Industrial Engineering and Operation Research -- regular points, linearly independence

Question

1. Find all regular points of the set {x ∈ R^4 | 2(x1 +x2 +x3)^3 + 3(x1 +x2 +x3)^2 = 1, x1^2 +x2^2 +x3^2 +x4 = 1}.

(( Just help me to look at the gradient of both constraints and determine if it is linearly independent.

Answer 1

Let f(x) = 2(x1 +x2 +x3)^3 + 3(x1 +x2 +x3)^2 - 1 and g(x) = x1^2 +x2^2 +x3^2 +x4 - 1 be the two constraints. Then, the gradients of these constraints are:

∇f(x) = [6(x1 +x2 +x3)^2 + 6(x1 +x2 +x3), 6(x1 +x2 +x3)^2 + 6(x1 +x2 +x3), 6(x1 +x2 +x3)^2 + 6(x1 +x2 +x3), 0]

∇g(x) = [2x1, 2x2, 2x3, 1]

To determine if these gradients are linearly independent, we can put them in a matrix and compute the determinant:

| 6(x1 +x2 +x3)^2 + 6(x1 +x2 +x3) 6(x1 +x2 +x3)^2 + 6(x1 +x2 +x3) 6(x1 +x2 +x3)^2 + 6(x1 +x2 +x3) 0 | | | | 2x1 2x2 2x3 1 |

Expanding the determinant along the first row, we get:

(6(x1 +x2 +x3)^2 + 6(x1 +x2 +x3)) [(2x2)(2x3) - 1] - (6(x1 +x2 +x3)^2 + 6(x1 +x2 +x3)) [(2x1)(2x3) - 1] + (6(x1 +x2 +x3)^2 + 6(x1 +x2 +x3)) [(2x1)(2x2) - 1]

Simplifying, we get:

72(x1 + x2 + x3)(x1^2 + x2^2 + x3^2) - 36(x1 + x2 + x3) - 12x1x2x3 + 3

Since we want the determinant to be nonzero for the gradients to be linearly independent, we need to solve the equation:

72(x1 + x2 + x3)(x1^2 + x2^2 + x3^2) - 36(x1 + x2 + x3) - 12x1x2x3 + 3 ≠ 0

Unfortunately, this equation is difficult to solve analytically, and we will need to resort to numerical methods or approximations.

However, we can make some observations about the problem. Note that the first constraint involves only the sum of x1, x2, and x3, and the second constraint involves only their squares. This suggests that the regular points will lie on the surface of a sphere centered at the origin, with the center of mass of x1, x2, and x3 as the point of intersection of the two constraints.