Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard $43 \mathrm{in}$. by $23 \mathrm{in}$. by cutting congruent squares from the corners and folding up the sides. Then find the volume. The dimensions of box of maximum volume are (Round to the nearest hundredth as needed. Use a comma to separate answers as needed.)

Step 1 ：Given a sheet of cardboard of dimensions 43 inches by 23 inches, we are to cut out congruent squares from each corner and fold up the sides to form an open box. The goal is to find the dimensions of the box that will give the maximum volume.

Step 2 ：The volume V of a box is given by the formula \(V = lwh\), where l is the length, w is the width, and h is the height. In this case, the length and width are given by the dimensions of the cardboard minus twice the size of the cut out square, and the height is given by the size of the cut out square.

Step 3 ：We can express the volume as a function of the size of the cut out square x: \(V(x) = (43-2x)(23-2x)x\). We want to find the maximum of this function.

Step 4 ：To find the maximum of a function, we can take the derivative and set it equal to zero. This will give us the x-values at which the function has a maximum or minimum. We can then test these x-values to see which one gives the maximum volume.

Step 5 ：Taking the derivative of the volume function \(V(x)\) and setting it equal to zero gives us the critical points \([11 - \sqrt{1389}/6, \sqrt{1389}/6 + 11]\).

Step 6 ：Testing these critical points, we find that the maximum volume occurs at the point \((11 - \sqrt{1389}/6, (1 + \sqrt{1389}/3)*(11 - \sqrt{1389}/6)*(\sqrt{1389}/3 + 21))\).

Step 7 ：Final Answer: The dimensions of the box that will give the maximum volume are approximately \(\boxed{11 - \sqrt{1389}/6}\) inches for the length and width, and \(\boxed{1 + \sqrt{1389}/3}\) inches for the height. The maximum volume is approximately \(\boxed{(1 + \sqrt{1389}/3)*(11 - \sqrt{1389}/6)*(\sqrt{1389}/3 + 21)}\) cubic inches.