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anonymous
 4 years ago
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
anonymous
 4 years ago
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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now 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
 4 years ago
Best ResponseYou've already chosen the best response.0and they get (x,x,0) as an absolute minima

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is this Calculus 2? I am doing single variable versions of this in Calculus 1 this week, too.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is calculus 3... calculus two is more into surface area and integrals... calc 3 is multivariable

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Right on. Bye the way, I thought you did great work helping that fellow rationalize the denom. Good teaching!
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