Find the minimum anx maximum values of $$f(x,y) = 3r^2 + 3y^2 + 2ry + 1$$ on the closed disk $$r^2 + y^2 \le 1$$

Question

anonymous · Answer

uh... is there an error? Should it be f(r,y) ? The equation :
$$y^2 \le 1 - r^2$$

is not a disk...  it is a band -1<=y<=1 as x is any value?

anonymous · Answer

well its what question says, i took that question from ODTU's (Thecnical University Of Orta Dogu) calculus final exam :D if it was wrong, then it would not be in the exam -.-' over 5 professors makes that exam can't do a mistake like that i guess o.O if they do thatn o.O :D

anonymous · Answer

It could be a printing error.. just a typo. This seems like a good problem.. maybe if the disk we'e defines as x^2 + y^2 <= r^2.

The way to solve it is to take partial derivatives and find where slope of the surface is 0. Second derivates tell us if its a max or min, so its a problem in taking implicit derivatives. I can't see how to proceed with it as is.

anonymous · Answer

hmm

anonymous · Answer

If I had to guess what this problem should be is: replace all the r's in the equation with x's and the equation of the disk should be x^2 + y^2 <= 1 ... this is a good little problem .. dont need r at all...