Interpretation of the Lagrange Multiplier

To interpret the Lagrange multiplier $\lambda$ consider the case where we have found that (x₀,y₀) is the optimum value of f(x,y) subject to the constraint $g(x,y) \le c$. Let's examine what happens if we vary the value of c (increase/decrease our budget.)

$$\frac{df}{fc} = \frac{\partial f}{\partial x} \frac{dx_0}{dc} + \frac{\partial f}{\partial y} \frac{dy_0}{dc}$$ (14)

Using the fact that $\mbox{grad}f = \lambda \mbox{grad}g$ at the point (x₀,y₀) we can rewrite this as

$$\frac{df}{dc} = \lambda \left( \frac{\partial g}{\partial x}... ...g}{\partial y} \frac{dy_0}{dc} \right) = \lambda \frac{dg}{dc}.$$ (15)

However at the critical point, g(x₀(c), y₀(c)) = c so dg/dc = 1 and hence, $df/dc = \lambda$. Thus $\lambda$ represents ``how much more bang you get for your buck.'' A more mathematical statement of this is

$$\frac{d}{dc} f(x_0(c), y_0(c)) = \lambda.$$ (16)