극값과 안장점

(a) For $x,y>0$,

$$\begin{aligned} f_x&=2-\frac{1}{x^2y}=0\iff y=\frac{1}{2x^2}\\ f_y&=4-\frac{1}{xy^2}=0 \end{aligned}$$

Substituting $y=\frac{1}{2x^2}$, from the first equation, into the second gives $4-4x^3=0$ which forces $x=1$, $y=\tfrac{1}{2}$. At $x=1$, $y=\tfrac{1}{2}$,

$$f\big(1,\tfrac{1}{2}\big)=2+2+2=6$$

(b) The second derivatives are

$$f_{xx}(x,y)=\frac{2}{x^3y}\qquad f_{xy}(x,y)=\frac{1}{x^2y^2}\qquad f_{yy}(x,y)=\frac{2}{xy^3}$$

In particular

$$f_{xx}\big(1,\tfrac{1}{2}\big)=4\qquad f_{xy}\big(1,\tfrac{1}{2}\big)=4\qquad f_{yy}\big(1,\tfrac{1}{2}\big)=16$$

Since $f_{xx}\big(1,\tfrac{1}{2}\big)f_{yy}\big(1,\tfrac{1}{2}\big)- f_{xy}\big(1,\tfrac{1}{2}\big)^2=4\times 16-4^2=48>0$ and $f_{xx}\big(1,\tfrac{1}{2}\big)=4>0$, the point $\big(1,\tfrac{1}{2}\big)$ is a local minimum.

(c) As $x$ or $y$ tends to infinity (with the other at least zero), $2x+4y$ tends to $+\infty$. As $(x,y)$ tends to any point on the first quadrant part of the $x$- and $y$–axes, $\frac{1}{xy}$ tends to $+\infty$. Hence as $x$ or $y$ tends to the boundary of the first quadrant (counting infinity as part of the boundary), $f(x,y)$ tends to $+\infty$. As a result $\big(1,\tfrac{1}{2}\big)$ is a global (and not just local) minimum for $f$ in the first quadrant. Hence $f(x,y)\ge f\big(1,\tfrac{1}{2}\big)=6$ for all $x,y>0$.