= Lagrange multiplier for quadratic cost
{c}
{title2=$\min_{x+y=c}(x^2+2y^2)=2c^2/3$}
Suppose two adjustable inputs must deliver a prescribed total $c$, and their combined cost is the <quadratic form> $f(x,y)=x^2+2y^2$. These are chosen, <dimensionless> quantities; the coefficient $2$ makes concentrating an input in $y$ more expensive. Initially allow all <real numbers> $x,y$, with the equality <optimization constraint> $g(x,y)=x+y=c$. The <optimization Lagrangian> with our chosen sign is
$$
L=x^2+2y^2-\lambda(x+y-c).
$$
The <stationary point> equations give $2x=\lambda$, $4y=\lambda$, and $x+y=c$. Consequently
$$
\boxed{x_*=\frac{2c}{3},\qquad y_* =\frac c3,\qquad \lambda_* =\frac{4c}{3},\qquad V(c)=\frac{2c^2}{3}.}
$$
For $c=3$, the unique <global minimum> is at $(2,1)$, with cost $6$ and <Lagrange multiplier> $4$. Equal allocation would cost $6.75$; assigning everything to $x$ would cost $9$. At the optimum, the two marginal costs agree: $\partial_x f=\partial_y f=4$. Moving a little input from one component to the other therefore has zero first-order effect on cost. When $c>0$, this solution also meets the extra nonnegativity <optimization constraints> $x,y\geq0$, so it remains optimal with those added.
\Image[/lagrange-quadratic-cost-tangent-contours.png]
{title=The first attainable cost ellipse touches $x+y=3$ at $(2,1)$}
{description=Original chosen <quadratic form> $x^2+2y^2$. Left: <level sets>, the equality <feasible set>, and the parallel <gradients> at the optimum. The blue arrow is $\nabla f/4$ so both arrows fit; it is not the full <gradient> of $f$. Right: substituting $y=3-x$ makes the unique <global minimum> visible on the entire feasible line, not just near its <stationary point>.}
Here the <Lagrange multiplier> equations locate a candidate, and a short independent check certifies it. On the entire <feasible set>, substitute $y=c-x$ and use <completing the square>:
$$
x^2+2(c-x)^2=3\left(x-\frac{2c}{3}\right)^2+\frac{2c^2}{3}.
$$
The squared term is nonnegative and vanishes only at $x=2c/3$. Thus the candidate is the unique <global minimum>; there is no <global maximum> on this unbounded <feasible set>, a <line>. This separates finding a <stationary point> from proving optimality. The geometric beauty is that equal marginal costs, parallel <gradients>, tangency of <level sets>, and <completing the square> identify the same allocation.
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