Minimum Distance Using Lagrange MultipliersDate: 11/17/1999 at 20:27:48 From: Gloria Compise Subject: Lagrange Multipliers I'm trying to solve the following problem. Find the minimum distance from (0,0,c) to the cone z^2 = (x^2/a^2)+(y^2/b^2). So far what I've done is, f(x,y,z) = x^2 + y^2 + (z-c)^2 = x^2 + y^2 + (x^2/a^2) + (y^2/b^2) + c^2 - 2c{(x^2/a^2) + (y^2/b^2)}^1/2 I know that I'm supposed to use Lagrange Multipliers, but I'm having a difficult time. Can you help me? Thank you. Date: 11/18/1999 at 12:08:01 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 our problem is to find stationary points on f(x,y,z) = x^2 + y^2 + (z-c)^2 subject to constraint g(x,y,z) = x^2/a^2 + y^2/b^2 - z^2 = 0 At stationary points of f(x,y,z) we have df = part(df/dx)*dx + part(df/dy)*dy + part(df/dz)*dz = 0 This implies that the vector [part(df/dx), part(df/dy), part(df/dz)] is perpendicular to the vector [dx, dy, dz]. Since g(x,y,z) = 0 we can write dg = part(dg/dx)*dx + part(dg/dy)*dy + part(dg/dz)*dz = 0 Thus the vector [part(dg/dx), part(dg/dy), part(dg/dz)] is also perpendicular to the vector [dx, dy, dz]. This implies that the vector [part(df/dx), part(df/dy), part(df/dz)] is parallel to the vector [part(dg/dx), part(dg/dy), part(dg/dz)] and that we can find a scalar k such that [part(df/dx), part(df/dy), part(df/dz)] - k[part(dg/dx), part(dg/dy), part(dg/dz)] = [0, 0, 0] This can be summarized by writing phi(x,y,z) = f(x,y,z) - kg(x,y,z) 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,z) = 0. This gives four equations to find x, y, z and k. k is the Lagrange multiplier and phi is the auxiliary function. Applying these ideas to our problem, we have f(x,y,z)= x^2 + y^2 +(z-c)^2 and g(x,y,z) = x^2/a^2 + y^2/b^2 -z^2 The auxiliary function is phi(x,y,z) = f(x,y,z) - kg(x,y,z) = x^2+y^2+(z-c)^2 - k(x^2/a^2 + y^2/b^2 - z^2) Then: part(d(phi)/dx) = 2x -k(2x/a^2) = 0 ......................(1) part(d(phi)/dy) = 2y -k(2y/b^2) = 0 ......................(2) part(d(phi)/dz) = 2(z-c) -k(-2z) = 0 ......................(3) g(x,y,z) = x^2/a^2 + y^2/b^2 - z^2 = 0 ............(4) Now we must solve (1), (2), (3) and (4) for k, x, y and z. From (1) and (2) x and y can take any values and we could have: From (1) 1 - k/a^2 = 0, so k = a^2 then from (2) y = 0 From (3) z-c + kz = 0, so z(1+k) = c, z = c/(1+k) z = c/(1+a^2) From (4) x^2/a^2 + y^2/b^2 - c^2/(1+k)^2 = 0 x^2/a^2 + 0 - c^2/(1+a^2)^2 = 0 x^2/a^2 = c^2/(1+a^2)^2 and taking square roots x/a = c/(1+a^2) x = ac/(1+a^2) So a point on the cone nearest to (0,0,c) is x = ac/(1+a^2) y = 0 z = c/(1+a^2). - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/