Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am asked to find all local extreme values & saddle points of

$$f(x,y) = 2x^2 + y^2 - xy - 7y + 8$$

$$f_x(x, y) = 4x-y, \qquad f_y(x,y) = 2y-x-7$$

$$f_x(x,y) = 0, \qquad y = 4x$$

$$f_y(x,y) = 0, \qquad 2y-x-7 = 0, \qquad x = 1$$

So I have a critical point at $(1,4)$. Then I use 2nd derivative test to check min/max

$$f_{xx}(x,y) = 4, \qquad f_{yy}(x,y) = 2, \qquad f_{xy}(x,y) = -1$$

$$H(x,y) = 4\times 2 + (-1)^2 = 9$$

$H(1,4) > 0$, $f_{xx} > 0$ so local min. Answer given is "Local min $-6$ at $(1,4)$". What does $-6$ refer to?

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

$$f(1,4)=2+16-4-28+8=26-32=-6$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.