), with \(x,y\geq0\), and let the budget be \(x+y\leq10\). Consider the interior, budget-exhausting tangency. Method: substitute the budget equality into utility and inspect the resulting one-variable function. This reveals why a first-order condition without concavity is insufficient.

On the budget line, \(y=10-x\), where \(0\leq x\leq10\). Utility along this line is \(g(x)=(x-5)^2+10-x\). Expanding gives \(g(x)=x^2-11x+35\), so \(g'(x)=2x-11\). Setting \(g'(x)=0\) gives \(x^*=11/2=5.5\), and then \(y^*=10-5.5=4.5\).

The marginal utilities at the candidate are \(U_x=2(x-5)=1\) and \(U_y=1\). Thus \(U_x/U_y=1=p_x/p_y\), and the multiplier is positive. But \(g''(x)=2>0\), so the stationary point is a minimum along the budget line, not a maximum. Its utility is \((0.5)^2+4.5=0.25+4.5=4.75\). The feasible endpoint \((0,10)\) has utility \((0-5)^2+10=25+10=35\), which is strictly greater. Here utility is not concave, so the global-optimality theorem does not apply. The common mistake is to treat the first-order condition, even with a positive multiplier, as sufficient without checking curvature.

Worked Example: A Concave Tangency with the Wrong Orientation

Consider \(U(x,y)=-(x+y)\), positive prices \(p_x=p_y=1\), and income \(m=10\). The bundle \((2,8)\) lies on the budget line, and its indifference curve is tangent to that line. Is it a maximum? Method: inspect both the multiplier sign and a feasible alternative. This avoids assuming that concavity alone validates every tangency.

The gradient is \(\nabla U=(-1,-1)\), so at \((2,8)\) it equals \(\lambda(1,1)\) with \(\lambda=-1\), not with a nonnegative multiplier. Utility is concave (it is linear), but it decreases as either quantity increases. The affordable bundle \((1,1)\) costs \(1+1=2\leq10\) and has utility \(-2\), whereas \(U(2,8)=-(2+8)=-10\). Since \(-2>-10\), the candidate is not a maximum. The indifference curve and budget line have matching slopes, but the preferred side points toward lower expenditure. The common mistake is to use the slope match while omitting the orientation check.

Common Exam Traps and a Reliable Check

The central distinction is between a first-order candidate and a global maximum. A slope match says that the budget line and indifference curve have the same slope at a smooth interior point. To upgrade that candidate to a global result, verify the assumptions that make every feasible alternative no better.

Assuming stationarity means maximum. A first derivative of zero can describe a minimum or another stationary point. Check concavity, or compare feasible values when concavity is absent.
Ignoring the sign of the multiplier. With positive prices, the standard sufficient condition uses \(\lambda\geq0\). A negative multiplier means the utility gradient points toward lower expenditure.
Using convex preferences without a supporting-side argument. Identify the preferred side of the boundary and show it lies outside the budget set. Convexity does not remove the need to establish the orientation.
Forgetting the feasible set. A global claim concerns every bundle satisfying the budget inequality and nonnegativity restrictions, not just bundles on the budget line near the candidate.

Key takeaway: For a differentiable, concave utility function, an affordable interior bundle that exhausts the budget and satisfies \(\nabla U=\lambda(p_x,p_y)\) with \(\lambda\geq0\) is a global maximum. The sign fixes the geometry; concavity makes the supporting-plane comparison valid everywhere in the feasible set.

Check Your Understanding

For each problem, check the budget and the conditions that support a global conclusion, not only the tangency equation.

A consumer has \(U(x,y)=\sqrt{xy}\), prices \(p_x=4\), \(p_y=2\), and income \(m=40\). Find the interior tangency and state why it is globally optimal. (Answer: \((x^*,y^*)=(5,10)\); \(U=5\sqrt{2}\), and concavity with a positive multiplier establishes global optimality.)
For \(U(x,y)=\ln x+\ln y\), prices \(p_x=1\), \(p_y=4\), and income \(m=20\), find the interior optimum and verify spending. (Answer: \((x^*,y^*)=(10,2.5)\); spending is \(10+4(2.5)=20\), and strict concavity establishes a unique global maximum.)
Let \(U(x,y)=(x-3)^2+y\) and \(x+y\leq8\). Find the interior stationary point on the budget line, then compare it with \((0,8)\). (Answer: the stationary point is \((3.5,4.5)\) with utility \(4.75\); \((0,8)\) has utility \(17\), so the stationary point is not a maximum.)
Let \(U(x,y)=-(2x+y)\), prices \((p_x,p_y)=(2,1)\), and income \(m=15\). At the budget-exhausting bundle \((5,5)\), identify the multiplier sign and give an affordable bundle with higher utility. (Answer: \(\nabla U=(-2,-1)=-1(2,1)\), so \(\lambda=-1\); \((1,1)\) costs \(3\) and has utility \(-3>-15=U(5,5)\).)
For \(U(x,y)=\ln x+3\ln y\), prices \(p_x=2\), \(p_y=1\), and income \(m=32\), solve for the tangency and explain the global conclusion. (Answer: \(y/(3x)=2\), so \(y=6x\); \(2x+6x=32\) gives \((x^*,y^*)=(4,24)\), a global maximum because utility is strictly concave and the multiplier is positive.)

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