To find the maximum rate of change of the function $f(x, y) = 4y x^{1/2}$ at the point $(4, 3)$, we need to use the gradient of the function. The gradient vector $\nabla f(x, y)$ points in the direction of the maximum rate of change of the function, and its magnitude gives the rate of change.

Step 1: Compute the gradient of $f(x, y)$

The gradient is a vector of partial derivatives of $f(x, y)$ with respect to both $x$ and $y$:

$\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$

Partial derivative with respect to $x$:

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( 4y x^{1/2} \right) = 4y \cdot \frac{1}{2} x^{-1/2} = \frac{2y}{\sqrt{x}}$

Partial derivative with respect to $y$:

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( 4y x^{1/2} \right) = 4x^{1/2}$

So, the gradient of $f(x, y)$ is:

$\nabla f(x, y) = \left( \frac{2y}{\sqrt{x}}, 4x^{1/2} \right)$

Step 2: Evaluate the gradient at the point $(4, 3)$

Now, substitute $x = 4$ and $y = 3$ into the gradient:

$\nabla f(4, 3) = \left( \frac{2(3)}{\sqrt{4}}, 4 \cdot \sqrt{4} \right)$

$= \left( \frac{6}{2}, 4 \cdot 2 \right)$

$= (3, 8)$

Step 3: Maximum rate of change

The magnitude of the gradient vector gives the maximum rate of change:

$\left| \nabla f(4, 3) \right| = \sqrt{3^2 + 8^2} = \sqrt{9 + 64} = \sqrt{73}$

So, the maximum rate of change is $\sqrt{73}$.

Step 4: Direction of maximum rate of change

The direction of the maximum rate of change is given by the unit vector in the direction of the gradient. The unit vector is:

$\mathbf{u} = \frac{\nabla f(4, 3)}{\left| \nabla f(4, 3) \right|}$

Substitute the values:

$\mathbf{u} = \frac{(3, 8)}{\sqrt{73}}$

Thus, the direction vector is:

$\mathbf{u} = \left( \frac{3}{\sqrt{73}}, \frac{8}{\sqrt{73}} \right)$

Final Answer:

• Maximum rate of change: $\sqrt{73}$
• Direction vector: $\left( \frac{3}{\sqrt{73}}, \frac{8}{\sqrt{73}} \right)$

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Would you like further details on this process, or have any questions? Here are some related questions you may find interesting:

1. How do you interpret the components of a gradient vector?
2. What is the significance of the direction of the gradient in optimization problems?
3. How would the gradient change if the function were different (e.g., $f(x, y) = x^2 + y^2$)?
4. How do we find the direction of the minimum rate of change using the gradient?
5. What happens to the maximum rate of change if we change the values of $x$ and $y$?

Tip: The gradient vector always points in the direction of the steepest ascent (maximum increase) for the function.