Lagrange Multipliers. All variables cancel

[NoOne0507]

Homework Statement

A cannonball is heated with with temperature distribution T(x,y,z)=60(y²+z²-x²).

The cannonball is a sphere of 1 ft with it's center at the origin

a) Where are the max and min temperatures in the cannonball, and where do they occur?

Homework Equations

\nablaf=λ\nablag

Where g is the the constraint and λ is the common ratio.

\nabla= f_x i + f_y j + f_z k

The Attempt at a Solution

The cannonball is the restraint so
λ\nablaf = 2xλ i + 2yλ j + 2zλ k
\nablaT = -120x i + 120y j + 120z k

2xλ = -120x, λ= -60
2yλ = 120y, λ = 60
2zλ = 120z, λ = 60

I don't know where to go from here. All the variables canceled so I can't relate them to each other, and the λs are different.

 

[Dick]

2*x*lambda=(-120*x) doesn't mean lambda=(-60). It means either x=0 OR lambda=(-60). You have to consider both cases.

 

[NoOne0507]

But couldn't x be any value? For x ≠ 0, λ=-60. And then when x = 0 we get 0=0 with no λ.

If you do that for all the values you can get x=y=z=0, plug into the constraint and get the center of the circle, with no value for λ. How would that help find the max and the min?

 

[Dick]

But couldn't x be any value? For x ≠ 0, λ=-60. And then when x = 0 we get 0=0 with no λ.

If you do that for all the values you can get x=y=z=0, plug into the constraint and get the center of the circle, with no value for λ. How would that help find the max and the min?

x=y=z=0 is an important point. You should keep it in mind. It's an interior critical point. But yes, if x is nonzero then lambda=(-60). What does that tell you about y and z? If x is zero then lambda could be 60. Still doesn't tell you anything about y and z. But now you have to use the constraint y^2+z^2-x^2=1. You have to think this all through. You don't necessarily get a finite number of extrema on the boundary. There might be a lot of them.

 

[NoOne0507]

Well if x=0 then the constraint becomes y^2+z^2=1, with λ=60. So would the maxima occur anywhere in the yz unit circle? Which would yield (0, y, ±√(1-y^2)) as the maxima?

And then the smallest λ would occur at x≠0, and y=z=0? Which would give Minima on
x: [-1,0) U (0,1]
y=0
z=0

Or am I off-base here?

 

[Dick]

Well if x=0 then the constraint becomes y^2+z^2=1, with λ=60. So would the maxima occur anywhere in the yz unit circle? Which would yield (0, y, ±√(1-y^2)) as the maxima?

And then the smallest λ would occur at x≠0, and y=z=0? Which would give Minima on
x: [-1,0) U (0,1]
y=0
z=0

Or am I off-base here?

Getting closer. So the points to think about are (1,0,0), (-1,0,0), (0,y,z) along the circle where y^2+z^2=1 and the interior critical point at (0,0,0). What is the value of T in each of those cases?

 

[NoOne0507]

T(x,y,z)=60(y^2 + z^2 - x^2)
T(0,0,0) = 0
T(1,0,0) = -60
T(-1,0,0) = -60
T(0,y,z) = 60(y^2+z^2), but since this is along a circle T(0,y,z) = 60

Which gives the minimum at (1,0,0) and (-1,0,0), and the max at (0,y,z) where y^2+z^2=1.

Okay, so from (almost) the beginning:
λ = -60, x≠0, y=z=0
λ = 60, x=0

Since the minima will occur at the smallest λ, and x≠0 provides that λ, but it also requires y=z=0 for that to occur. This leaves T(x,0,0) = -60x² on -1≤x≤1 because of the unit sphere restriction. x=0 is a critical point, and the end points need to be checked, so T(-1,0,0) = T(1,0,0) = -60, and T(0,0,0)=0 The absolute min occurs at (-1,0,0) and (1,0,0). While (0,0,0) is a relative max (or the absolute max on the interval).

The maxima occur at x=0 since λ =60 when x=0. This leaves T(0,y,z)=60(y²+z²), with the restriction y²+z²≤1. The maxima are clearly the endpoints, y²+z²=1. So the maximum value for T occurs along that circle, which gives T(0,y,z)=60, where y²+z²=1.

Hey, it makes sense. Thank you.

 

[Ray Vickson]

T(x,y,z)=60(y^2 + z^2 - x^2)
T(0,0,0) = 0
T(1,0,0) = -60
T(-1,0,0) = -60
T(0,y,z) = 60(y^2+z^2), but since this is along a circle T(0,y,z) = 60

Which gives the minimum at (1,0,0) and (-1,0,0), and the max at (0,y,z) where y^2+z^2=1.

Okay, so from (almost) the beginning:
λ = -60, x≠0, y=z=0
λ = 60, x=0

Since the minima will occur at the smallest λ, and x≠0 provides that λ, but it also requires y=z=0 for that to occur. This leaves T(x,0,0) = -60x² on -1≤x≤1 because of the unit sphere restriction. x=0 is a critical point, and the end points need to be checked, so T(-1,0,0) = T(1,0,0) = -60, and T(0,0,0)=0 The absolute min occurs at (-1,0,0) and (1,0,0). While (0,0,0) is a relative max (or the absolute max on the interval).

The maxima occur at x=0 since λ =60 when x=0. This leaves T(0,y,z)=60(y²+z²), with the restriction y²+z²≤1. The maxima are clearly the endpoints, y²+z²=1. So the maximum value for T occurs along that circle, which gives T(0,y,z)=60, where y²+z²=1.

Hey, it makes sense. Thank you.

Of course, this problem is easy to solve without calculus. T = 60(y^2+z^2-x^2) <= 60(y^2+z^2+x^2) <= 60 in the sphere x^2+y^2+z^2<=1. Furthermore, we reach T=60 at any point on y^2+z^2=1, x=0. Also: T >= 60*(-x^2) >= -60, and we reach T=-60 at y=z=0, x=+-1. It is nice to see that the Lagrangian multiplier method dies, indeed, give such results.

RGV

 

[HallsofIvy]

Personally, I wouldn't use "Lagrange Multipliers" for this at all. The temperature function is T(x,y,z)= 60(y^2+ z^2- x^2)= 60(x^2+ y^2+ z^2- 2x^2) which in spherical coordinates is T(\rho, \theta, \phi)= 60(\rho^2- 2\rho^2sin^2\phi cos^2(\theta)). On the surface of the sphere, \rho= 1 so that is T(\theta, \phi)= 60(1- 2sin^2(\phi)cos^2(\theta).