In Lagrange Multipliers you maximized a function along a curve by lining up two gradients. The consumer's problem from The Best Affordable Bundle is exactly that: maximize $U(x, y)$ along the budget line. The method gives the same bundle as substitution, plus one more number with a meaning of its own.
Key idea
Form $\mathcal{L} = U(x, y) - \lambda(p_1x + p_2y - m)$ and set its three partial derivatives to zero: $$MU_x = \lambda p_1, \qquad MU_y = \lambda p_2, \qquad p_1x + p_2y = m.$$ The multiplier $\lambda = \dfrac{MU_x}{p_1} = \dfrac{MU_y}{p_2}$ is the marginal utility of income: the extra utility from one more dollar.
Why it works
At the best bundle the indifference curve and the budget line are tangent, so their normal vectors point the same way: $\nabla U = (MU_x, MU_y)$ is parallel to $(p_1, p_2)$, the gradient of the budget. "Parallel" means one is a multiple of the other: $\nabla U = \lambda(p_1, p_2)$. Dividing the first two conditions gives $MU_x/MU_y = p_1/p_2$, the tangency rule. And $\lambda$ is utility per dollar on either good. If income rises by \$1, spending it on either good adds about $\lambda$ to utility, so $dU^*/dm = \lambda$ (this is the envelope theorem).
Worked example: Solving with λ
Maximize $U = xy$ subject to $2x + 4y = 80$.
- $\mathcal{L} = xy - \lambda(2x + 4y - 80)$.
- $\partial\mathcal{L}/\partial x = y - 2\lambda = 0$, so $y = 2\lambda$. $\partial\mathcal{L}/\partial y = x - 4\lambda = 0$, so $x = 4\lambda$.
- Budget: $2(4\lambda) + 4(2\lambda) = 16\lambda = 80$, so $\lambda = 5$.
- $x = 20$, $y = 10$. Check the meaning: the maximum utility is $U^* = m^2/32$, so $dU^*/dm = m/16 = 5 = \lambda$.
Answer: $(20, 10)$ with $\lambda = 5$.
Worked example: Log utility
Maximize $U = \ln x + 2\ln y$ subject to $3x + y = 90$. Find $\lambda$.
- FOCs: $\dfrac{1}{x} = 3\lambda$ and $\dfrac{2}{y} = \lambda$, so $3x = \dfrac{1}{\lambda}$ and $y = \dfrac{2}{\lambda}$.
- Budget: $\dfrac{1}{\lambda} + \dfrac{2}{\lambda} = 90$, so $\lambda = \dfrac{3}{90} = \dfrac{1}{30}$.
- $x = 10$, $y = 60$. Another \$1 raises utility by about $1/30$.
Answer: $\lambda = 1/30$.
Common mistake
Solving the first two conditions and stopping. $MU_x = \lambda p_1$ and $MU_y = \lambda p_2$ are two equations in three unknowns $(x, y, \lambda)$: they only tell you the ratio of $x$ to $y$. The third condition, $\partial\mathcal{L}/\partial\lambda = 0$, is the budget itself, and it pins down the scale. Check: your bundle must cost exactly $m$.
The size of $\lambda$ depends on how utility is scaled (double $U$ and $\lambda$ doubles), so compare $\lambda$s only for the same utility function. The same method works with more goods: one condition per good plus the budget. Nothing later in this course needs it; the substitution method always works too.