Answer to Question #138350 in Calculus for Promise Omiponle

Question

The function f(x,y)=xy(1−6x−5y) has 4 critical points. List them and select the type of critical point.

Points should be entered as ordered pairs and listed in increasing lexicographic order. By that we mean that (x,y) comes before (z,w) if x<z or if x=z and y<w.

First point ___of type ____

Second point___ of type ___

Third point ___ of type ___

Fourth point ___ of type___

Answer 1

$f(x,y)=xy(1−6x−5y)=xy-6x^2y-5xy^2.$

The first order partial derivatives of $f(x,y)$ are:

$f_x(x,y)=y-12xy-5y^2=y(1-12x-5y),$

$f_y(x,y)=x-6x^2-10xy=x(1-6x-10y).$

To find the critical points, we solve $f_x(x,y)=f_y(x,y)=0.$

We have:

$y(1-12x-5y)=0$ and $x(1-6x-10y)=0.$

The first equation is satisfied if either $y=0$ or $1-12x-5y=0.$

The second equation is satisfied if either $x=0$ or $1-6x-10y=0.$

Hence there are four possibilities:

1) $y=0, x=0$ and we have the critical point $(0,0)$ .

2) $y=0, 1-6x-10y=0.$ In this case: $1-6x-12 \cdot 0=0, 1-6x=0, 6x=1, x=\frac{1}{6}.$ We have the critical point $(\frac{1}{6},0).$

3) $x=0, 1-12x-5y=0.$ In this case: $1-12\cdot 0-5y=0, 1-5y=0, 5y=1, y=\frac{1}{5}.$ We have the critical point $(0, \frac{1}{5}).$

4) $1-12x-5y=0, 1-6x-10y=0.$

$12x+5y=1, 6x+10y=1.$

We multiply the first equation by 2 and subtract the second equation: $(24x+10y)-(6x+10y)=2-1, 18x=1, x=\frac{1}{18}.$

Then we have from the first equation:

$12\cdot \frac{1}{18}+5y=1, \frac{2}{3}+5y=1, 5y=\frac{1}{3}, y=\frac{1}{15}.$

We have the critical point $(\frac{1}{18},\frac{1}{15}).$

All four points in increasing lexicographic order: $(0,0), (0,\frac{1}{5}), (\frac{1}{18}, \frac{1}{15}), (\frac{1}{6},0).$

To find the nature of the critical points, we can apply the second derivative test.

$f_{xx}(x,y)=-12y, f_{yy}(x,y)=-10x, f_{xy}(x,y)=1-12x-10y.$

$D=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2=120xy-(1-12x-10y)^2.$

At the point $(0,0): D=0-(1-0-0)^2=0-1^2=-1<0,$ hence $(0,0)$ is a saddle point.

At the point $(0,\frac{1}{5}): D=0-(1-10 \cdot \frac{1}{5})^2=0-(1-2)^2=-1<0,$ hence $(0,\frac{1}{5})$ is a saddle point of $f.$

At the point $(\frac{1}{18}, \frac{1}{15}): D=120\cdot \frac{1}{18}\cdot \frac{1}{15}-(1-12\cdot \frac{1}{18}-10\cdot \frac{1}{15})^2=$

$=\frac{4}{9}-(1-\frac{2}{3}-\frac{2}{3})^2=\frac{4}{9}-(-\frac{1}{3})^2=\frac{4}{9}-\frac{1}{9}=\frac{3}{9}= \frac{1}{3}>0.$

As $f_{xx}(\frac{1}{18}, \frac{1}{15})=-12 \cdot \frac{1}{15}=-\frac{4}{5}<0,$ hence there is a local maximum at $(\frac{1}{18}, \frac{1}{15}).$

At the point $(\frac{1}{6},0): D=0-(1-12 \cdot \frac{1}{6})^2=0-(1-2)^2=-1<0,$ hence $(\frac{1}{6},0)$ is a saddle point of $f.$

Answer.

First point $(0,0)$ of type saddle.

Second point $(0,\frac{1}{5})$ of type saddle.

Third point $(\frac{1}{18}, \frac{1}{15})$ of type local maximum.

Fourth point $(\frac{1}{6},0)$ of type saddle.