Question

maximum and minimum values f(x, y)=x^3+y^3+3xy​

Answer 1

Explanation:

You correctly found

fx =3x2−3y fy =−3x+3y2

and identified critical points as (0,0) and (1,1).

We can calculate also the second derivatives: fxx=6x, fxy=−3, fxy=6y. We have

D1 =6x D2 =| 6x −3 −3 6y |=36xy−9

We see that for x=y=1 we get D1>0 and D2>0. So the corresponding matrix is positive definite and this is a local minimum.

For x=y=0 we have D1=0, so we cannot say from this whether it is maximum or minimum. But it is relatively easy to see that it is neither of them, since we can find close to (0,0) points with positive values and points with negative values. (It suffices to check f(ε,0)=ε3 for small ε.)

We can also ask Wolfram Alpha to check for local extrema of x^3+y^3-3xy to verify whether our calculations are correct.