Find the maximum rate of change of f at the origin The directional derivatives of $f$ at the origin in the directions of  $\overrightarrow v =\langle1,-1\rangle$ and $\overrightarrow w =\langle\sqrt{3}, 1\rangle$ are $-\sqrt{2}$ and $4 + 3\sqrt{3}$ respectively. Find the
maximum rate of change of $f$ at the origin.
Can someone help me answer? Or give me an idea or the steps I should do to answer this? 
Thank you! 
 A: The direction derivate along the vector $\langle a,b \rangle$ at origin is given as $$\langle a,b \rangle \cdot \langle f_x,f_y \rangle_{(0,0)} = af_x(0,0) + bf_y(0,0)$$
The direction derivate along the direction $\langle a,b \rangle$ at origin is given as $$\dfrac{\langle a,b \rangle}{\Vert \langle a,b \rangle \Vert} \cdot \langle f_x,f_y \rangle_{(0,0)} = \dfrac{af_x(0,0) + bf_y(0,0)}{\langle a,b \rangle}$$
You are given that $$\dfrac{f_x(0,0) - f_y(0,0)}{\sqrt{2}} = -\sqrt{2}$$ and $$\dfrac{\sqrt{3} f_x(0,0) + f_y(0,0)}2 = 4 + 3\sqrt{3}$$
Solving for $f_x(0,0)$ and $f_y(0,0)$ gives us $$f_x(0,0) = 6; \,\,\,\, f_y(0,0) = 8$$
The maximum rate of change is along $\vec{\nabla} f$ i.e. along the direction $\dfrac{\langle f_x,f_y \rangle}{\Vert \langle f_x,f_y \rangle \Vert}$. Hence, the maximum rate of change at the origin is $$\langle f_x,f_y \rangle \cdot \dfrac{\langle f_x,f_y \rangle}{\Vert \langle f_x,f_y \rangle \Vert} = \Vert \langle f_x,f_y \rangle \Vert = \sqrt{f_x(0,0)^2 + f_y(0,0)^2}$$ which gives us the answer as $10$.
