Maximization of f(x,y) over a ConstraintDate: 03/04/99 at 15:38:44 From: Skylar Cox Subject: Maximization of f(x,y) over a constraint I have to find the maximum value of f(x,y) = xy^2 on the ellipse x^2/a^2 + y^2/b^2 = 1. I know that the gradient of the function f is equal to lambda times the gradient of g (ellipse). I get a couple of equations and I am having trouble finding any points that will satisfy them. Please help. Thanks. Date: 03/04/99 at 19:17:53 From: Doctor Anthony Subject: Re: Maximization of f(x,y) over a constraint The general problem is to find stationary points of f(x,y) subject to constraint g(x,y) = 0 [note that the constraint must be written in this form]. So for our problem g(x.y) = x^2/a^2 + y^2/b^2 - 1 At stationary points of f(x,y) we have: df = part(df/dx)*dx + part(df/dy)*dy = 0 This implies that the vector [part(df/dx), part(df/dy)] is perpendicular to the vector [dx, dy] Since g(x,y) = 0 we can write dg = part(dg/dx)*dx + part(dg/dy)*dy = 0 Thus the vector [part(dg/dx), part(dg/dy)] is also perpendicular to the vector [dx, dy]. This implies that the vector [part(df/dx), part(df/dy)] is parallel to the vector [part(dg/dx), part(dg/dy)] and that we can find a number 'k' such that [part(df/dx), part(df/dy)] - k[part(dg/dx), part(dg/dy)] = [0, 0] This can be summarized by writing phi(x,y) = f(x,y) - kg(x,y) Then f(x,y) will have a stationary point subject to constraint g(x,y) = 0 when part(d(phi)/dx) = 0, part(d(phi)/dy) = 0 and g(x,y) = 0 This gives three equations to find x, y and k. k is the Lagrange multiplier and phi is the auxiliary function. Applying these ideas to our problem, we have: f(x,y)= xy^2 and g(x,y) = x^2/a^2 + y^2/b^2 - 1 The auxiliary function is phi(x,y) = f(x,y) - kg(x,y) = xy^2 - k(x^2/a^2 + y^2/b^2 - 1) Then: part(d(phi)/dx) = y^2 - 2kx/a^2 = 0 (1) part(d(phi)/dy) = 2xy - 2ky/b^2 = 0 (2) g(x,y) = x^2/a^2 + y^2/b^2 - 1 = 0 (3) Solving (1), (2) and (3) for k, x and y we get From (2) x = k/b^2 and then from (1) y = +-k.sqrt(2)/(ab) Putting these values into (3) k^2/(a^2.b^4) + 2k^2/(a^2.b^4) = 1 3k^2 = a^2.b^4 k = +- a.b^2/sqrt(3) So, x = +-ab^2/(b^2.sqrt(3)) = +-a/sqrt(3) y = +-a.b^2.sqrt(2)/(ab.sqrt(3) = +- b.sqrt(2/3) and then max value of f(xy) = xy^2 = [a/sqrt(3)][b^2.(2/3)] = 2ab^2/[3sqrt(3)] - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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