Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.
A) (1, 0); relative maximum value: f(1, 0) = 2
B) (0, 0); relative minimum value: f(0, 0) = 5
C) (0, - 1); saddle point: f(0, - 1) = - 5
D) there are no critical points

Question

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. 
A) (1, 0); relative maximum value: f(1, 0) = 2 
B) (0, 0); relative minimum value: f(0, 0) = 5 
C) (0, - 1); saddle point: f(0, - 1) = - 5 
D) there are no critical points

Quizplus · Accepted Answer

To find the critical points, we need to find where the gradient of the function is equal to zero or undefined. 
Taking the partial derivatives of the given function with respect to x and y:
$f_x(x,y) = 6x^2 - 6y$ and $f_y(x,y) = -6x + 6y$
Setting $f_x(x,y) = 0$ and $f_y(x,y) = 0$ simultaneously, we get $x=0$ and $y=0$.
Hence the critical point is (0,0).

To determine the nature of the critical point, we need to find the second-order partial derivatives:
$f_{xx}(x,y) = 12x$,
$f_{xy}(x,y) = f_{yx}(x,y) = -6$,
$f_{yy}(x,y) = 6$

At the critical point (0,0), 
$f_{xx}(0,0) = 0$,
$f_{xy}(0,0) = f_{yx}(0,0) = -6$,
$f_{yy}(0,0) = 6$

The discriminant is $D = f_{xx}(0,0)f_{yy}(0,0) - (f_{xy}(0,0))^2 = 0 - 36 < 0$. So, we cannot use the second derivative test to classify the nature of the critical point.

But we can see that $f(x,y)$ takes its minimum value at the critical point (0,0), which is $f(0,0) = 5$. So, the answer is choice letter B.

Find the Critical Point(s) of the Function