Find the extrema of the multivariable function $f(x,y)=x^2y^3(6-x-y)$

Question

Find the extrema of the multivariable function $f(x,y)=x^2y^3(6-x-y)$

From $$\frac{\partial}{\partial x}f=xy^3(12-3x-3y)=0 \\ \frac{\partial}{\partial y}f=x^2y^2(18-3x-4y)=0 $$ I was able to find the stationary points which are $A=(2,3)$, $B=(0,y)$, $C=(x,0)$ for $x,y \in \mathbb R$.

Now from $$D=f_{xx}(a,b)f_{yy}(a,b)-f_{xy}^2(a,b)$$ I have $D(2,3) \gt 0$ and $f_{xx}(2,3) \lt 0$ so there is a relative maximum at $A(2,3)$

Now for the points, $B$ and $C$ I have $D=0$ and this is where I'm stuck a little and I would appreciate some help.

In the book they take $$\Delta f(0,y)=f( \Delta x, y+\Delta y)-f(0, \Delta y )=\Delta x^2(y+\Delta y)^3(6-\Delta x-y-\Delta y)$$

For very small values of $\Delta x$ and $\Delta y$ we have $$\Delta f(0,y) \le 0$$ for $- \infty \lt y \lt 0$ or $6 \lt y \lt + \infty$ and $$\Delta f(0,y) \ge 0$$ for $0 \lt y \lt 6$

so in the point $B(0,y)$ for $- \infty \lt y \lt 0$ or $6 \lt y \lt + \infty$ f has minimum, and in $B(0,y)$ for $0 \lt y \lt 6$ f has maximum.

Can somebody explain why is the last part true?

Answer 1

Through the second partial derivative test, I first confirmed that $(2,3)$ is a local maximum.

Starting from

$$f(x,y)=x^2y^3(6-x-y)$$

the partial derivatives are

$$f_x=x y^3 (-3 x - 2 y + 12) \\ f_y=-x^2 y^2 (3 x + 4 y - 18)\\ f_{xx}=-2 y^3 (3 x + y - 6) \\f_{yy}= -6 x^2 y (x + 2 y - 6) \\ f_{xy}=x y^2 (-9 x - 8 y + 36) $$

If we set $f_x=0$ then

$$x y^3 (-3 x - 2 y + 12)=0\implies y = 6-\frac{3x}{2} ~\text{ or }~ x=y=0$$

where

$$y=6-\frac{3x}{2}\tag{1}$$ $$x=y=0\tag{2}$$

If we set $f_y=0$ then

$$-x^2 y^2 (3 x + 4 y - 18)=0\implies y = \frac{9}{2}-\frac{3x}{4} ~\text{ or }~ x=y=0$$

where

$$y = \frac{9}{2}-\frac{3x}{4}\tag{3}$$

So, equating $(1)$ and $(3)$ forms

$$6-\frac{3x}{2}=\frac{9}{2}-\frac{3x}{4} \implies x=2$$

which from $(1)$ gives us the critical points $(0,0)$ and $(2,3)$

We now need to apply the second partial derivative test. The test tells us that provided $f(x, y)$ is a differentiable real function of two variables whose second partial derivatives exist we can compute

$$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\Big(f_{xy}(x,y)\Big)^2$$

and that

1. If $D(a,b)>0$ and $f_{xx}(a,b)>0$ then $(a,b)$ is a local minimum of $f$.
2. If $D(a,b)>0$ and $f_{xx}(a,b)<0$ then $(a,b)$ is a local maximum of $f$.
3. If $D(a,b)<0$ then $(a,b)$ is a saddle point of $f$.
4. If $D(a,b)=0$ then the second derivative test is inconclusive, and the point $(a, b)$ could be any of a minimum, maximum or saddle point.

We found before that

$$f_{xx}=-2 y^3 (3 x + y - 6) \\f_{yy}= -6 x^2 y (x + 2 y - 6) \\ f_{xy}=x y^2 (-9 x - 8 y + 36)$$

Thus, if we compute $D(x,y)$ for our four critical points we have

$$D\big(0,0\big)=(0)(0)-(0)^2=0$$ $$D\big(2,3)=(-162)(-144)-(-108)^2=11664 > 0$$

Therefore, by the second partial derivative test,

1. $(0,0)$ is inconclusive.
2. $(2,3)$ is a local maximum since $D\big(2,3\big) > 0$ and $f_{xx}(2,3)=-162<0$.

Substituting $(2,3)$ back into $f(x,y)$ forms the local maximum of $f(2,3)=108$. This cannot be a global maximum as $f(-2,3)=540$. Also, $f(0,0)=0$ and $f(-1,-1)=-8$ so $(0,0)$ cannot be a global minimum.

To find the global extremum, I evaluated the limit as $x$ and $y$ approach plus or minus infinity. As $f(x,y)$ isn't bounded by a rectangular region in $\mathbb R^2$, I need to check what happens at the four end points.

$$\lim_{x\to-\infty}f(x,y)=\lim_{x\to-\infty}x^2y^3(6-x-y)=y^3(\infty)$$

$$\lim_{x\to\infty}f(x,y)=\lim_{x\to\infty}x^2y^3(6-x-y)=y^3(-\infty)$$

$$\lim_{y\to-\infty}f(x,y)=\lim_{y\to-\infty}\frac{e^{-y}}{\frac{1}{y}}=x^2(-\infty)$$

$$\lim_{y\to\infty}f(x,y)=\lim_{y\to\infty}\frac{e^{-y}}{\frac{1}{y}}=x^2(-\infty)$$

So, it is clear that our critical points won't be larger or smaller than these and therefore there is no global extremum.