Solve the Problem $\mathrm { A } = \pi \mathrm { D } ^ { 2 } / 4$

Solve the problem.
-The cross-sectional area of a cylinder is given by $\mathrm { A } = \pi \mathrm { D } ^ { 2 } / 4$ , where $\mathrm { D }$ is the cylinder diameter. Find the tolerance range of $\mathrm { D }$ such that $| \mathrm { A } - 10 | < 0.01$ as long as $\mathrm { D } _ { \min } < \mathrm { D } < \mathrm { D } _ { \max }$ .

A) $\mathrm { D } _ { \min } = 3.567 , \mathrm { D } _ { \max } = 3.570$
B) $\mathrm { D } _ { \min } = 3.567 , \mathrm { D } _ { \max } = 3.578$
C) $\mathrm { D } _ { \min } = 3.558 , \mathrm { D } _ { \max } = 3.570$
D) $\mathrm { D } _ { \min } = 3.558 , \mathrm { D } _ { \max } = 3.578$