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.