Find Min of f(x,y,z)=x^8+y^8+z^8 on x^4+y^4+z^4=4

[jenc305]

Find the points at which the function f(x,y,z)=x^8+y^8+z^8 achieves its minimum on the surface x^4+y^4+z^4=4.

I know
8x^7=(lamda)4x^3
8y^7=(lamda)4y^3
8z^7=(lamda)4z^3
x^4+y^4+z^4=4

Case1: x not equal to 0, y not equal to 0, and z not equal to 0
I get 3(4th root of 4/3 to the eigth)?

I'm I doing this right?

 

[HallsofIvy]

Are you sure this doesn't belong in the homework section?

Since grad f always points in the direction of fastest increase of f, to get to a minimum, you should go in the exact opposite direction. Of course, if you are required to stay on the surface x⁴+ y⁴+z⁴= 4, you can't do that. What you can do is go opposite to the component tangent to the surface. That works until you get to a point where grad f is perpendicular to the surface and has no component tangent- that's when you are at the minimum point on that surface.
If you let g(x,y,z)= x⁴+ y⁴+ z⁴ then the surface is a level surface of g- and grad g is perpendicular to the surface: we must have grad f parallel to grad g which means grad f= &lamda; grad g for some number &lamda;. That's the idea of the LaGrange multiplier.

Yes, your formulas are correct!

If x is NOT 0, then, dividing the first equation by 8x³, we have
x⁴= (1/2)λ. Similarly for y not 0 and z not 0: which means that, whatever &lamda; is, x= y= z so the x⁴+ y⁴+ z⁴=
3x⁴= 4 so x= y= z= (4/3)^1/4. The value you give is the value at that point- you were asked for the points themselves.

However, it is not necessarily true that "either x,y,z are all 0 or they are all non-zero"! If x= 0, y= 0 but z is NOT 0, then z= 4^1/4 so (0, 0, 4^1/4) is a possible place for a minimum. If x= 0 but y and z are not 0, then as before we can show that y= z so 2y⁴= 4 and (0, 2^1/4,2^1/4) is possible minimum. The same is true of "y and z 0 but not x", "y 0 but not x or z", etc.

 

[M98Ranger]

Yes, you are on the right track. To find the minimum of a function on a given surface, you can use the method of Lagrange multipliers. This involves solving a system of equations where the gradient of the function is equal to a scalar multiple of the gradient of the constraint (in this case, x^4+y^4+z^4=4).

In this case, we have the following equations:

8x^7 = λ4x^3
8y^7 = λ4y^3
8z^7 = λ4z^3
x^4 + y^4 + z^4 = 4

To simplify the equations, we can divide both sides by 4:

2x^7 = λx^3
2y^7 = λy^3
2z^7 = λz^3
x^4 + y^4 + z^4 = 1

Now, we can solve for λ in terms of x, y, and z:

λ = 2x^4 / x^3 = 2x
λ = 2y^4 / y^3 = 2y
λ = 2z^4 / z^3 = 2z

Since λ is the same for all three equations, we can equate them:

2x = 2y = 2z

This implies that x = y = z. Substituting this into the constraint equation, we get:

3x^4 = 4
x = (4/3)^(1/4)

Therefore, the minimum of the function f(x,y,z) = x^8 + y^8 + z^8 on the surface x^4 + y^4 + z^4 = 4 occurs at the point (x,y,z) = ((4/3)^(1/4), (4/3)^(1/4), (4/3)^(1/4)). Plugging this into the function, we get the minimum value of f(x,y,z) = 3(4/3)^(2/4) = 3(4/3)^(1/2) ≈ 3.266.

Note that this is the minimum value when x, y, and z are not equal to 0. If any of them is 0, then the minimum value of the function would be 0.