A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 inches by 10 inches. Six squares will be cut from the cardboard: one square will be cut from each of the corners, and one square will be cut from the middle of each of the 10 -inch sides. The remaining cardboard will be folded to form the box and its lid.  Letting \( x \) represent the side-lengths (in inches) of the squares, 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.

We are tasked with finding the value of \( x \) that maximizes the volume of the box and the maximum volume itself.

### Step 1: Understanding the problem

The box is created by cutting six squares from a 3-inch by 10-inch piece of cardboard. These squares have a side length of \( x \), where:
- One square is cut from each corner (four squares total),
- One square is cut from the middle of each 10-inch side (two additional squares).

After cutting, the remaining cardboard will be folded to form a box with a hinged lid. The dimensions of the resulting box will be:
- The length of the box: \( 10 - 2x \),
- The width of the box: \( 3 - x \),
- The height of the box: \( x \) (from the depth created by folding the flaps).

### Step 2: Formula for the volume of the box

The volume \( V \) of the box is the product of its length, width, and height:
\[
V(x) = (10 - 2x)(3 - x)(x)
\]
This expression represents the volume as a function of \( x \).

### Step 3: Maximize the volume

We need to find the value of \( x \) that maximizes the volume. To do this, we'll differentiate \( V(x) \) with respect to \( x \) and find its critical points.

The volume function is:
\[
V(x) = x(10 - 2x)(3 - x)
\]
Expanding this:
\[
V(x) = x \left(30 - 10x - 6x + 2x^2 \right)
\]
\[
V(x) = x(30 - 16x + 2x^2)
\]
\[
V(x) = 30x - 16x^2 + 2x^3
\]

Now, take the derivative of \( V(x) \) with respect to \( x \):
\[
V'(x) = 30 - 32x + 6x^2
\]
To find the critical points, set \( V'(x) = 0 \):
\[
6x^2 - 32x + 30 = 0
\]
We will now solve this quadratic equation using the quadratic formula:
\[
x = \frac{-(-32) \pm \sqrt{(-32)^2 - 4(6)(30)}}{2(6)}
\]
\[
x = \frac{32 \pm \sqrt{1024 - 720}}{12}
\]
\[
x = \frac{32 \pm \sqrt{304}}{12}
\]
\[
x = \frac{32 \pm 17.44}{12}
\]

So the two potential solutions for \( x \) are:
\[
x = \frac{32 + 17.44}{12} \approx 4.12, \quad x = \frac{32 - 17.44}{12} \approx 1.22
\]

### Step 4: Determine the maximum volume

Since \( x = 4.12 \) is too large (it would result in negative dimensions for the box), we discard it and focus on \( x \approx 1.22 \).

Substitute \( x = 1.22 \) into the volume equation to find the maximum volume:
\[
V(1.22) = (10 - 2(1.22))(3 - 1.22)(1.22)
\]
\[
V(1.22) = (10 - 2.44)(3 - 1.22)(1.22)
\]
\[
V(1.22) = (7.56)(1.78)(1.22)
\]
\[
V(1.22) \approx 16.44 \, \text{cubic inches}
\]

### Conclusion:
- The value of \( x \) that maximizes the volume is approximately \( x = 1.22 \) inches.
- The maximum volume is approximately \( 16.44 \) cubic inches.

Would you like a deeper explanation of any step, or have any further questions?

---

Here are five related questions:
1. How would the volume change if the dimensions of the cardboard were altered?
2. What if a different number of squares were cut from the cardboard?
3. How does increasing the side length \( x \) beyond the feasible range affect the box's shape?
4. How could you use calculus to verify whether a critical point is a maximum or minimum?
5. How can this problem be solved graphically instead of algebraically?

**Tip:** Always check the physical feasibility of your solutions when solving optimization problems, especially when the solution involves dimensions or geometry.

A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 inches by 10 inches. Six squares will be cut from the cardboard: one square will be cut from each of the corners, and one square will be cut from the middle of each of the 10-inch sides. The remaining cardboard will be folded to form the box and its lid. Letting x represent the side-lengths (in inches) of the squares, 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.

This problem focuses on maximizing the volume of a box made from a 3-inch by 10-inch cardboard by cutting squares from the corners. Using algebra and calculus, including optimization through derivatives, we find that the optimal side-length x is approximately 1.22 inches, giving a maximum volume of 16.44 cubic inches. Suitable for high school students in Grades 10-12.

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