Answer to Question #117910 in Calculus for Olivia

Question

Let f(x,y)=x^3+y^3−12xy, then, Select one:
a. f has a local maximum at (4,4)

b. f has a local minimum at (0,0)

c. f has a saddle at (0,0)

d. f has a saddle at (4,4)

Answer 1

$Since\ f(x,y)=x^3+y^3-12xy, \text{we have}\\ f_x=\frac{\partial f}{\partial x}=3x^2-12y,\\ f_y=\frac{\partial f}{\partial y}=3y^2-12x,\\ f_{xx}=\frac{\partial^2 f}{\partial x^2}=6x,\\ f_{yy}=\frac{\partial^2 f}{\partial y^2}=6y,\\ f_{xy}=\frac{\partial^2 f}{\partial y\partial x}=-12,\\ and\\ f_{yx}=-12.\\ Putting \ f_{x}=f_{y}=0, \\ \text{we obtain the critical points}\\ x=0,y =0 \ and\ x=4,y =4\\ at \ (x,y)=(0,0), \text{we get}\\ f_{xx} f_{yy}-f_{xy}^2=0-144=-144<0.\\ \text{This implies that} \\ f \ \text{has a saddle at } (0,0).\\ \text{So, the correct answer is}\ \ c$