Question

Find the maximum rate of change of f at the given point and the direction in which it occurs. f(x, y) = 4y sqrt (x) , (4, 5) maximum rate of change = ___________ direction vector = _________________

Answer 1

Answer:

Remember that the maximum rate of change of f at a point u is the length of the of the gradient vector evaluate in u, and the direction in which it occurs is in direction of the gradient vector evaluate in u.

The gradient vector of f is

$\triangledown f(x,y)=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})=(2yx^{-\frac{1}{2}}, 4\sqrt x)$

Then, the maximum rate of change is$|\triangledown f(4,5)|=|(2*5*4^{-\frac{1}{2}}, 4\sqrt 4)|=|(5,8)|=\sqrt{5^2+8^2}=\sqrt89$ in the direction of (5,8).