Lagrange Multiplier. Dealing with f(x,y) =xy^2

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The discussion focuses on applying the Lagrange Multiplier method to the function f(x,y) = xy² under a specific constraint. Participants clarify that the equations derived from the gradients, such as 5 = 2λx and -3 = 2λy, are essential for finding critical points. The confusion arises when dealing with products of variables, as in the case of f(x,y) = xy², where the correct approach involves calculating the partial derivatives ∂f/∂x and ∂f/∂y. This method ensures accurate identification of maximum and minimum values subject to constraints.

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King_Silver
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Given a question like this:
Findhe maximum and minimum of [PLAIN]http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers_files/eq0043M.gif[PLAIN]http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers_files/empty.gif subject to the constraint [PLAIN]http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers_files/eq0044M.gif[PLAIN]http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers_files/empty.gif.

I know that 5 = 2λx, -3 = 2λy.

However, if I am given a question where f(x,y) = xy2 how would I write this?
There is no sign separating the x and the y so I cannot simply presume what sign it is.

Would 1 = 2λx and 1 = 2λy be incorrect?
If so, why? and how is this sort of question dealt with when the x and y are being multiplied together instead of being subtracted? cheers
 
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For the Lagrange method you look at the gradient of ##f - \lambda g##.
So all you have to do is determine the partial derivatives ##\partial f\over \partial x ## and ##\partial f\over \partial y ## . These aren't constants any more in this case.
 
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