Deriving the Tangency Rule with Lagrange Multipliers

Why derive the tangency rule?

The tangency rule describes a consumer’s interior optimum when choosing between two goods. Its familiar geometric interpretation is that the indifference curve and budget line have the same slope at the chosen bundle. The Lagrange multiplier method gives an algebraic derivation: it incorporates the budget constraint directly into the optimization problem and produces a first-order condition for each choice variable.

The derivation is conditional, not universal. It applies to a differentiable utility function at an optimum where both goods are consumed, the budget binds, and both marginal utilities are positive. If a consumer chooses zero units of a good, or if utility is not differentiable at the choice, the interior equations need not describe the optimum.

Definition: Let $x$ and $y$ be quantities of two goods, with prices $p_x>0$ and $p_y>0$, and income $I>0$. An interior choice has $x>0$ and $y>0$. The marginal utilities are $U_x=\partial U/\partial x$ and $U_y=\partial U/\partial y$. When $U_y>0$, define the marginal rate of substitution of $x$ for $y$ as $MRS_{xy}=U_x/U_y$: the quantity of $y$ the consumer is locally willing to give up for one more unit of $x$, holding utility constant.

Assumptions for the Lagrange derivation

The consumer chooses nonnegative quantities to maximize $U(x,y)$, subject to the budget inequality $p_xx+p_yy\le I$. For the interior derivation, assume $U$ is differentiable at the optimum, the optimum has $x>0$ and $y>0$, and $U_x>0$ and $U_y>0$ there. Also assume preferences are strictly increasing in both goods. That last assumption matters: with positive prices, a bundle that leaves income unspent can be improved by buying a little more of a good. Thus an optimum spends the full budget:

$$ p_xx+p_yy=I. $$

Differentiability and interiority by themselves would not imply this equality. For example, utility could be highest near a preferred bundle even when additional goods reduce utility. The full-budget conclusion here follows from strict monotonicity, not simply from using a Lagrangian.

The equality constraint has gradient $(p_x,p_y)$, which is nonzero because prices are positive. Therefore the usual first-order Lagrange condition can be applied at an interior constrained optimum. Define the Lagrangian, using the convention that the multiplier term is income minus expenditure:

\mathcal{L}(x,y,\lambda) =U(x,y)+\lambda(I-p_xx-p_yy).

Here $\lambda$ is an auxiliary variable enforcing the budget equality. The first-order conditions are obtained by differentiating with respect to $x$, $y$, and $\lambda$. The condition for $\lambda$ reproduces the budget constraint; the other two balance each marginal utility against the corresponding price, with the same multiplier.

\begin{aligned} \frac{\partial \mathcal{L}}{\partial x} &=U_x-\lambda p_x=0,\\ \frac{\partial \mathcal{L}}{\partial y} &=U_y-\lambda p_y=0,\\ \frac{\partial \mathcal{L}}{\partial \lambda} &=I-p_xx-p_yy=0. \end{aligned}

Dividing the first-order conditions

The first two equations give $U_x=\lambda p_x$ and $U_y=\lambda p_y$. Since $U_y>0$ and $p_y>0$, the second equation also implies $\lambda>0$. Dividing the first equation by the second cancels the common multiplier:

\frac{U_x}{U_y} = \frac{\lambda p_x}{\lambda p_y} = \frac{p_x}{p_y}.

Theorem: At a differentiable interior optimum with positive marginal utilities and a binding budget, the necessary first-order condition is $$ MRS_{xy}=\frac{U_x}{U_y}=\frac{p_x}{p_y}. $$ This is a necessary condition for an interior optimum. By itself, it does not establish that a candidate is a maximum, that the maximum is global, or that an interior optimum exists.

The same equality has a slope interpretation. Along a differentiable indifference curve, utility is constant, so a small change in the bundle satisfies $dU=U_x\,dx+U_y\,dy=0$. If $U_y>0$, its slope is $dy/dx=-U_x/U_y$. The budget line has slope $-p_x/p_y$. The first-order condition makes these slopes equal. This gives a brief geometric justification for why the Lagrange equations express tangency.

Write the constrained problem.
Specify the utility function, positive prices, income, and the budget constraint.

Check the conditions.
For this derivation, confirm that the optimum is interior, utility is differentiable there, both marginal utilities are positive, and the budget binds. Strictly increasing preferences ensure the budget binds in the stated setting.

Form and differentiate the Lagrangian.
Use $\mathcal{L}=U+\lambda(I-p_xx-p_yy)$, then set all three first derivatives equal to zero.

Divide and interpret.
Divide the two choice-variable equations to obtain $U_x/U_y=p_x/p_y$, then check whether the candidate satisfies the budget and is in fact a maximum.

Worked applications

Worked Example: Solving an Interior Choice from the Lagrangian

A consumer has utility $U(x,y)=\ln(x+1)+\ln(y+1)$, income $I=9$, and prices $p_x=1$, $p_y=2$. Assume quantities can be any nonnegative real numbers. Find the interior candidate using the Lagrange conditions, and check that it is the unique maximum.

Method and assumptions. Both logarithm arguments are positive for nonnegative quantities. The utility is strictly increasing because $U_x=1/(x+1)>0$ and $U_y=1/(y+1)>0$, so the budget binds. It is differentiable, and we first solve the interior first-order conditions.

Set up the Lagrangian.

\mathcal{L} =\ln(x+1)+\ln(y+1)+\lambda(9-x-2y).

Differentiate. The first-order conditions are $1/(x+1)-\lambda=0$, $1/(y+1)-2\lambda=0$, and $9-x-2y=0$. Dividing the first equation by the second gives

\frac{1/(x+1)}{1/(y+1)} =\frac{1}{2}, \qquad \frac{y+1}{x+1}=\frac{1}{2}.

Cross-multiplying gives $2y+2=x+1$, so $x=2y+1$. Substitute into the budget: $(2y+1)+2y=9$, hence $4y=8$, $y=2$, and $x=5$.

Check. Expenditure is $1(5)+2(2)=9$, exactly the income. The marginal utilities are $U_x=1/6$ and $U_y=1/3$, so $U_x/U_y=(1/6)/(1/3)=1/2=p_x/p_y$. The Hessian is diagonal with entries $-1/(x+1)^2$ and $-1/(y+1)^2$, both negative throughout the feasible region; utility is strictly concave. Therefore this interior candidate is the unique global maximum. The method avoids treating a first-order condition alone as proof of a maximum.

Worked Example: Keeping the Price Ratio in the Correct Order

Let $U(x,y)=\ln x+2\ln y$, with income $I=20$, $p_x=2$, and $p_y=1$. Find the interior candidate and verify the tangency condition.

Method and assumptions. On $x>0,y>0$, utility is differentiable and strictly increasing, with positive marginal utilities $U_x=1/x$ and $U_y=2/y$. The budget binds. Form

\mathcal{L}=\ln x+2\ln y+\lambda(20-2x-y).

Differentiation gives $1/x-2\lambda=0$, $2/y-\lambda=0$, and $20-2x-y=0$. Dividing the first choice equation by the second yields

\frac{1/x}{2/y} =\frac{2}{1}, \qquad \frac{y}{2x}=2.

Thus $y=4x$. Substituting in $2x+y=20$ gives $2x+4x=20$, so $x=10/3$ and $y=40/3$.

Check. Expenditure is $2(10/3)+40/3=60/3=20$. At this bundle, $U_x=3/10$ and $U_y=3/20$, and their ratio is $(3/10)/(3/20)=2=p_x/p_y$. The utility has negative second derivatives in both variables and no cross-partial term, so it is strictly concave on the positive quadrant; the candidate is the unique maximum. The calculation avoids reversing the relative-price ratio: with $MRS_{xy}=U_x/U_y$, the comparison is with $p_x/p_y$, not $p_y/p_x$.

Worked Example: Applying the Conditions with Shifted Utility

Suppose $U(x,y)=\ln x+\ln(y+2)$, income is $I=11$, and prices are $p_x=3$, $p_y=1$. Find the interior candidate and verify it.

Method and assumptions. For $x>0,y\ge0$, the marginal utilities are $U_x=1/x>0$ and $U_y=1/(y+2)>0$. Utility is differentiable at any interior bundle and strictly increasing, so the budget binds. The Lagrangian is

\mathcal{L}=\ln x+\ln(y+2)+\lambda(11-3x-y).

Differentiating gives $1/x-3\lambda=0$, $1/(y+2)-\lambda=0$, and $11-3x-y=0$. Divide the first choice equation by the second:

\frac{1/x}{1/(y+2)} =\frac{3}{1}, \qquad \frac{y+2}{x}=3.

Therefore $y+2=3x$, or $y=3x-2$. The budget condition gives $3x+(3x-2)=11$, so $6x=13$, $x=13/6$, and $y=3(13/6)-2=9/2$. Both quantities are positive, as required for an interior solution.

Check. Expenditure is $3(13/6)+9/2=13/2+9/2=11$. The marginal utilities are $U_x=6/13$ and $U_y=1/(13/2)=2/13$, so $U_x/U_y=3=p_x/p_y$. The Hessian is diagonal with negative entries $-1/x^2$ and $-1/(y+2)^2$, making utility strictly concave; thus the candidate is the unique maximum. The shift inside the logarithm changes the derivative to $1/(y+2)$; replacing it incorrectly with $1/y$ would give the wrong condition and choice.

Limits of the condition and a common exam trap

The tangency equation is a first-order condition for an interior optimum, not a complete solution procedure for every consumer problem. If the optimum lies at $x=0$ or $y=0$, the interior derivation does not apply directly. It also cannot be divided as shown if a marginal utility is zero. A kink in utility can prevent differentiability, and a tangency candidate may fail to be a maximum when preferences are not suitably shaped. In such cases, the boundary or nonsmooth conditions must be considered rather than assuming an interior tangency.

A separate trap is to claim that every interior optimum spends all income merely because the Lagrangian contains a budget equality. In the original problem, the budget is an inequality. The equality is justified here by the assumption of strictly increasing preferences. For instance, with $I=10$ and $U(x,y)=-(x-1)^2-(y-1)^2$, the unconstrained preferred bundle $(1,1)$ is interior and spends $2$ when both prices are $1$; extra consumption lowers utility. This utility is not strictly increasing, so the full-budget argument and the equality-constrained derivation do not follow from interiority alone.

A reliable exam solution states the assumptions, writes the Lagrangian with a consistent sign convention, derives all three first-order equations, and checks the candidate against the budget and the relevant maximum conditions. This procedure is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.

Check Your Understanding

For each problem, use the stated assumptions to decide whether the interior Lagrange derivation applies, then calculate or interpret the requested condition.

A differentiable utility function has $U_x=6$ and $U_y=3$ at an interior optimum. The prices are $p_x=4$ and $p_y=2$. Does the tangency condition hold? (Answer: Yes; $U_x/U_y=6/3=2=p_x/p_y=4/2$.)
Let $U(x,y)=\ln(x+2)+\ln(y+1)$, $p_x=1$, $p_y=2$, and $I=10$. Find the interior candidate from the first-order conditions. (Answer: $x=5,\ y=5/2$.)
Let $U(x,y)=\ln(2x+1)+2\ln(y+1)$, $p_x=2$, $p_y=1$, and $I=11$. Find the interior candidate. (Answer: $x=5/3,\ y=23/3$.)
A consumer has $U(x,y)=-(x-1)^2-(y-1)^2$, prices $p_x=p_y=1$, and income $I=10$. The bundle $(1,1)$ is interior. Can one infer that the budget binds and apply the equality-constrained tangency derivation there? (Answer: No; the bundle spends $2<10$, and this utility is not strictly increasing.)
At a candidate interior bundle, suppose $U_x=0$, $U_y=4$, and both prices are positive. Which step in the usual derivation of $U_x/U_y=p_x/p_y$ fails, and does the stated tangency equality hold? (Answer: Dividing the first-order conditions would imply $0=p_x/p_y$, which is impossible for positive prices; the stated positive-marginal-utility interior derivation does not apply.)

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