what is the optimal condition of u(x1,x2) = 3x1+ 2x2
The optimal condition for a function u(x1, x2) = 3x1 + 2x2 occurs when the partial derivatives with respect to both x1 and x2 are equal to zero.
Mathematically, this can be expressed as:
∂u/∂x1 = 3 = 0
∂u/∂x2 = 2 = 0
Since the partial derivatives are constants and non-zero, there are no values of x1 and x2 that satisfy this condition. Therefore, the given function u(x1, x2) = 3x1 + 2x2 does not have an optimal condition.
To find the optimal condition of u(x1, x2) = 3x1 + 2x2, we need to determine the values of x1 and x2 that maximize or minimize the function.
To find the optimal condition, we can use the method of calculus called the first partial derivatives.
1. Compute the partial derivative of u with respect to x1:
∂u/∂x1 = 3
2. Compute the partial derivative of u with respect to x2:
∂u/∂x2 = 2
3. Set the partial derivatives equal to zero and solve for x1 and x2:
∂u/∂x1 = 3 = 0
∂u/∂x2 = 2 = 0
Since these equations do not have any solutions, we can conclude that there is no critical point or extreme value for u(x1, x2) = 3x1 + 2x2. Hence, there is no specific optimal condition for this function.