극값과 안장점

The region of interest is

$$D=\big\{ \ (x,y,z)\ \big|\ x\ge 0,\ y\ge 0,\ z\ge 0,\ 2x+y+z=5\ \big\}$$

First observe that, on the boundary of this region, at least one of $x$, $y$ and $z$ is zero. So $f(x,y,z)=x^2 y^2 z$ is zero on the boundary. As $f$ takes values which are strictly bigger than zero at all points of $D$ that are not on the boundary, the minimum value of $f$ is $0$ on

$$\partial D = \big\{ \ (x,y,z)\ \big|\ x\ge 0,\ y\ge 0,\ z\ge 0,\ 2x+y+z=5,\ \text{at least one of $x,y,z$ zero}\ \big\}$$

The maximum value of $f$ will be taken at a critical point. On $D$

$$\begin{aligned} f &= x^2 y^2 (5-2x-y) =5x^2 y^2 - 2x^3 y^2 -x^2 y^3 \end{aligned}$$

So the critical points are the solutions of

$$\begin{aligned} 0&=f_x(x,y) = 10xy^2 -6x^2y^2 -2xy^3 \\ 0&=f_y(x,y) = 10x^2y -4x^3y -3x^2y^2 \end{aligned}$$

or, dividing by the first equation by $xy^2$ and the second equation by $x^2y$, (recall that $x,y\ne 0$)

$$\begin{aligned} 10 -6x -2y &=0 &\qquad\text{or}\qquad 3x+y&=5 \\ 10 -4x -3y &=0 &\qquad\text{or}\qquad 4x+3y&=10 \end{aligned}$$

Substituting $y=5-3x$, from the first equation, into the second equation gives

$$4x+3(5-3x)=10 \implies -5x +15 =10 \implies x=1,\ y=5-3(1)=2$$

So the maximum value of $f$ is $(1)^2(2)^2(5-2-2)=4$ at $(1,2,1)$.