When I use the Lagrange method, how do I know when the multiplier wont be one of the FOC's?
In my Economics textbook (Varian), the author sometimes doesn't use dL/dLambda (=0) as one of the FOC's. I dont get why?
In my attached example, why dont the FOC's become
dL/dx1 = 0
dL/dG = 0
dL/dLambda = 0
dL/dMu = 0
Example without lambda:
Thanks for input!
FOC stands for First-Order Condition. FOCs are necessary conditions derived from first derivatives in constrained maximization/minimization problems. Consider for example the problem
$$\max f(x) \quad s.t. \quad g(x)=0$$
where $f\colon \mathbb {R} ^{n}\rightarrow \mathbb {R} $ is the objective function, and $\displaystyle g\colon \mathbb {R} ^{n}\rightarrow \mathbb {R} $ is the constraint function.
Under some regularity conditions, the Lagrange Multipliers Theorem tells us that a necessary condition for a point $x^*$ to be a solution of the above problem is that there exists some $\lambda\in\mathbb R$ such that
$$\nabla f(x^*)=\lambda\nabla g(x^*)$$
where $\nabla$ denotes the gradient.
In economics textbooks this is usually done by defining the Lagrangian function
$$\mathcal L(x,\lambda):=f(x)-\lambda g(x)$$
so that the previous FOC can be written as $\nabla_x \mathcal L(x^*,\lambda) =0$.
Now, the constraint condition $g(x^*)=0$ is obviously also a necessary condition, and setting the derivative of the Lagrangian w.r.t. $\lambda$ equal to zero gives precisely this constraint condition.
So the bottom line is that the derivative of $\mathcal L(x,\lambda)$ w.r.t. $\lambda$ doesn't give you any additional information, since we already know that the constraint must be satisfied at the optimum. (although this condition is often used with the other FOCs when solving for $x^*$).