The problem concerns analyzing the function $F(x, y) = 4z : y + z^2$ and involves calculating the approximate change in the function value and determining the relative percentage error. Let me extract the details and solve the problem step by step.

Question Breakdown:

1. Part a) Approximate the change in the function $F(x, y)$ at point $(2, 2)$, given that $x$ and $y$ increase by 10%. You are given: $\frac{\partial F}{\partial x}(2, 2) = 12, \quad \frac{\partial F}{\partial y}(2, 2) = 8$

   • Use these partial derivatives and the 10% increase in $x$ and $y$ to compute the approximate change in $F(x, y)$.

2. Part b) Compute the relative percentage error between the actual change in function value (denoted as $\Delta F(2, 2) = 21$) and the approximation calculated in part (a).

Solution for Part a)

We approximate the change in the function using the formula: $\Delta F \approx \frac{\partial F}{\partial x}(x, y) \cdot \Delta x + \frac{\partial F}{\partial y}(x, y) \cdot \Delta y$ Here, $\Delta x = 0.1x$ and $\Delta y = 0.1y$ because $x$ and $y$ are increased by 10%. At $(x, y) = (2, 2)$: $\Delta x = 0.1 \cdot 2 = 0.2, \quad \Delta y = 0.1 \cdot 2 = 0.2$ Substitute these values and the partial derivatives: $\Delta F \approx 12 \cdot 0.2 + 8 \cdot 0.2 = 2.4 + 1.6 = 4.0$

The approximate function change is: $\Delta F_{\text{approx}} = 4.0$

Solution for Part b)

The relative percentage error is computed as: $\text{Relative error} = \frac{\lvert \Delta F_{\text{actual}} - \Delta F_{\text{approx}} \rvert}{\Delta F_{\text{actual}}} \cdot 100$ Given $\Delta F_{\text{actual}} = 21$ and $\Delta F_{\text{approx}} = 4.0$, we substitute: [ \text{Relative error} = \frac{\lvert 21 - 4.0 \rvert}{21} \cdot 100 = \frac{17}{21