# Lagrange Multiplier Problem with trivial gradient of $g$

Consider the Lagrange Multiplier problem where we want to maximize $f(x,y) = x$ subject to $g(x,y) = y^2+x^4-x^3 = 0$

Now, setting up $\nabla f(x,y) = \lambda \nabla g(x,y)$, we obtain $$\langle 1, 0 \rangle = \lambda \langle 4x^{3}-3x^{2},2y \rangle \\ \implies \begin{cases} 1& = &\lambda(4x^{3}-3x^{2}) \\ 0 & = &\lambda 2y\end{cases}$$

However, I am confused about 3 things: 1) How to proceed at this point if $\lambda = 0$. 2) If $\lambda \neq 0$, then $y = 0$, and we must have both that $x^{4} - x^{3} = 0$ and $4x^{3}-3x^{2} = 0$. But, in the case where $x \neq 0$ here, we would have that both $x = 1$ and $x = 3/4$ at the same time, which we can't have. So, I'm assuming then that if $\lambda \neq 0$, then $(0,0)$ is the only critical point? 3) I was told that this problem is a case where the gradient of $g$ is trivial. Trivial as in $\nabla g(x,y) = (0, 0)$? How is it trivial? Because the only critical point is $(0,0)$?

• what is the definition of your function $f(x,y)$? Nov 21 '17 at 11:23
• @DrSonnhardGraubner I'm sorry. I made a typo. It's fixed now.
– user100463
Nov 21 '17 at 12:48

2) If lambda is nonzero then y is 0. But $4x^3-3x^2 = 1/\lambda$ rather than 0. To proceed, we solve $x^4-x^3=0$, yielding $x=0$ or $x=1.$ If $x$ is 0, there is no lambda that satisfies the last equation, and if $x=1$, then $\lambda=1.$ This is the only solution.
3) Yes, trivial means the gradient of $g$ vanishes. There is no good reason to use the word trivial here--in fact the point of the article is that this "trivial" case does, in fact, occur.
The point of the article is that the Lagrange multiplier method only finds solutions where the gradient of g does not vanish, so you also need to check the points where the gradient of $g$ is 0 in addition to the solutions of the equations we were just looking at. In this example, that is $(0,0)$, and the minimum turns out to be one of those points, which is why it is a good example for the article.