How Can Lagrange Multipliers Be Used to Solve Optimization Problems in Calculus?

[Callisto]

This problem was given in my calc class during the semester,
"Find the lowest point on the intersection of the sphere x^2+y^2 +z^2 = 30 and the cone 2*x^2 +y^2 = c^2". I don't know how to solve this problem with lagrange multipliers.
How is it done?
Thanks!

Callisto

 

[marlon]

This problem was given in my calc class during the semester,
"Find the lowest point on the intersection of the sphere x^2+y^2 +z^2 = 30 and the cone 2*x^2 +y^2 = c^2". I don't know how to solve this problem with lagrange multipliers.
How is it done?
Thanks!

Callisto

first of all you need to determin of which function you are going to search the extrema (minima or maxima, each of these two can be searched separately as you know)...Can you find this function ? If that is done just apply the algorithm of Lagrangian multiplyers

regards
marlon

 

[HallsofIvy]

I assume the "c²" was a typo and the second equation is actually 2x²+ y²= z². You want to find the "lowest" point on the curve of intersection which means that you want to minimize z itself with the two constraints:
x²+ y²+ z²= 30 and 2x²+ y²= z².

Then if these as level surfaces of the functions g(x,y,z= x²+ y²+ z²= 30 and h(x,y,z)= 2x²+ y²- z²= 0. The gradient vector of the first is 2xi+ 2yj+ 2zk and the gradient vector of the second is 4xi+ 2yj- 2zk. The gradient of the object function f(x,y,z)= z is just k.
Using "Lagrange multipliers", we try to make that gradient a linear combination of the other two: k= λ(2xi+ 2yj+ 2zk)+ ρ(4xi+ 2yj- 2zk). That has 5 "unknowns", x, y, z, λ, and ρ (although you only need to find x, y, and z) and we have 5 equations:
the i component: 0= 2λx+ 4ρx
the j component: 0= 2λy+ 2ρy
the k component:1= 2λx- 2ρz
the equation of the sphere: x²+ y²+ z²d= 30
the equation of the cone: 2x²+ y²= z².

Solve those equations for x, y, and z.

 

[Callisto]

thanks for your replies'

from 0 = 2*lambda*x+ 4*rho*x
= 2*x(lambda +2*rho) = 0 when x = 0
if x = 0 then y^2 = z^2 so z = y = +/-sqrt(15)
so the min value is -sqrt(15),
that must mean the max value is sqrt(15)

is this right?

 

[marlon]

I assume the "c²" was a typo and the second equation is actually 2x²+ y²= z². You want to find the "lowest" point on the curve of intersection which means that you want to minimize z itself with the two constraints:
x²+ y²+ z²= 30 and 2x²+ y²= z².

Then if these as level surfaces of the functions g(x,y,z= x²+ y²+ z²= 30 and h(x,y,z)= 2x²+ y²- z²= 0. The gradient vector of the first is 2xi+ 2yj+ 2zk and the gradient vector of the second is 4xi+ 2yj- 2zk. The gradient of the object function f(x,y,z)= z is just k.
Using "Lagrange multipliers", we try to make that gradient a linear combination of the other two: k= λ(2xi+ 2yj+ 2zk)+ ρ(4xi+ 2yj- 2zk). That has 5 "unknowns", x, y, z, λ, and ρ (although you only need to find x, y, and z) and we have 5 equations:
the i component: 0= 2λx+ 4ρx
the j component: 0= 2λy+ 2ρy
the k component:1= 2λx- 2ρz
the equation of the sphere: x²+ y²+ z²d= 30
the equation of the cone: 2x²+ y²= z².

Solve those equations for x, y, and z.

ok this makes my post look quite stupid :)

i was under the impression that we were not allowed to give complete answers, or have the PF-rules changed ?

marlon