Lagrange Multipliers and Lambda We know that: > The upshot of all this is the following: at a local maximum, the gradient of $f$ and the gradient of $g$ are pointing in the same direction. In other words, they are proportional. In other words, there's some constant $\lambda$ such that the gradient of $f$ is $\lambda$ times the gradient of $g$. That's it. This $\lambda$ is your Lagrange multiplier. Why is it that $\nabla f = \lambda\nabla g$, not $\lambda \nabla f = \nabla g?$ If $\nabla f $ is proportinal to $\nabla g$, that is to say that $\nabla g$ is proportional to $\nabla f$. So what's stopping us from placing lambda in front of $\nabla f$?

Remember "λ" is defined to be a constant.
And as such either 1/λ and λ has no specified differences.
Say in your case ∇f=λ∇g. If we let λ' = λ, its correct to establish λ'∇f=∇g

Besides there is one little reminder, in order to prevent confusion. We usually retain the habit of using **∇f=λ∇g** when we look for the **extreme points of 'f'**