We are tasked with finding the dimensions and maximum volume of an open rectangular box that can be made from a sheet of cardboard by cutting congruent squares from the corners and folding up the sides.

Let’s break this problem into steps:

Part 1: Dimensions of the Box

1. Start with the given dimensions of the cardboard: 11 inches by 7 inches.
2. Cutting squares from each corner: Suppose the side length of the squares cut from the corners is $x$ inches.
3. After cutting and folding, the dimensions of the box will be:
   • Length = $11 - 2x$ (since squares are cut from both sides along the length),
   • Width = $7 - 2x$,
   • Height = $x$ (since the height of the box is equal to the side length of the squares).

Part 2: Volume of the Box

The volume $V(x)$ of the open box is given by: $V(x) = (11 - 2x)(7 - 2x)(x)$ We want to maximize this volume. To find the value of $x$ that maximizes the volume, we take the derivative of the volume function and find its critical points.

Step 1: Write the volume function

$V(x) = (11 - 2x)(7 - 2x)(x)$

Step 2: Expand the volume function

$V(x) = x(77 - 36x + 4x^2) = 4x^3 - 36x^2 + 77x$

Step 3: Differentiate $V(x)$ to find critical points

$V'(x) = 12x^2 - 72x + 77$

Step 4: Solve $V'(x) = 0$

We solve the quadratic equation $12x^2 - 72x + 77 = 0$ to find the critical points.

Let me calculate these critical points.The critical points are:

$x_1 = 3 - \frac{\sqrt{93}}{6} \quad \text{and} \quad x_2 = 3 + \frac{\sqrt{93}}{6}$

Since $x$ represents the side length of the square cut from the corners, we only consider values of $x$ between 0 and $\min\left(\frac{11}{2}, \frac{7}{2}\right)$, which is 3.5. So, we evaluate which of these critical points falls within this range.

Let's compute the approximate values of these critical points.The two critical points are approximately:

$x_1 \approx 1.39 \quad \text{and} \quad x_2 \approx 4.61$

Since $x_2 = 4.61$ is outside the feasible range (as $x$ must be less than 3.5), we discard it. Thus, the critical point to consider is $x_1 \approx 1.39$.

Part 3: Find the dimensions of the box

Using $x = 1.39$, we calculate the dimensions of the box:

• Length = $11 - 2(1.39) \approx 11 - 2.78 = 8.22$ inches,
• Width = $7 - 2(1.39) \approx 7 - 2.78 = 4.22$ inches,
• Height = $x = 1.39$ inches.

Thus, the dimensions of the box are approximately $8.22$ inches by $4.22$ inches by $1.39$ inches.

Part 4: Maximum Volume

Now, let’s compute the maximum volume by substituting $x = 1.39$ into the volume formula:

$V = (11 - 2x)(7 - 2x)(x)$

Let me calculate the maximum volume now.The dimensions of the box of maximum volume are approximately:

$8.22 \, \text{in.} \times 4.22 \, \text{in.} \times 1.39 \, \text{in.}$

Part 2: Maximum Volume

The maximum volume of the box is approximately:

$48.22 \, \text{in.}^3$

Would you like more details or further explanation?

Here are 5 related questions to expand your understanding:

1. How does changing the size of the cut-out squares affect the volume of the box?
2. Why can we ignore one of the critical points in this problem?
3. How would the maximum volume change if the dimensions of the cardboard were different?
4. Can the derivative test be used for other optimization problems involving dimensions?
5. How would the result differ if we made a closed box instead of an open one?

Tip: When solving optimization problems, always check whether the critical points fall within the allowable range.