Answer
$u_{max}= \lt -\dfrac{1}{\sqrt 2}, \dfrac{1}{\sqrt 2} \gt$
and $u_{min}= \lt \dfrac{1}{\sqrt 2}, -\dfrac{1}{\sqrt 2} \gt$
$D_u f_{max} =\sqrt 2$
and $D_u f_{min} =- \sqrt 2$
Work Step by Step
$ f_x= 2x+y$ and $ f_y= x+2y $
Now, $f_x (-1,1) = 2x+y=2(-1)+1=-1$ and $f_y (-1,1) =x+2y=-1+2(1)=1$
$u_{max}=\dfrac{\nabla f (-1,1)}{|\nabla f (-1,1) |}= \lt -\dfrac{1}{\sqrt 2}, \dfrac{1}{\sqrt 2} \gt$
and $u_{min}=- u_{max}= \lt \dfrac{1}{\sqrt 2}, -\dfrac{1}{\sqrt 2} \gt$
We know that $D_u f = \nabla f \cdot u$
Therefore, $D_u f_{max} =\nabla f (-1,1) \cdot u_{max} =\lt -1, 1 \gt \cdot \lt -\dfrac{1}{\sqrt 2}, \dfrac{1}{\sqrt 2} \gt =\sqrt 2$
and $D_u f_{min} =\nabla f (-1,1) \cdot u_{min} =\lt -1, 1 \gt \cdot \lt \dfrac{1}{\sqrt 2}, -\dfrac{1}{\sqrt 2} \gt =- \sqrt 2$
