example needing two Lagrange multipliers
Find the semi-axes of the ellipse of intersection
, formed when the plane z=x+y intersects the ellipsoid
x24+y25+z225=1. |
Let (x,y,z) be any point of the ellipsoid. The square (http://planetmath.org/SquareOfNumber) x2+y2+z2 of the distance of this point from the midpoint (http://planetmath.org/Midpoint3) (0, 0, 0) has under the constraints
{g:=x24+y25+z225-1= 0,h:=x+y-z= 0 | (1) |
the minimum and maximum values at the end points of the semi-axes of the ellipse. Since we have two constraints, we must take equally many Lagrange multipliers, λ and μ. A necessary condition of the extremums
of
f:=x2+y2+z2 |
is that in to (1), also the equations
{∂f∂x+λ∂g∂x+μ∂h∂x= 2x+12xλ+μ= 0,∂f∂y+λ∂g∂y+μ∂h∂y= 2y+25yλ+μ= 0,∂f∂z+λ∂g∂z+μ∂h∂z= 2z+225zλ-μ= 0, | (2) |
are satisfied. I.e., we have five equations (1), (2) and five unknowns λ, μ, x, y, z.
The equations (2) give
x=-2μλ+4,y=-5μ2λ+10,z=25μ2λ+50, |
which expressions may be put into the equation h=0, and so on. One obtains the values
λ1=-10,λ2=-7517,μ1=±6√519,μ2=±14017√646 |
with which the extremum points (x,y,z) can be evaluated. The corresponding values of fare 10 and 7517, whence the major semi-axis is √10 and the minor semi-axis 5√25517.
Title | example needing two Lagrange multipliers |
---|---|
Canonical name | ExampleNeedingTwoLagrangeMultipliers |
Date of creation | 2013-03-22 18:48:18 |
Last modified on | 2013-03-22 18:48:18 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 7 |
Author | pahio (2872) |
Entry type | Example |
Classification | msc 51N20 |
Classification | msc 26B10 |
Synonym | using Lagrange multipliers to find semi-axes |
Related topic | ExampleOfLagrangeMultipliers |
Related topic | ExampleOfUsingLagrangeMultipliers |