How do Lagrange multipliers solve maximizing x*y subject to x + y = 10?

## The problem A constrained optimization problem has two parts: an **objective** you want to push as high (or as low) as possible, and a **constraint** limiting which points you may consider. In `maximize x*y subject to x + y = 10`, the objective is `f(x, y) = x*y` and the constraint is the line `g(x, y) = x + y = 10`. Only points on that line are *feasible*; a huge product at a point off the line is worthless. The unconstrained instinct - set both partial derivatives of the objective to zero - fails immediately. `df/dx = y` and `df/dy = x`, so it would demand `x = y = 0`, which is not even on the line. The constraint has to enter the equations themselves, not act as a filter applied afterwards. ## The Lagrangian Lagrange's method introduces one extra unknown, the **multiplier** `lambda`, for each equality constraint, and folds objective and constraint into one function. Rewrite the constraint so one side is zero (`x + y - 10 = 0`) and attach it with weight `lambda`: `L(x, y, lambda) = x*y - lambda*(x + y - 10)` Now treat `L` as an ordinary unconstrained function of all three variables and set every partial derivative to zero: - `dL/dx = y - lambda = 0` - `dL/dy = x - lambda = 0` - `dL/dlambda = -(x + y - 10) = 0` The third equation is the constraint itself; that is precisely why `lambda` is made a variable rather than a fixed number. The first two give `x = lambda` and `y = lambda`. Substituting into the third, `2*lambda = 10`, so `lambda = 5`, `x = y = 5`, and the maximum product is `25`. ## What the stationarity equations mean Written together, the first two conditions are `df/dx = lambda * dg/dx` and `df/dy = lambda * dg/dy`: at the optimum the objective's gradient equals a scalar multiple of the constraint's gradient, and `lambda` is that scalar. The geometric content is that at any feasible point where the two gradients are *not* proportional, there is still a direction that stays on the constraint and increases the objective, so you cannot be at the best feasible point. The search stops only when the objective's push has been fully absorbed by the constraint. ## Confirming it is a maximum Stationarity is a first-order condition: it produces candidates, not verdicts. Here it can be settled directly. On the line, `y = 10 - x`, so the product is `x*(10 - x) = -x^2 + 10*x`, a downward-opening parabola peaking at `x = 5`. It has no minimum at all - push `x` up and `y` down and the product runs to minus infinity - so `(5, 5)` with value 25 is the maximum. In bigger problems you check a second-order condition restricted to directions that stay on the constraint surface, or you invoke concavity of the objective over a convex feasible set. ## Why bother, when substitution works? On two variables and one linear constraint, substitution is faster. It stops being an option when the constraint cannot be solved for any single variable in closed form, when there are several constraints that interact, or when the problem is symmetric and substitution destroys that symmetry (here the symmetric answer `x = y` falls out of the equations rather than being spotted). The multiplier form also keeps the problem in a shape a numerical solver can attack directly, and it produces `lambda` as a by-product, which carries real information about how binding the constraint is. ## Several constraints, and the sign convention With several equality constraints you add one multiplier each: `L = f - sum_j lambda_j * (g_j - c_j)`, and you set the partial derivative with respect to every variable and every multiplier to zero. For equality constraints the sign convention is cosmetic - writing `+ lambda` instead of `- lambda` only flips the sign of the recovered multiplier, and the same optimal point solves both. The convention starts to matter for inequality constraints, where multipliers are required to be nonnegative and the sign encodes which side the constraint pushes from. ## The fine print The method assumes the objective and constraint are differentiable near the candidate, and it needs a **constraint qualification**: roughly, the constraint gradients must not degenerate at the point (for a single constraint, `dg` must not vanish there). Without it, a genuine optimum can fail to satisfy the stationarity equations at all. And stationary points of `L` are candidates for maxima, minima or neither, so the second-order or convexity check is not optional bookkeeping.