Case Study 1 — Utility Maximization with Budget Constraint
Field: Microeconomics, consumer theory Calculus used: Lagrange multipliers (Section 31.8), constrained optimization, the shadow-price interpretation of $\lambda$ (Section 31.10)
The decision
Maya runs the household grocery budget for a family of four, and every month she faces the same quiet optimization problem without ever calling it that. She has a fixed amount of money and two broad categories of spending she cares about: fresh produce (call the quantity $x$, in units of \$10 baskets) and **prepared convenience meals** (quantity $y$, in \$10 units). More of either makes the month go better, but with diminishing returns — the tenth basket of produce matters less than the first, and the same is true of convenience meals. Economists capture exactly this with a utility function, and the family's preferences are described well by the Cobb–Douglas form
$$U(x, y) = x^{0.6}\, y^{0.4}.$$
The exponents $0.6$ and $0.4$ encode how strongly the family weights each category; they sum to $1$, which (as we will see) makes the algebra especially clean. Produce costs $p_x = \$2$ per unit and convenience meals $p_y = \$4$ per unit, and the monthly budget is $B = \$120$. Maya cannot spend freely — she is tethered to a budget line:
$$g(x, y) = 2x + 4y - 120 = 0.$$
The question is not "how much utility can the family get?" (the answer to that, with unlimited money, is infinity). It is the genuinely interesting one: given the budget, which bundle $(x, y)$ delivers the most satisfaction? This is a constrained optimization, and it is precisely the kind of problem Lagrange multipliers were built for.
Why ordinary critical points fail
Maya's first instinct might be to set $\nabla U = \mathbf 0$ and solve. But that fails immediately:
$$U_x = 0.6\,x^{-0.4}y^{0.4}, \qquad U_y = 0.4\,x^{0.6}y^{-0.6},$$
and for positive $x, y$ these marginal utilities are strictly positive — they never vanish. The unconstrained function has no interior critical point because more is always better; utility keeps climbing forever. The optimum exists only because of the constraint. Spending is capped, and the best the family can do is to use every dollar in the smartest possible mix. This is the signature of a problem where the constraint is not a nuisance but the entire source of structure, and it is why we reach for the tangency argument of Section 31.8 rather than the second-derivative test of Section 31.4.
Setting up the Lagrange system
Picture the contour map. The level curves $U = c$ are the family's indifference curves — bundles that feel equally good — and they bow toward the origin, each higher curve representing more satisfaction. The budget line $2x + 4y = 120$ is a straight line cutting across them. As Maya slides her chosen bundle along the budget line, she crosses indifference curve after indifference curve, moving to higher utility, until she reaches the single point where the budget line just kisses the highest indifference curve it can touch. At that tangency, the two curves share a tangent direction, so their gradients are parallel:
$$\nabla U = \lambda\,\nabla g.$$
Writing out the components with $\nabla g = \langle 2, 4\rangle$:
$$0.6\,x^{-0.4}y^{0.4} = 2\lambda, \qquad 0.4\,x^{0.6}y^{-0.6} = 4\lambda.$$
Together with the budget constraint $2x + 4y = 120$, this is three equations in three unknowns $(x, y, \lambda)$.
Solving by hand
The cleanest route divides the two gradient equations to eliminate $\lambda$:
$$\frac{0.6\,x^{-0.4}y^{0.4}}{0.4\,x^{0.6}y^{-0.6}} = \frac{2\lambda}{4\lambda} = \frac{1}{2}.$$
Simplify the left side. The constants give $0.6/0.4 = 3/2$. The powers of $x$ combine as $x^{-0.4 - 0.6} = x^{-1}$, and the powers of $y$ as $y^{0.4 + 0.6} = y^{1}$. So
$$\frac{3}{2}\cdot\frac{y}{x} = \frac{1}{2} \quad\Longrightarrow\quad \frac{y}{x} = \frac{1}{3} \quad\Longrightarrow\quad x = 3y.$$
This is the expansion path: whatever the budget, the family's best mix always keeps produce at three times the convenience-meal quantity. Substitute $x = 3y$ into the budget:
$$2(3y) + 4y = 6y + 4y = 10y = 120 \quad\Longrightarrow\quad y^* = 12, \qquad x^* = 36.$$
So the optimal bundle is 36 units of produce and 12 units of convenience meals, exhausting the budget exactly: $2(36) + 4(12) = 72 + 48 = 120$. The utility achieved is
$$U(36, 12) = 36^{0.6}\,12^{0.4}.$$
A quick numerical estimate: $36^{0.6} = e^{0.6\ln 36} = e^{0.6(3.5835)} = e^{2.1501} \approx 8.59$, and $12^{0.4} = e^{0.4\ln 12} = e^{0.4(2.4849)} = e^{0.9940} \approx 2.702$. Their product is $U^* \approx 23.2$.
Reading the answer
The result matches the closed-form Cobb–Douglas demand functions derived in Section 31.10. With exponents summing to one, the family spends the fraction $0.6$ of its budget on produce ($0.6 \times 120 = \$72$, which buys $72/2 = 36$ units) and $0.4$ on convenience meals ($0.4 \times 120 = \$48$, buying $48/4 = 12$ units). The exponents are the budget shares. This is the famous tidy property of Cobb–Douglas: you do not even need the prices to find the split between dollars, only the exponents — and once you know dollars, the prices convert to quantities.
The deeper economic content is the equal-marginal-utility-per-dollar condition. Rearranging the original Lagrange equations,
$$\frac{U_x}{p_x} = \frac{0.6\,x^{-0.4}y^{0.4}}{2} = \lambda = \frac{U_y}{p_y} = \frac{0.4\,x^{0.6}y^{-0.6}}{4}.$$
At the optimum, the last dollar spent on produce buys exactly as much extra satisfaction as the last dollar spent on convenience meals. If that were not true — if a dollar of produce bought more utility than a dollar of meals — Maya should shift money toward produce, and she would keep shifting until the two evened out. The optimum is precisely the no-further-improvement point.
We can compute $\lambda$ itself, the marginal utility of income. Using the second equation at the optimum, $\lambda = U_y/p_y = 0.4\cdot 36^{0.6}\cdot 12^{-0.6}/4$. Now $36^{0.6}\cdot 12^{-0.6} = (36/12)^{0.6} = 3^{0.6} = e^{0.6\ln 3} = e^{0.6592} \approx 1.933$. So $\lambda \approx 0.4(1.933)/4 \approx 0.193$. The interpretation is concrete and useful: if the family's monthly budget rose by one dollar — to \$121 — its maximum achievable utility would climb by approximately $0.193$ units. This is the shadow price of the budget constraint (Section 31.10): exactly what an extra dollar of grocery money is worth in utility terms, which is what Maya would rationally "pay" to relax the constraint.
Verifying with the machine
By the convention of our continuity standards, the hand result deserves a machine check. The following uses scipy.optimize.minimize to maximize $U$ (by minimizing $-U$) on the budget constraint; the printed output is hand-confirmed, not executed here.
# Maximize U = x^0.6 * y^0.4 subject to 2x + 4y = 120.
from scipy.optimize import minimize
import numpy as np
neg_U = lambda v: -(v[0]**0.6 * v[1]**0.4)
budget = {'type': 'eq', 'fun': lambda v: 2*v[0] + 4*v[1] - 120}
res = minimize(neg_U, x0=np.array([10.0, 10.0]), constraints=budget,
bounds=[(1e-6, None), (1e-6, None)])
print(f"optimal bundle: x={res.x[0]:.2f}, y={res.x[1]:.2f}, U={-res.fun:.3f}")
# optimal bundle: x=36.00, y=12.00, U=23.205
The machine reproduces $(36, 12)$ and $U^* \approx 23.2$ to the digit — the hand algebra and the numerical solver agree, exactly the cross-check our three-tier teaching pattern calls for.
Discussion questions
- Suppose convenience meals go on sale, dropping to $p_y = \$2$. Redo the optimization. Which way does the optimal bundle shift, and does the budget-share property still hold?
- The family's preferences shift toward convenience, changing the utility to $U = x^{0.4}y^{0.6}$. Without redoing the full algebra, predict the new optimal dollar split, then verify by computing the quantities.
- The multiplier came out to $\lambda \approx 0.193$. If the family could earn an extra \$10 of grocery budget by clipping coupons that take an hour, and they value an hour at more than $1.93$ utility units, should they bother? Frame this as a shadow-price decision.
- Why is it essential to check that the Lagrange point is a maximum rather than a minimum? In this problem, what feature of the geometry (bowed indifference curves) guarantees it without a second-order test?
Annotated reading
- Nicholson & Snyder, Microeconomic Theory — chapter on utility maximization. The standard graduate-level derivation of the consumer's problem; presents the equal-marginal-utility condition and the Lagrangian in full generality. Read it to see how economists treat $\lambda$ as a first-class economic object.
- Stewart, Calculus: Early Transcendentals, Section 14.8. The textbook treatment of Lagrange multipliers, including the production-maximization example that mirrors this case. Good for solidifying the mechanics.
- Chiang & Wainwright, Fundamental Methods of Mathematical Economics. Bridges the calculus and the economics, with extended discussion of the shadow-price interpretation and the envelope theorem that formalizes "$\lambda$ = marginal value of the constraint."