MATH1120: Mathematics for Engineering, Science and Technology 2

We located the critical points of this function in an earlier example. We found that there are 4 critical points at

To determine the nature of these critical points, in this example we will use the second derivative test. To this end, note that

Thus

Now, applying the second derivative test:

• At $(0,0)\text{,}$ $D=20>0\text{,}$ $f_{xx}=10>0$ and so $(0,0)$ is a local minimum.

• At $(-{\displaystyle \frac{5}{3}},0)\text{,}$ $D={\displaystyle \frac{40}{3}}>0\text{,}$ $f_{xx}=-10<0$ and so $(-{\displaystyle \frac{5}{3}},0)$ is a local maximum.

• At $(-1,2)\text{,}$ $D=-16<0$ and so $(-1,2)$ is a saddle point.

• At $(-1,-2)\text{,}$ $D=-16<0$ and so $(-1,-2)$ is also a saddle point.

This is in agreement with the conclusions we made on the nature of these critical pointson the basis of the level curves of the function.

Here

For this function both partial derivatives are undefined at $(x,y)=(0,0)$ and so this point will be a critical point. However we cannot use the second derivative test to determine the nature of this critical point. In this case we can see from the graph of the function that the critical point at $(x,y)=(0,0)$ is a local maximum.