2 Location of stationary points

As we said in the previous subsection, the tangent plane to the surface $z=f\left(x,y\right)$ is horizontal at a stationary point. A condition which guarantees that the function $f\left(x,y\right)$ will have a stationary point at a point $\left(x_0,y_0\right)$ is that, at that point both $f_x=0$ and $f_y=0$ simultaneously.

Task!

Verify that $\left(0,2\right)$ is a stationary point of the function $f\left(x,y\right)=8x^2+6y^2-2y^3+5$ and find the stationary value $f\left(0,2\right)$ .

First, find $f_x$ and $f_y$ :

Answer

Now find the values of these partial derivatives at $x=0,y=2$ :

Answer

Example 9

Find a second stationary point of $f\left(x,y\right)=8x^2+6y^2-2y^3+5$ .

Solution

$f_x=16x$ and $f_y\equiv 6y\left(2-y\right)$ . From this we note that $f_x=0$ when $x=0$ , and $f_x=0$ and when $y=0$ , so $x=0,y=0$ i.e. $\left(0,0\right)$ is a second stationary point of the function.

It is important when solving the simultaneous equations $f_x=0$ and $f_y=0$ to find stationary points not to miss any solutions. A useful tip is to factorise the left-hand sides and consider systematically all the possibilities.

Example 10

Locate the stationary points of

$f\left(x,y\right)=x^4+y^4-36xy$

Solution

First we write down the partial derivatives of $f\left(x,y\right)$

$\frac{\partial f}{\partial x}=4x^3-36y=4\left(x^3-9y\right)\frac{\partial f}{\partial y}=4y^3-36x=4\left(y^3-9x\right)$

Now we solve the equations $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$ :

Now substitute from (iii) into (i)

Now, using (iii): when $y=0,x=0$ , when $y=3,x=3$ , and when $y=-3,x=-3$ .

The stationary points are $\left(0,0\right),\left(-3,-3\right)$ and $\left(3,3\right)$ .

Task!

Locate the stationary points of

$f\left(x,y\right)=x^3+y^2-3x-6y-1.$

First find the partial derivatives of $f\left(x,y\right)$ :

Answer

Now solve simultaneously the equations $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$ :

Answer