Solution:When the number of variables in an objective function is greater than two, it's no longer possible to graph the function.
Assume that
y = f (x₁, x₂..........xₙ) and the constraint (x₁, x₂.........xₙ) =0
The second partial derivative is f (= ∂²y/∂xᵢ ∂ⱼ)
where i,j = 1,2,3,....n.
The satisfaction of the first-order condition identifies some values as stationary values of the objective function. If, at a stationary value of y, we find that d₂y is positive definite, this is sufficient to establish that
value of y as a minimum.
Similarly, the negative definiteness of d₂y is a sufficient condition for the stationary value to be a maximum.
This raises the question of how to express d²y and how to determine its positive or negative definiteness when the function has three variables. For this purpose, the Hessian matrix is used, which is as follows:
Whose leading principal minors can be written as follows:
Thus, based on the determinant criteria for positive and negative definiteness, we can state the second-order sufficient condition for a maximum as follows:

Using this condition, we must evaluate all the leading principal minors at the stationary point, where f₁ = f₂ = f₃ = 0.