Step 1 :Let's denote the side length of the squares cut from the corners as \(x\).
Step 2 :The volume of the box can be represented by the function \(V(x) = x(2-2x)(9-2x)\).
Step 3 :To find the maximum volume, we need to find the maximum value of this function.
Step 4 :This can be done by finding the derivative of the function, setting it equal to zero, and solving for \(x\).
Step 5 :The derivative of the volume function is \(-2x(2 - 2x) - 2x(9 - 2x) + (2 - 2x)(9 - 2x)\).
Step 6 :The critical points of the function are \(\frac{11}{6} - \frac{\sqrt{67}}{6}\) and \(\frac{\sqrt{67}}{6} + \frac{11}{6}\).
Step 7 :The maximum volume is \(\left(-\frac{5}{3} + \frac{\sqrt{67}}{3}\right)\left(\frac{11}{6} - \frac{\sqrt{67}}{6}\right)\left(\frac{\sqrt{67}}{3} + \frac{16}{3}\right)\).
Step 8 :After simplifying, we find that the value of \(x\) that maximizes the volume is approximately \(0.59\) inches.
Step 9 :Substituting this value back into the volume function, we find that the maximum volume is approximately \(4.19\) cubic inches.
Step 10 :\(\boxed{\text{Final Answer: The value of } x \text{ that maximizes the volume is approximately } 0.59 \text{ inches and the maximum volume is approximately } 4.19 \text{ cubic inches.}}\)