The area of a circle is to be computed from a measured value of its diameter. Estimate the maximum permissible percentage error in the diameter measurement if the percentage error in the area must be kept within \( \pm 5 \% \). Enter the exact answer. Maximum Error \( = \pm \)

To estimate the maximum permissible percentage error in the diameter measurement, we need to consider the relationship between the diameter and the area of a circle. The formula for the area of a circle is given by: \[ A = \pi \times \left( \frac{d}{2} \right)^2 \] where: - \( A \) is the area of the circle, - \( d \) is the diameter of the circle, - \( \pi \) is the mathematical constant approximately equal to 3.14159. Given that the percentage error in the area must be kept within \( \pm 5\% \), we can express this as: \[ \frac{\Delta A}{A} \times 100\% \leq 5\% \] where \( \Delta A \) is the maximum permissible error in the area. Substitute the formula for the area of a circle into the above inequality: \[ \frac{\Delta A}{\pi \times \left( \frac{d}{2} \right)^2} \times 100\% \leq 5\% \] Simplify the inequality: \[ \frac{\Delta A}{\frac{\pi d^2}{4}} \times 100\% \leq 5\% \] \[ \frac{4 \Delta A}{\pi d^2} \times 100\% \leq 5\% \] \[ \frac{4 \Delta A}{\pi d^2} \leq 0.05 \] Now, we need to find the maximum permissible percentage error in the diameter measurement, which is denoted as \( \Delta d \). We can express this as: \[ \frac{\Delta d}{d} \times 100\% \] Since the diameter is related to the area through the formula \( A = \pi \times \left( \frac{d}{2} \right)^2 \), we can express the diameter in terms of the area: \[ d = 2 \sqrt{\frac{A}{\pi}} \] Substitute this expression for \( d \) into the inequality: \[ \frac{4 \Delta A}{\pi \times \left( 2 \sqrt{\frac{A}{\pi}} \right)^2} \leq 0.05 \] \[ \frac{4 \Delta A}{\pi \times 4 \times \frac{A}{\pi}} \leq 0.05 \] \[ \frac{\Delta A}{A} \leq 0.05 \] Therefore, the maximum permissible percentage error in the diameter measurement is \( \pm 5\% \).