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Math Help - Lagrange Multiplier

  1. #1
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    Lagrange Multiplier

    Hey I got a question on the Lagrange Multiplier. Let's say the constraint is x+y=3 and the function is f(x,y) = 3x^2+y^2. Now if we think about this geometrically, there will just be a line diagonal line on the surface of f. Therefore, no gradient vector of f will have the same direction as the gradient vector of x+y=3. Does that mean there is no maximum or minimum? Does this also mean that the constraint has to be a closed function? (Like a circle or something instead of a line)?
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  2. #2
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    Quote Originally Posted by lilaziz1 View Post
    Hey I got a question on the Lagrange Multiplier. Let's say the constraint is x+y=3 and the function is f(x,y) = 3x^2+y^2. Now if we think about this geometrically, there will just be a line diagonal line on the surface of f.
    This statement is wrong. In fact, it doesn't even make sense. There is NO line "on the surface of f". In fact, there is no "surface" defined by f(x, y)= 3x^2+ y^2 You seem to be assuming that they are talking about the paraboloid "z= f(x,y)" but that is not suggested here. This is just a problem in the xy-plane. You are asked to find points on the line g(x, y)= x+ y= 3 such that f(x,y) has maximum or minimum values. There is no reason to interpret f "geometrically".
    \nabla f= 6x\vec{i}+ 2y\vec{j}= \lambda \nabla g= \lambda(\vec{i}+ \vec{j})

    That is, we must have 6x= \lambda and 2y= \lambda. Dividing the second equation by the first, \frac{2y}{6x}= \frac{y}{3x}= 1 so y= 3x. x+ y= x+ 3x= 4x= 3 so x= 3/4 and y= 9/4. (3/4, 9/4) is the only critical point for f(x,y) on this line.

    Therefore, no gradient vector of f will have the same direction as the gradient vector of x+y=3. Does that mean there is no maximum or minimum? Does this also mean that the constraint has to be a closed function? (Like a circle or something instead of a line)?
    In order to be sure that a given function has both maximum and minimum values on a set, that set must be a "closed, bounded" set but that is not what is happening here.
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  3. #3
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    According to Stewart, the constraint g(x,y) = k is projected on z = f(x,y): http://i180.photobucket.com/albums/x...Multiplier.png The first picture (on the left) consists of the constraint g(x,y) = k and a level curve of f(x,y). As you can see, there are two points where the gradient vector has the same direction as the level curve's gradient vector; thus, those two points are min./max. points (min if you look @ the 3D). Now on the 3D drawing on the right, the drawing consists of g(x,y) = k projected/slapped onto f(x,y). The orange plane is just where the level curve is in the 3D drawing. That is the "geometric" representation.

    Now as you can see, g(x,y) = k is some sort of closed function (meaning it's a circle/eclipse). Now if g(x,y) = k was a line, like x+y=3 for example, it would just be a line running across the f(x,y) in the 3D perspective when it is projected/slapped onto the function. When that happens, there would not be any point on any level curve that would have the same gradient vector as the gradient vector of g(x,y) = k.

    Finally, this brings me to my initial questions: Does that mean there is no maximum or minimum? Does this also mean that the constraint has to be a closed function? (Like a circle or something instead of a line)?
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