Consider the following. \[ \begin{array}{r}f(x, y)=8 x-6 y \\ \text { (a) Find } f(1,8) \text { and } f(1.1,8.05) \text { and calculate } \Delta z \\ f(1,8)=\square \\ f(1.1,8.05)=\square \\ \text { (b) Use the total differential } d z \text { to approximate } \Delta z \\ d z=\end{array} \]

To solve the problem step-by-step, let's first calculate \( f(1, 8) \) and \( f(1.1, 8.05) \). 1. **Calculate \( f(1, 8) \)**: \[ f(1, 8) = 8(1) - 6(8) = 8 - 48 = -40 \] 2. **Calculate \( f(1.1, 8.05) \)**: \[ f(1.1, 8.05) = 8(1.1) - 6(8.05) = 8.8 - 48.3 = -39.5 \] Now we can calculate \( \Delta z \): \[ \Delta z = f(1.1, 8.05) - f(1, 8) = -39.5 - (-40) = 0.5 \] Thus, we have \( f(1, 8) = -40 \), \( f(1.1, 8.05) = -39.5 \), and \( \Delta z = 0.5 \). (b) Now we will find the total differential \( dz \) to approximate \( \Delta z \). To find \( dz \), we need to find the partial derivatives: \[ \frac{\partial f}{\partial x} = 8, \quad \frac{\partial f}{\partial y} = -6 \] Now let's compute \( dz \): \[ dz = \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y \] where \( \Delta x = 1.1 - 1 = 0.1 \) and \( \Delta y = 8.05 - 8 = 0.05 \). Plugging in the values: \[ dz = (8)(0.1) + (-6)(0.05) = 0.8 - 0.3 = 0.5 \] In conclusion: \[ f(1, 8) = -40, \quad f(1.1, 8.05) = -39.5, \quad \Delta z = 0.5, \quad dz = 0.5 \]