Answer
$3.384\leq x \leq 3.387$
Work Step by Step
Step 1. Identify the given conditions: Ideal area is $9 in^2$, ideal diameter is $c=3.385in$, allowable error for area is $0.01in^2$, formula of area is $A=\pi(x/2)^2$, and we are looking for the $x$ interval which satisfies the inequality $|A-9|\leq 0.01$
Step 2. Start from the inequality above: $-0.01\leq A-9 \leq 0.01$ or $8.99\leq A \leq9.01$
Step 3. Use the area formula: $8.99\leq \pi(x/2)^2 \leq9.01$ or $\frac{35.96}{\pi}\leq x^2 \leq \frac{36.04}{\pi}$
Step 4. As $x$ represents the diameter and $x\geq0$, we have $\sqrt {\frac{35.96}{\pi}}\leq x \leq \sqrt {\frac{36.04}{\pi}}$
Step 5. Approximate the values: $\sqrt {\frac{35.96}{\pi}}\approx3.38326$ and $ \sqrt {\frac{36.04}{\pi}}\approx3.38702$
Step 6. Thus we can safely use the values that $3.384\leq x \leq 3.387$
