6. A manufacturer wants to design an open box having a square base and a surface area of 48 square inches. What dimensions will produce a box with maximum volume? What is the maximum volume?

## Step1<br />The problem is about optimizing the volume of an open box with a square base. The volume of a box is given by the product of its length, width, and height. Since the box has a square base, the length and width are equal, and we denote them as \(x\). The height of the box is denoted as \(h\).<br /><br />## Step2<br />The surface area of the box is given by the sum of the areas of the base and the four sides. The base is a square with area \(x^2\), and the four sides are rectangles with area \(xh\) each. Therefore, the total surface area is \(x^2 + 4xh\).<br /><br />## Step3<br />The problem states that the total surface area is 48 square inches. Therefore, we can set up the equation \(x^2 + 4xh = 48\).<br /><br />## Step4<br />We can solve this equation for \(h\) to express it in terms of \(x\). Doing so gives us \(h = \frac{48 - x^2}{4x}\).<br /><br />## Step5<br />The volume of the box is given by the product of its length, width, and height, which is \(x^2h\). Substituting \(h\) from the previous step, we get the volume as a function of \(x\): \(V(x) = x \cdot \frac{48 - x^2}{4x}\).<br /><br />## Step6<br />To find the maximum volume, we need to find the maximum value of \(V(x)\). This can be done by differentiating \(V(x)\) with respect to \(x\) and setting the derivative equal to zero to find the critical points.<br /><br />## Step7<br />Solving the equation from Step 6 gives us the value of \(x\) that gives the maximum volume. Substituting this value back into the volume function gives us the maximum volume.