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, 5) and (2, 5); saddle point: f(1, 5) = - 46; relative minimum value: f(2, 5) = - 47
B)(2, 5); relative maximum value: f(2, 5) = 47
C)(1, 5); saddle point: f(1, 5) = - 46
D)(1, 5) and (2, 5); relative maximum value: f(1, 5) = - 47; relative minimum value: f(2, 5) = 46
E)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, 5) and (2, 5); saddle point: f(1, 5) = - 46; relative minimum value: f(2, 5) = - 47 
B)(2, 5); relative maximum value: f(2, 5) = 47 
C)(1, 5); saddle point: f(1, 5) = - 46 
D)(1, 5) and (2, 5); relative maximum value: f(1, 5) = - 47; relative minimum value: f(2, 5) = 46 
E)there are no critical points

Quizplus · Accepted Answer

The function has two critical points at (1, 5) and (2, 5) where the first partial derivatives are both zero.

To classify the nature of each point using the second derivative test, we need to compute the Hessian matrix of second partial derivatives:

H(x, y) = [6x 2y; 2y 6x]

At (1, 5), we have H(1, 5) = [6 10; 10 6], which has eigenvalues λ1 = 16 and λ2 = -4. Since the eigenvalues have opposite signs, this critical point is a saddle point.

At (2, 5), we have H(2, 5) = [12 10; 10 12], which has eigenvalues λ1 = 22 and λ2 = 2. Since λ1 > 0 and λ2 > 0, this critical point is a relative minimum.

Therefore, the correct answer is A, which states that (1, 5) is a saddle point with f(1, 5) = -46, and (2, 5) is a relative minimum with f(2, 5) = -47.

Find the Critical Point(s) of the Function