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anonymous
 5 years ago
Find the local maximum and minimum values and saddle point(s) of the function.
f(x, y) = x3y + 12x2  8y
anonymous
 5 years ago
Find the local maximum and minimum values and saddle point(s) of the function. f(x, y) = x3y + 12x2  8y

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got fx(x,y)= 3x^2y+24 fy(xy)= x^38

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0fxx (x,y) = 6xy+ 24 fyy(x,y)= 3x^2 so x=0,8, 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What do you mean x=0, 8, 2? What is that the result of?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0setting the fx and fy =0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am not sure of what i am suppose to do really past finding the partial derv.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1fx=3x^2y+24x (this was the only mistake i seen from above so far) fy=x^38 fxx=6xy+24 fyy=3x^2

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1i think you meant that since you did fxx right though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0From fx and fy, you should set each to 0 and solve as simultaneous equations and you get (x,y) critical points.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is where i got the values for x. those are the critical points, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Critical points are from fx and fy. Set each to 0; solve as simultaneous equations for (x,y) critical points.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1fx=0 implies 3x^2y+24x=0 3x(xy+8)=0 3x=0 => x=0 xy+8=0 => x=8/y fy=0 implies x^38=0 when x=2 x=2 => 2=8/y =>y=4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Evaluate fxx at critical points Evaluate fxy at critical points Evaluate fyy at critical points

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let's call above evaluations A (fxx) B (fxy) C (fyy)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1it looks that the only only critical number is (2,4)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If AC(B)^2>0 and A>0 local min If ACB^2>0 and A<0 local max If ACB^2<0 saddle point
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