You have a budget, a utility function, and exactly one way to maximise: the Lagrange method. Constrained optimisation is the workhorse of intermediate microeconomics, and it is where many students first hit a wall. If you are studying in London, a mathematical economics tutor London students trust will walk you through the logic until the Lagrangian feels like second nature.

1 · The problem constrained optimisation solves

Every economic agent faces a constraint. A consumer has income M. A firm has a production budget. A government has a revenue target. The objective — utility, profit, output — must be maximised subject to that limit.

Write the problem formally:

max f(x, y) subject to g(x, y) = c

Here f is the objective function. g(x, y) = c is the constraint. The variables x and y are choice variables. The parameter c is the fixed resource.

Without the constraint, you would set partial derivatives to zero and solve. With the constraint, you must respect the boundary. The Lagrange method converts the constrained problem into an unconstrained one by introducing a new variable — the Lagrange multiplier λ.

2 · The Lagrangian — one function, three variables

The Lagrangian L is the objective plus the constraint multiplied by λ:

L(x, y, λ) = f(x, y) − λ(g(x, y) − c)

The minus sign is conventional. Some textbooks use a plus sign. The first-order conditions are the same either way — just check your department’s notation. Marking schemes are strict about this.

Now treat L as an unconstrained function of three variables. Take three partial derivatives and set each to zero:

∂L/∂x = 0, ∂L/∂y = 0, ∂L/∂λ = 0

The third condition, ∂L/∂λ = 0, just recovers the constraint: g(x, y) = c. That is the point. The Lagrangian embeds the constraint into the optimisation.

3 · What the Lagrange multiplier means

λ is not just a mathematical trick. It has an economic interpretation: the marginal value of relaxing the constraint.

If the constraint is a budget, λ is the marginal utility of income. If the constraint is a production target, λ is the marginal cost of the target. A one-unit increase in c raises the maximised value of f by λ.

This is the envelope theorem in action. Examiners love asking for the interpretation of λ in words. Practise saying it: “The Lagrange multiplier measures how much the objective would increase if the constraint were loosened by one unit.”

4 · The tangency condition

The first-order conditions imply something geometric. Rearrange ∂L/∂x = 0 and ∂L/∂y = 0 to get:

f_x / f_y = g_x / g_y

The ratio of marginal benefits equals the ratio of marginal costs. At the optimum, the slope of the indifference curve (or isoquant) equals the slope of the budget line (or isocost). The two curves are tangent.

This tangency condition is the intuition behind the algebra. The Lagrangian just formalises it.

5 · Second-order conditions — checking you found a maximum

The first-order conditions give candidate points. They could be maxima, minima, or saddle points. For a maximum, the bordered Hessian must be positive definite (for a minimum, negative definite). The test involves taking second partials of the Lagrangian and checking the sign of a determinant.

The full matrix is:

H_B = [[0, g_x, g_y], [g_x, L_xx, L_xy], [g_y, L_yx, L_yy]]

For two variables and one constraint, the condition is that the determinant of H_B is positive. That is the second-order sufficient condition. You do not need to compute it in every problem, but you should know it exists. Examiners sometimes ask: “State the second-order condition for a maximum.”

Worked example — utility maximisation with a budget

A consumer has utility U(x, y) = x^0.5 y^0.5. The price of x is p_x = 4, the price of y is p_y = 2, and income is M = 60. Maximise utility subject to the budget.

Step 1 — Write the constraint. g(x, y) = 4x + 2y = 60.

Step 2 — Write the Lagrangian. L = x^0.5 y^0.5 − λ(4x + 2y − 60).

Step 3 — First-order conditions. ∂L/∂x = 0.5x^−0.5 y^0.5 − 4λ = 0. ∂L/∂y = 0.5x^0.5 y^−0.5 − 2λ = 0. ∂L/∂λ = −(4x + 2y − 60) = 0.

Step 4 — Solve for x and y. From the first two conditions, divide the first by the second: (0.5x^−0.5 y^0.5) / (0.5x^0.5 y^−0.5) = (4λ)/(2λ) → y/x = 2 → y = 2x.

Step 5 — Substitute into the budget. 4x + 2(2x) = 60 → 8x = 60 → x = 7.5. Then y = 2(7.5) = 15.

Step 6 — Find λ. From ∂L/∂x: 0.5(7.5)^−0.5 (15)^0.5 = 4λ. Compute: 0.5 × √(15/7.5) = 0.5 × √2 ≈ 0.5 × 1.414 = 0.707. So λ = 0.707/4 ≈ 0.177.

Step 7 — Interpret. The consumer buys 7.5 units of x and 15 units of y, achieving utility U = √(7.5 × 15) = √112.5 ≈ 10.61. The Lagrange multiplier λ ≈ 0.177 means that one extra unit of income would raise utility by about 0.177 units.

Struggling to see why the tangency condition is the same as the first-order conditions? That connection is where the algebra meets the geometry, and it is exactly what a one-on-one mathematical economics tutor London students recommend works through with you. Book a trial session.

Practice

Q1. A firm produces output Q(K, L) = K^0.4 L^0.6. The rental rate of capital is r = 5, the wage is w = 10, and the firm has a production budget of C = 200. Use the Lagrange method to find the cost-minimising combination of K and L.

Q2. A consumer has utility U(x, y) = x + ln(y). Prices are p_x = 2, p_y = 1, and income is M = 10. Set up the Lagrangian and find the optimal x and y.

Q3. For the utility function U(x, y) = x^0.5 y^0.5 with p_x = 4, p_y = 2, and M = 60 (the worked example), verify the second-order sufficient condition by computing the bordered Hessian determinant.

Answers: Q1: K = 16, L = 12. Q2: λ = 0.5, y = 2, x = 4. Q3: g_x = 4, g_y = 2, L_xx = −0.25x^−1.5 y^0.5 = −0.25(7.5)^−1.5(15)^0.5 ≈ −0.0471, L_yy = −0.25x^0.5 y^−1.5 = −0.25(7.5)^0.5(15)^−1.5 ≈ −0.0118, L_xy = 0.25x^−0.5 y^−0.5 = 0.25(7.5)^−0.5(15)^−0.5 ≈ 0.0236. Determinant = 0(…) − 4(4×(−0.0118) − 2×0.0236) + 2(4×0.0236 − 2×(−0.0471)) = −4(−0.0471 − 0.0471) + 2(0.0943 + 0.0943) = −4(−0.0943) + 2(0.1886) = 0.3771 + 0.3771 = 0.754 > 0, confirming a maximum.

Key takeaways

• The Lagrange method converts a constrained optimisation problem into an unconstrained one by introducing the multiplier λ.
• The first-order conditions are three partial derivatives set to zero. The third condition recovers the constraint.
• The Lagrange multiplier λ measures the marginal value of relaxing the constraint — the shadow price.
• The tangency condition f_x/f_y = g_x/g_y is the geometric equivalent of the first-order conditions.
• Second-order conditions use the bordered Hessian. For a maximum with one constraint and two variables, the determinant must be positive.

Why London students choose our mathematical economics tutoring

• One-on-one format: every session is private and built around your course, from your lecture slides to your problem sets and your department’s preferred Lagrangian sign convention.
• Intermediate-level specialists: our tutors teach constrained optimisation as your university teaches it, from the basic tangency diagram through the envelope theorem and the bordered Hessian.
• Exam-first preparation: sessions work through past papers with marking schemes in view, because the “interpret λ” question is where intermediate marks are won and lost.

FAQ

Q: What is the Lagrange method in economics?
A: A technique for maximising or minimising a function subject to a constraint. You form the Lagrangian L = f(x, y) − λ(g(x, y) − c), take three partial derivatives, set them to zero, and solve for x, y, and λ.

Q: What does the Lagrange multiplier mean?
A: It is the shadow price of the constraint. If the constraint is a budget, λ is the marginal utility of income. It tells you how much the objective would increase if the constraint were relaxed by one unit.

Q: How do I know if I found a maximum or a minimum?
A: Check the second-order condition using the bordered Hessian. For a maximum with one constraint and two variables, the determinant of the bordered Hessian must be positive. For a minimum, it must be negative.

Q: Do I need to compute the bordered Hessian in every problem?
A: No. Many exam questions only ask you to set up the Lagrangian and find the first-order conditions. But you should know the second-order condition exists and be able to state it.

Q: What if my Lagrangian has a plus sign instead of a minus?
A: The first-order conditions are the same either way. The Lagrange multiplier will have the opposite sign, but its interpretation (marginal value of relaxing the constraint) is unchanged. Check your department’s convention.

Q: Can the Lagrange method handle more than two variables?
A: Yes. For n variables and m constraints, you form the Lagrangian with m multipliers and take n + m partial derivatives. The bordered Hessian grows to (n + m) × (n + m).

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