Final answer:
To guarantee a local maximum at (0, 0, 0) for the function f with given second derivatives, the variable 'a' must be greater than -12. This is determined by ensuring that the Hessian matrix is negative definite at that point.
Step-by-step explanation:
The question addresses finding the range of values for the variable a that ensures a local maximum for a differentiable function f(x, y, z) given the second partial derivatives at a critical point (0,0,0).
The key to solving this problem is applying the second partial derivative test for functions of multiple variables. A critical point (0,0,0) is a local maximum if the Hessian matrix at that point is negative definite. The Hessian matrix H for f at (0,0,0) is given by:
H = \begin{bmatrix} -2 & 1 & a \\ 1 & -2 & 1 \\ a & 1 & -2 \end{bmatrix}
To have a negative definite matrix, all the leading principal minors of H need to be negative. The first leading principal minor is -2 (which is negative), the second is (-2) * (-2) - 1 * 1 = 4 - 1 = 3 (which is positive, but we need the even-order minors to be positive for a negative definite matrix), and the third leading principal minor is the determinant of H.
The determinant of H is \Delta = (-2)(-2)(-2) + 2(a)(1) - (1)(1)(a) - (1)(-2)(1) - (-2)(1)(1) - (a)(1)(1). Simplifying, we get \Delta = -8 + 2a - a - 2 - 2 - a = -12 - a. For H to be negative definite, \Delta must be negative, which implies a > -12.
Therefore, the range of values for a that guarantees (0, 0, 0) is a point of local maximum is a > -12.