For function $f ( x , y ) = 8 - \frac { x } { 8 } - \frac { y } { 7 } , \text { find the maximum value of the directional }$ derivative at (3,2).
A) $\sqrt { 15 } / 8$
B) $\sqrt { 113 } / 56$
C) $\sqrt { 113 } / 224$
D) $\sqrt { 113 } / 112$
E) $\sqrt { 113 } / 168$

Question

For function $f ( x , y ) = 8 - \frac { x } { 8 } - \frac { y } { 7 } , \text { find the maximum value of the directional }$ derivative at (3,2). 
A) $\sqrt { 15 } / 8$
B) $\sqrt { 113 } / 56$
C) $\sqrt { 113 } / 224$
D) $\sqrt { 113 } / 112$
E) $\sqrt { 113 } / 168$

Quizplus · Accepted Answer

The maximum value of the directional derivative of a function at a point is given by the magnitude of the gradient vector at that point. First, we find the partial derivatives of $f(x, y)$ with respect to $x$ and $y$, respectively:$$f_x = -\frac{1}{8}, \quad f_y = -\frac{1}{7}.$$Thus, the gradient vector of $f$ at any point $(x, y)$ is $\nabla f = \left(-\frac{1}{8}, -\frac{1}{7}\right)$. The magnitude of this gradient vector is$$\|\nabla f\| = \sqrt{\left(-\frac{1}{8}\right)^2 + \left(-\frac{1}{7}\right)^2} = \sqrt{\frac{1}{64} + \frac{1}{49}} = \sqrt{\frac{113}{64 \cdot 49}} = \sqrt{\frac{113}{3136}} = \frac{\sqrt{113}}{56}.$$Therefore, the maximum value of the directional derivative at the point (3,2) is $\frac{\sqrt{113}}{56}$.

For Function f(x,y)=8−x8−y7, find the maximum value of the directional f ( x , y ) = 8 - \frac { x } { 8 } - \frac { y } { 7 } , \text { find the maximum value of the directional }f(x,y)=8−8x​−7y​, find the maximum value of the directional

For Function $f ( x , y ) = 8 - \frac { x } { 8 } - \frac { y } { 7 } , \text { find the maximum value of the directional }$