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
[tex]\triangledown f(x,y)=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})=(2yx^{-\frac{1}{2}}, 4\sqrt x)[/tex]
Then, the maximum rate of change is[tex]|\triangledown f(4,5)|=|(2*5*4^{-\frac{1}{2}}, 4\sqrt 4)|=|(5,8)|=\sqrt{5^2+8^2}=\sqrt89[/tex] in the direction of (5,8).
The maximum rate of change of f(x, y) at the given point (4, 5) and the direction is [tex]\sqrt{89}[/tex].
It is used in vector calculus and differential geometry. The function depends on two or more two variables. Then to differentiate with respect to x then we consider all the variables as a constant other than x.
The function is shown below.
[tex]\rm f(x, y) = 4y \sqrt{x}[/tex]
Then find the maximum rate of change of f(x, y) at the given point (4, 5) and the direction.
Then we know that
[tex]\rm \bigtriangledown f(x, y) = \dfrac{\partial f}{\partial x} , \dfrac{\partial f}{\partial y} = 2y x^{- \frac{1}{2}} , 4\sqrt{x}[/tex]
Then the maximum rate of change will be
[tex]\rm \bigtriangledown f(4, 5) = 2*5 4^{- \frac{1}{2}} , 4\sqrt{4} = |(5, 8)| = \sqrt{5^2 + 8^2} = \sqrt{89}[/tex]
Thus, the maximum rate of change of f(x, y) at the given point (4, 5) and the direction is [tex]\sqrt{89}[/tex].
More about the partial differentiation link is given below.
https://brainly.com/question/6758337
