Critical point, Local max or min?

Question

Let z=(x²+y²)sin(2x+y). Show that (0,0 is a critical point. Is it a local max or min? 

Answer 1

For a critical point, both partial derivatives must be zero.

So, you have

∂z/∂x = 2(x²+y²)cos(2x+y) + 2x sin(2x+y)

∂z/∂y = (x²+y²)cos(2x+y) + 2y sin(2x+y)

Clearly, (0,0) satisfies the criterion.

Now, what happens near (0,0)?

z(0,0) = 0

since both x and y, as well as sin(2x+y) increase as x,y increase, z increases

But, as x and y decrease, sin(2x+y) decreases.

So, (0,0) is not a minimum nor a maximum.

In fact, there are several criteria for determining extrema at critical points

Suppose (a,b) is a critical point of f(x,y)

Let fxx, fxy, fyy be the second partial derivatives

Let D = fxx(a,b) fyy(a,b) − [fxy(a,b)]²

We then have the following classifications of the critical point.

If D>0 and fxx(a,b)>0 then there is a relative minimum at (a,b)

If D>0 and fxx(a,b)<0 then there is a relative maximum at (a,b)

If D<0 then the point (a,b) is a saddle point.

If D=0 then the point (a,b) may be a relative minimum, relative maximum or a saddle point. Other techniques would need to be used to classify the critical point.