Lagrange MultipliersDate: 01/28/2001 at 04:56:22 From: li Subject: Lagrange Multipliers The temperature of a point(x,y,z) on the unit sphere is given by T(x,y,z)=xy+yz. Using Lagrange multipliers, find the temperature of the hottest point on the sphere. Date: 01/28/2001 at 06:03:18 From: Doctor Mitteldorf Subject: Re: Lagrange Multipliers Dear Li, The idea of Lagrange multipliers is this: For globabl maximization, you would set all the partial derivatives = 0 (dT/dx = 0, dT/dy = 0, dT/dz = 0). For Lagrange optimization, you set the vector of partial derivatives (dT/dx, dT/dy, dT/dz) equal to a constant multiple lambda times a vector that is orthogonal to your surface at every point. In this case, the vector (x,y,z) will do just fine. So you have the equation of the sphere plus three more equations in the four variables x,y,z, lambda. The three equations are dT/dx = lambda x dT/dy = lambda y dT/dz = lambda z - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 01/28/2001 at 08:14:51 From: Doctor Anthony Subject: Re: Lagrange Multipliers The general problem is to find stationary points of f(x,y,z) subject to constraint g(x,y,z) = 0 [Note that the constraint must be written in this form]. So for our problem g(x,y,z) = x^2 + y^2 + z^2 - 1 and the required equation is phi(x,y) = f(x,y) - kg(x,y) where k is the Lagrange multiplier and phi is the auxiliary function. Then f(x,y,z) will have a stationary point subject to constraint g(x,y,z) = 0 when part(d(phi)/dx) = 0, part(d(phi)/dy) = 0, part(d(phi)/dz) = 0 and g(x,y) = 0 This gives four equations to find x, y, z and k. Applying these ideas to our problem, we have f(x,y,z) = xy + yz and g(x,y,z) = x^2 + y^2 + z^2 - 1 The auxiliary function is phi(x,y,z) = f(x,y,z) - kg(x,y,z) = xy + yz - k[x^2 + y^2 + z^2 - 1] Then: part(d(phi)/dx) = y - 2kx = 0 (1) part(d(phi)/dy) = x + z - 2ky = 0 (2) part(d(phi)/dz) = y - 2kz = 0 (3) g(x,y,z) = x^2 + y^2 + z^2 - 1 = 0 (4) From (1) and (3) z = x and then from (2) 2x - 2ky = 0 so ky = x and returning to (1) y - 2k(ky) = 0 -> 1 - 2k^2 = 0 k^2 = 1/2 -> k = +-sqrt(1/2) We have x = z = ky and substituting into (4) k^2.y^2 + y^2 + k^2.y^2 - 1 = 0 y^2(2k^2 + 1) - 1 = 0 y^2(1 + 1) - 1 = 0 2y^2 = 1 y^2 = 1/2 Then x^2 = 1/4 and z^2 = 1/4 So the stationary points will be at [+-1/2, +-1/sqrt(2), +-1/2] It is obvious that the temperature, xy+yz, will be maximum if all signs are positive or all signs are negative. So max temp = (1/2)(1/sqrt(2)) + (1/sqrt(2))(1/2) = 1/sqrt(2) - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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