A manufacturer cuts squares from the corners of a rectangular piece of sheet metal that measures 2 inches by 9 inches. (See Figure 1.) The manufacturer then folds the metal upward to make an open-topped box. (See Figure 2.) Letting $x$ represent the side-lengths (in inches) of the squares, use the ALEKS graphing calculator to find the value of $x$ that maximizes the volume enclosed by this box. Then give the maximum volume. Round your responses to two decimal places.
Figure 1
Figure 2
Value of $x$ that maximizes volume: in Maximum volume:
in $^{3}$

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 ：\(\boxed{\text{Final Answer: The value of } x \text{ that maximizes the volume is approximately } 0.59 \text{ inches and the maximum volume is approximately } 4.19 \text{ cubic inches.}}\)