anonymous
  • anonymous
Find absolute extrema of f(x,y)= x^2+2xy+y^2, bound by |x|<=2, |y|<=1 Find absolute extrema of f(x,y)= x^2+2xy+y^2, bound by |x|<=2, |y|<=1 @Mathematics
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
now i got the maximum by putting my lines x=-2,2 and y=-1,1 however when i try to find the functions zeroes i get x=-y... am i supposed to put that into the equation and evaluate so.... x^2+2(-y)y+y^2=x^2?
anonymous
  • anonymous
and they get (x,-x,0) as an absolute minima
anonymous
  • anonymous
just wondering if i did right in which the function x^2 will always be positive... and what would happen if i got something that didn't exactly get 0?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Is this Calculus 2? I am doing single variable versions of this in Calculus 1 this week, too.
anonymous
  • anonymous
this is calculus 3... calculus two is more into surface area and integrals... calc 3 is multivariable
anonymous
  • anonymous
Right on. Bye the way, I thought you did great work helping that fellow rationalize the denom. Good teaching!
anonymous
  • anonymous
thanks

Looking for something else?

Not the answer you are looking for? Search for more explanations.