Question Video: Finding and Classifying Critical Points of a Function of Two Variables

Video Transcript

Find all stationary points of the function 𝑓 of 𝑥, 𝑦 equals 𝑥 cubed plus three 𝑥 squared plus 𝑦 cubed minus three 𝑦 squared, stating whether they are minima, maxima, or saddle points.

Recall that a stationary point of a function 𝑓 of two variables 𝑥 and 𝑦 is found by setting 𝜕𝑓 by 𝜕𝑥 and 𝜕𝑓 by 𝜕𝑦 equal to zero. In this case, differentiating partially with respect to 𝑥, treating 𝑦 as a constant, gives us three 𝑥 squared plus six 𝑥. We can take out common factor of three 𝑥 to give three 𝑥 times 𝑥 plus two. Now we set this expression equal to zero. Solving for 𝑥 gives two solutions: 𝑥 equals zero or 𝑥 equals negative two. Similarly, differentiating partially with respect to 𝑦 gives three 𝑦 squared minus six 𝑦. Taking out a common factor of three 𝑦 gives three 𝑦 times 𝑦 minus two. And set this equal to zero. Solving this for 𝑦 gives two solutions: 𝑦 equals zero or 𝑦 equals two.

Any combination of these solutions for 𝑥 and 𝑦 will be a stationary point of 𝑓. So for 𝑥 equals zero, we could have 𝑦 equals zero or 𝑦 equals two. Therefore, two stationary points are zero, zero and zero, two. Likewise, for 𝑥 equals negative two, we could have 𝑦 equals zero or 𝑦 equals two, so the other two stationary points are negative two, zero and negative two, two. Next, we need to determine the nature of the stationary points. We do this using the second partial derivative test by evaluating a quantity sometimes known as the discriminant: 𝐷 equals 𝜕 two 𝑓 by 𝜕𝑥 squared times 𝜕 two 𝑓 by 𝜕𝑦 squared minus 𝜕 two 𝑓 by 𝜕𝑥𝜕𝑦 all squared.

The value of 𝐷 determines the nature of the stationary point. If 𝐷 is greater than zero, then the stationary point is either a maximum or a minimum. We can determine if it’s a minimum if 𝜕 two 𝑓 by 𝜕𝑥 squared is greater than zero. We can tell if it’s a maximum if 𝜕 two 𝑓 by 𝜕𝑥 squared is less than zero. Note that our choice of 𝑥 here is merely arbitrary because it’s the first variable. We could just as easily use 𝑦 if it turns out to be simpler. In this case, the problem is fairly symmetrical, so it doesn’t matter. And finally, if 𝐷 is less than zero, then the stationary point is a saddle point. If 𝐷 is equal to zero, then the test is inconclusive and we need to use another test. So let’s begin by finding 𝜕 two 𝑓 by 𝜕𝑥 squared and 𝜕 two 𝑓 by 𝜕𝑦 squared.

We already found 𝜕𝑓 by 𝜕𝑥 earlier to be three 𝑥 squared plus six 𝑥. So 𝜕 two 𝑓 by 𝜕𝑥 squared is just 𝜕 by 𝜕𝑥 of three 𝑥 squared plus six 𝑥. And this differentiates to six 𝑥 plus six. Likewise, we found 𝜕𝑓 by 𝜕𝑦 earlier to be three 𝑦 squared minus six 𝑦. So 𝜕 two 𝑓 by 𝜕𝑦 squared is just 𝜕 by 𝜕𝑦 of three 𝑦 squared minus six 𝑦, which comes to six 𝑦 minus six. And finally, 𝜕 two 𝑓 by 𝜕𝑥𝜕𝑦 is 𝜕 by 𝜕𝑦 of three 𝑥 squared plus six 𝑥, which has no dependence on 𝑦. So this comes to zero. So substituting these expressions into our equation for 𝐷, we get 𝐷 equals six 𝑥 plus six times six 𝑦 minus six minus zero squared. This simplifies to 36 times 𝑥 plus one times 𝑦 minus one.

We now need to evaluate this expression at each of the stationary points to determine their nature. As a quick aside, remember that six 𝑥 plus six is equal to 𝜕 two 𝑓 by 𝜕𝑥 squared, as we will need this in a moment. So to begin with, 𝐷 evaluated at the stationary point zero, zero is equal to 36 times zero plus one times zero minus one, which comes to negative 36. This is less than zero, so the stationary point at zero, zero is a subtle point. Next, we have 𝐷 evaluated at the stationary point zero, two which equals 36 times zero plus one times two minus one, which equals 36. This is greater than zero, so this stationary point is either a minimum or a maximum.

To determine which, we need to evaluate 𝜕 two 𝑓 by 𝜕𝑥 squared. 𝜕 two 𝑓 by 𝜕𝑥 squared is given by six 𝑥 plus six. So at the point zero, two, this evaluates to six times zero plus six, which is equal to six. This is greater than zero, so the stationary point at zero, two is a minimum. Next, we have 𝐷 evaluated at the stationary points negative two, zero, which comes to 36 times negative two plus one times zero minus one, which equals 36. This is once again greater than zero. So once again, we have either a minimum or a maximum and must evaluate 𝜕 two 𝑓 by 𝜕𝑥 squared. So this comes to six times negative two plus six, which equals negative six. This is less than zero, so the stationary point at negative two, zero is a maximum.

And finally, we have 𝐷 evaluated at the stationary point negative two, two which equals 36 times negative two plus one times two minus one, which equals negative 36. This is less than zero, so the stationary point at negative two, two is a saddle point. So to summarize, we have stationary points at zero, zero; zero, two; negative two, zero; and negative two, two, which are a saddle point, a local minimum, a local maximum, and a saddle point, respectively.