To find the maximum rate of change of the function $f(x, y, z) = 7x + 9yz$ at the point $(3, 7, -1)$, we use the gradient of the function $\nabla f(x, y, z)$. The gradient points in the direction of the maximum rate of change, and its magnitude gives the rate of change.

Step 1: Compute the partial derivatives

We need to compute the partial derivatives of $f(x, y, z)$ with respect to $x$, $y$, and $z$:

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(7x + 9yz) = 7$ $\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(7x + 9yz) = 9z$ $\frac{\partial f}{\partial z} = \frac{\partial}{\partial z}(7x + 9yz) = 9y$

Step 2: Evaluate the gradient at the point (3, 7, -1)

Now, substitute $x = 3$, $y = 7$, and $z = -1$ into the gradient components:

$\frac{\partial f}{\partial x} \bigg|_{(3,7,-1)} = 7$ $\frac{\partial f}{\partial y} \bigg|_{(3,7,-1)} = 9(-1) = -9$ $\frac{\partial f}{\partial z} \bigg|_{(3,7,-1)} = 9(7) = 63$

So, the gradient vector at the point $(3, 7, -1)$ is:

$\nabla f(3, 7, -1) = \langle 7, -9, 63 \rangle$

Step 3: Maximum rate of change

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

$|\nabla f(3, 7, -1)| = \sqrt{7^2 + (-9)^2 + 63^2} = \sqrt{49 + 81 + 3969} = \sqrt{4099}$

$|\nabla f(3, 7, -1)| \approx 64.03$

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:

$\mathbf{u} = \frac{\nabla f(3, 7, -1)}{|\nabla f(3, 7, -1)|} = \frac{1}{\sqrt{4099}} \langle 7, -9, 63 \rangle$

Thus, the maximum rate of change occurs in the direction of $\langle 7, -9, 63 \rangle$, and the unit vector in that direction is:

$\mathbf{u} \approx \langle 0.11, -0.14, 0.98 \rangle$

Final Answer:

• Maximum rate of change: $\approx 64.03$
• Direction: $\langle 7, -9, 63 \rangle$ or the unit vector $\mathbf{u} \approx \langle 0.11, -0.14, 0.98 \rangle$

Would you like further details or clarification?

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Here are 5 related questions:

1. What is the geometric interpretation of the gradient?
2. How would the rate of change differ if evaluated at a different point?
3. What is the significance of the unit vector in this context?
4. How can the gradient be applied to optimize functions in multiple dimensions?
5. What is the role of partial derivatives in understanding changes in multivariable functions?

Tip: The gradient not only provides the direction of steepest ascent but also the steepness itself (magnitude).