Lagrange Multipliers
- Thread starter afung22
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Why are you finding all those?
From f(x,y) and the contstraint, create:
\(\displaystyle H(x,y,\lambda) = f(x,y)-\lambda (x^{2}+y^{2}-4)\)
Find \(\displaystyle H_{x}\), \(\displaystyle H_{y}\), and \(\displaystyle H_{\lambda}\)
Solve for \(\displaystyle x\), \(\displaystyle y\), and \(\displaystyle \lambda\) so that the three partial derivatives are all simultaneously zero (0).
From f(x,y) and the contstraint, create:
\(\displaystyle H(x,y,\lambda) = f(x,y)-\lambda (x^{2}+y^{2}-4)\)
Find \(\displaystyle H_{x}\), \(\displaystyle H_{y}\), and \(\displaystyle H_{\lambda}\)
Solve for \(\displaystyle x\), \(\displaystyle y\), and \(\displaystyle \lambda\) so that the three partial derivatives are all simultaneously zero (0).
HallsofIvy
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What your professor keeps telling you is equivalent to what tkhunny is saying.
To find maximum of minimum of f(x,y) subject to the constraint g(x,y)= constant, you can set the derivatives of \(\displaystyle f- \lambd a\) and set the equal to 0 or set the derivatives of f and \(\displaystyle \lambda g\) equal to each other.
\(\displaystyle (f- \lambda g)_x= f_x- \lambda g_x= 0\) is the same as \(\displaystyle f_x= \lambda g_x\).
Geometrically, this is saying that the gradient of f is parallel to the gradient of g.
To find maximum of minimum of f(x,y) subject to the constraint g(x,y)= constant, you can set the derivatives of \(\displaystyle f- \lambd a\) and set the equal to 0 or set the derivatives of f and \(\displaystyle \lambda g\) equal to each other.
\(\displaystyle (f- \lambda g)_x= f_x- \lambda g_x= 0\) is the same as \(\displaystyle f_x= \lambda g_x\).
Geometrically, this is saying that the gradient of f is parallel to the gradient of g.
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