Question

Find the maximum rate of change of f at the given point and the direction in which it occurs. f(p, q) = 8qe−p + 4pe−q, (0, 0) maximum rate of change $$ Correct: Your answer is correct. direction $$

Answer 1

Answer:

Step-by-step explanation:

$Given \ f (p,q) = 8qe^{-p} + 4pe^{-q} \ where \ P_o(0,0) \\ \text{thus the gradient of f is: } \\ \\ \bigtriangledown f(p,q) = \Big(\dfrac{\partial f}{\partial p}, \dfrac{\partial f}{\partial q} \Big) \\ \\ \dfrac{\partial f}{\partial p} = -8qe^{-p} + 4pe^{-q} \\ \\ \dfrac{\partial f}{\partial p} = 8qe^{-p} - 4pe^{-q} \\ \\ Then: \bigtriangledown f(p.q) = (-4qe^{-p}+ 8qe^{-q}, 4qe^{-p}- 8qe^{-q}) \\ \\ f(0,0) = (-4*(0)e^{-0}+ 8*(0)e^{-(0)}, 4*(0)e^{-(0)}- 8*(0)e^{-(0)}) \\ \\ = (0+8,4-0) = (8.4)$

$\Big| \Big | \bigtriangledown f(0,0) = \sqrt{(8)^2+4^2} \\ \\ = \sqrt{64+16} \\ \\ = \sqrt{80}$

$\mathbf{the \ direction \ of \ maximum \ change \ is }= \mathbf{\sqrt{80}} \\ \\ \mathbf{direction }(8,4)$