Find the critical points of the following functions. Use the Second Derivative Test to determine (if possible) whether each critical point corresponds to a local maximum, a local minimum, or a saddle point. If the Second Derivative Test is inconclusive, determine the behavior of the Junction at the critical points. f(x, y)=x^4y^2

In a function of two variables, the critical points are these in which both derivatives are 0, and if:

the point is a saddle point.

If the above equation is larger than zero, the critical points can be either a local maximum or a local minimum.

If it is equal to zero, no conclusions can be made.

First, let's find all these derivatives.

df/dx = 4*(x^3)*(y^2)

This is zero when x or y are equal to zero.

df/dy = 2*(x^4)*(y)

This is equal to zero when either x or y are equal to zero.

So in both cases, the points of the form:

(x, 0) and (0, y) are zeros of the first derivative, so these are the critical points.

Now let's look at the second derivatives:

Then, the equation to see if the points are saddle points or not are:

In any of the previous found points (the ones of the form (x, 0) and (0, y)) the above equation will be equal to zero, then none of these points is a saddle point. More than that, in the particular case where the above equation is equal to zero, the critical points can be either: A relative minimum, a relative maximum, or a saddle point (so we need more analysis to find it out)

But in this case is ratter simple, because both powers in f(x, y) are even, we can conclude that the function is even and symmetric about x= 0 and y = 0, and the function is also positive, so all the points:

(0, y) and (x, 0) are the minimum values that f(x, y) can take.

Then the critical points are relative minimums.

answer: 1 + 1 = 11

step-by-step explanation: