Given the expression \( f(x, y, z) = x^{-2} y^3 z^{-2} \), we aim to determine its maximum fractional error. The general formula for the fractional error in a function \( f(x, y, z) \) is:\[\text{fractional error in } f = \left| \frac{\partial f}{\partial x} \cdot \frac{\Delta x}{x} \right| + \left| \frac{\partial f}{\partial y} \cdot \frac{\Delta y}{y} \right| + \left| \frac{\partial f}{\partial z} \cdot \frac{\Delta z}{z} \right|\]For \( f(x, y, z) = x^{-2} y^3 z^{-2} \), the fractional errors in \( x \), \( y \), and \( z \) are 0.1, 0.2, and 0.5, respectively.
Calculating the partial derivatives:
- \( \frac{\partial f}{\partial x} = -2x^{-3} \)
- \( \frac{\partial f}{\partial y} = 3y^2 \)
- \( \frac{\partial f}{\partial z} = -2z^{-3} \)
The total fractional error is:\[\text{fractional error} = \left| -2x^{-3} \cdot \frac{\Delta x}{x} \right| + \left| 3y^2 \cdot \frac{\Delta y}{y} \right| + \left| -2z^{-3} \cdot \frac{\Delta z}{z} \right|\]Assuming the variables x, y, and z are positive, and considering the magnitudes of the terms, the formula simplifies to:\[\text{fractional error} = 2 \cdot \frac{\Delta x}{x} + 3 \cdot \frac{\Delta y}{y} + 2 \cdot \frac{\Delta z}{z}\]Substituting the given fractional errors:\[\text{fractional error} = 2 \cdot 0.1 + 3 \cdot 0.2 + 2 \cdot 0.5 = 0.2 + 0.6 + 1.0 = 1.8.\]Thus, the correct answer is \( 1.8 \).