# How to find the point in whose direction the directional derivative is max without a unit vector but with restriction?

Given surface $z=3-3x^2-y^2$ find the point and the direction in which the directional derivative $D_uf$ is max on the circle $x^2+y^2=1$.

$\nabla f_z=\langle -6x,-2y\rangle$.

Because we're looking for a point which belongs to the circle we need to solve the system: $$\begin{cases} -3x^2-y^2=-3\\x^2+y^2=1 \end{cases}$$ so we get $y=0$ and $x=\pm1$. This means there're two points $(1,0)$ and $(-1,0)$ where the $D_uf$ is max.

Then we get the gradient in those points to be $\langle\pm6,0\rangle$ which means that the unit vector in which direction $D_uf$ is max is $\frac{\nabla f}{||\nabla f||}=\langle \pm1,0\rangle$.

The official solution regards this to be a finding extrema points problem, specifically the circle is regarded to be a restriction. Would it be correct to solve such exercise and not mention at all that we're finding function max/min? Can we just make do with the fact that point needs to belong to the circle so we solve the system?

• Note that there was a small error in my answer. I just fixed it. Aug 8 '17 at 18:28

I think there's been a misinterpretation of the problem. First of all, to be clear since this isn't written in the post, we have a function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $$f(x,y) = 3 - 3x^2 - y^2 \, .$$ The problem is to find the point in the plane such that $\lvert \nabla f \rvert$ is maximized subjected to the constraint that $x^2+y^2=1$.
However, you've written one of your constraint equations as $$3 - 3x^2 - y^2 = 0$$ which could also be written as $f(x,y)=0$. That's a much stronger constraint than what's in the problem, and I don't think it's what the problem intends.
There are a few ways to solve the problem. We could use Lagrange multipliers or we can parametrize the circle and use simple calculus. Let's do that latter. The thing we're trying to maximize is $$\lvert \nabla f \rvert = \lvert -6x\hat{x} -2y\hat{y} \rvert = \sqrt{36x^2 + 4y^2} \, .$$ To simplify our life, let's maximize $g(x,y) \equiv \lvert \nabla f \rvert^2$ instead: $$g(x,y) = 36x^2 + 4y^2 \, .$$ Now normally to find an extremum we'd compute $\partial g / \partial x$ and $\partial g / \partial y$, set them to zero, and solve for $x$ and $y$. However, we can't do that here because we have a constraint. Fortunately, it's easy to respect the constraint by putting $y^2 = 1 - x^2$, giving $$g(x) = 36x^2 + 4(1 - x^2) = 4 - 32x^2$$ so $$dg/dx =0 \longrightarrow x = 0 \, .$$