라그랑주 승수법

(a) $f(x, y) = x^2 +y^2$ is the square of the distance from the point $(x,y)$ to the origin. There are points on the curve $xy=1$ that have either $x$ or $y$ arbitrarily large and so whose distance from the origin is arbitrarily large. So $f$ has no maximum on the curve. On the other hand $f$ will have a minimum, achieved at the points of $xy=1$ that are closest to the origin.

(b) On the curve $xy=1$ we have $y=\frac{1}{x}$ and hence $f=x^2+\frac{1}{x^2}$. As

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left(x^2+\frac{1}{x^2}\right) =2x-\frac{2}{x^3} =\frac{2}{x^3}(x^4-1) \end{aligned}$$

and as no point of the curve has $x=0$, the minimum is achieved when $x=\pm 1$. So the minima are at $\pm (1,1)$, where $f$ takes the value $2$.