Gradient Calculation for Function

The question asks for the magnitude of the maximum rate of change of the given quadratic function $f(x_1,x_2) = x_1^2+2x_2^2+3x_1 + 3x_2 + x_1x_2 +1$ at the specific point $(1,1)$. The maximum rate of change of a scalar function of multiple variables at a given point occurs in the direction of the gradient vector, and its magnitude is equal to the magnitude of this gradient vector.

Partial Derivative Computation

To find the gradient vector, $ \nabla f $, we need to calculate the partial derivatives of the function $f(x_1, x_2)$ with respect to each variable, $x_1$ and $x_2$.

• Partial derivative with respect to $x_1$:

  We differentiate $f(x_1, x_2)$ with respect to $x_1$, treating $x_2$ as a constant:

  $ \frac{\partial f}{\partial x_1} = \frac{\partial}{\partial x_1} (x_1^2+2x_2^2+3x_1 + 3x_2 + x_1x_2 +1) $ $ \frac{\partial f}{\partial x_1} = 2x_1^{2-1} + 0 + 3x_1^{1-1} + 0 + 1 \cdot x_2 + 0 $ $ \frac{\partial f}{\partial x_1} = 2x_1 + 3 + x_2 $
• Partial derivative with respect to $x_2$:

  Similarly, we differentiate $f(x_1, x_2)$ with respect to $x_2$, treating $x_1$ as a constant:

  $ \frac{\partial f}{\partial x_2} = \frac{\partial}{\partial x_2} (x_1^2+2x_2^2+3x_1 + 3x_2 + x_1x_2 +1) $ $ \frac{\partial f}{\partial x_2} = 0 + 2 \cdot 2x_2^{2-1} + 0 + 3x_2^{1-1} + x_1 \cdot 1 + 0 $ $ \frac{\partial f}{\partial x_2} = 4x_2 + 3 + x_1 $

Evaluating Derivatives at the Point

Now, we substitute the coordinates of the given point $(1,1)$ into the partial derivative expressions we found.

• Evaluating $ \frac{\partial f}{\partial x_1} $ at $(1,1)$: $ \frac{\partial f}{\partial x_1}(1,1) = 2(1) + 3 + (1) $ $ \frac{\partial f}{\partial x_1}(1,1) = 2 + 3 + 1 = 6 $
• Evaluating $ \frac{\partial f}{\partial x_2} $ at $(1,1)$: $ \frac{\partial f}{\partial x_2}(1,1) = 4(1) + 3 + (1) $ $ \frac{\partial f}{\partial x_2}(1,1) = 4 + 3 + 1 = 8 $

The gradient vector $ \nabla f $ at the point $(1,1)$ is therefore $ \nabla f(1,1) = \left( \frac{\partial f}{\partial x_1}(1,1), \frac{\partial f}{\partial x_2}(1,1) \right) = (6, 8) $.

Magnitude Calculation

The maximum rate of change is the magnitude of the gradient vector $ \nabla f(1,1) $. The magnitude of a vector $ \mathbf{v} = (v_1, v_2) $ is calculated using the formula $ ||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2} $.

Applying this formula to our gradient vector $(6, 8)$: $ ||\nabla f(1,1)|| = \sqrt{(6)^2 + (8)^2} $ $ ||\nabla f(1,1)|| = \sqrt{36 + 64} $ $ ||\nabla f(1,1)|| = \sqrt{100} $ $ ||\nabla f(1,1)|| = 10 $

Thus, the magnitude of the maximum rate of change of the function $f(x_1, x_2)$ at the point $(1,1)$ is 10.