2.5 Just in case ...

... someone has forgotten how to find the extremum of a many-variable function under additional constraints. To remind you: the method of undetermined Lagrange multipliers waits to be applied here.

Let $f(x,y)=x^{2}+y^{2}$ be the given function. Of course, this is the equation of a parabolid with its tip - or minimum - at the origin. However, let $g(x,y) \equiv x-y-1=0$ be the constraint equation, meaning that we don't search for the global minimum but for the minimum along the line $y=x-1$. There are two ways to go about it. The simple but inconvenient way is to substitute $y=x-1$ in $f(x,y)$, thus rendering $f$ a function of $x$ only. Equating the derivative $df/dx$ to zero we find the locus of the conditional minimum, $x=1/2$ and $y=-1/2$. The process of substitution is, in general, tedious.

A more elegant method is this: defining a (undetermined) Lagrange multiplier $\alpha$, find the minimum of the function $f(x,y)-\alpha g(x,y)$ according to

$$\frac{\partial f}{\partial x} - \alpha \frac{\partial g}{\partial x} =0$$ (2.53)

$$\frac{\partial f}{\partial y} - \alpha \frac{\partial g}{\partial y} =0$$ (2.54)

Eliminating $\alpha$ we find the solution without substituting anything. In our case

$$2x-\alpha =0$$ (2.55)

$$2y+\alpha =0$$ (2.56)

$\Longrightarrow$$x+y=0$, and from $g(x,y)=0$: $(x,y)=(1/2,-1/2)$.