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I am completely struck with the problem: Let $f$ be a function of three variables having continuously partial derivatives. For each direction vector $h=(h_1,h_2,h_3)$ such that $h_1^1+h_2^2+h_3^3=1$, Let $D_hf(x,y,z)$ be the directional derivative of $f$ along $h$ at $(x,y,z)$. For a point $(x_0,y_0,z_0)$ is not zero, mmaximize $D_hf(x_0,y_0,z_0)$ as a function off $h$

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up vote 2 down vote accepted

The maximum of the directional derivative at a point is reached in the direction of the gradient of the function at that point. $$ D_hf(x_0,y_0,z_0)=h\cdot\nabla f(x_0,y_0,z_0)=\|\nabla f(x_0,y_0,z_0)\|\cos(\langle\nabla f(x_0,y_0,z_0),h\rangle). $$ The maximum is reached when $\cos(\langle\nabla f(x_0,y_0,z_0),h\rangle)=1$, that is, when $h$ is in the direction of $\nabla f(x_0,y_0,z_0)$, so that $$ h=\frac{\nabla f(x_0,y_0,z_0)}{\|\nabla f(x_0,y_0,z_0)\|}. $$

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What you need is the following fact from Schwarz's inequality:

$$ |\langle x,y\rangle| \leq \|x\| \|y\| $$

with equality if and only if $x \parallel y$.

Hence since

$$ D_h f = \langle h, \nabla f\rangle \leq \|\nabla f\| \|h\| $$

we have that the maximizer only occurs when $h \parallel \nabla f $. (And obviously you want to choose the direction where the inner product is positive.)

In other words,

$$ h = \frac{ \nabla f}{\|\nabla f\|} $$

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