a rectangular box has a volume of 144 cubic inches, a surface area of 192 square inches, and a height one inch greater than its width. what is the distance from one corner of the box to the diagonally opposite corner?

To solve the problem, we need to find the dimensions of the rectangular box and then calculate the length of the space diagonal.

Let's denote:
- \( l \) as the length,
- \( w \) as the width,
- \( h \) as the height.

From the problem statement, we have:
1. \( l \times w \times h = 144 \) (volume)
2. \( 2(lw + lh + wh) = 192 \) (surface area)
3. \( h = w + 1 \) (height is one inch greater than width)

First, let's express everything in terms of \( l \) and \( w \):
- Substitute \( h \) in the volume equation:
  \[
  l \times w \times (w + 1) = 144
  \]
  \[
  lw(w + 1) = 144
  \]
  \[
  lw^2 + lw = 144
  \]

- Next, substitute \( h \) in the surface area equation:
  \[
  2(lw + l(w + 1) + w(w + 1)) = 192
  \]
  Simplify:
  \[
  2(lw + lw + l + w^2 + w) = 192
  \]
  \[
  2(2lw + l + w^2 + w) = 192
  \]
  \[
  4lw + 2l + 2w^2 + 2w = 192
  \]
  Divide by 2:
  \[
  2lw + l + w^2 + w = 96
  \]

Now we have two equations:
1. \( lw^2 + lw = 144 \)
2. \( 2lw + l + w^2 + w = 96 \)

Let's express \( l \) in terms of \( w \) using the first equation:
  \[
  l(w^2 + w) = 144
  \]
  \[
  l = \frac{144}{w^2 + w}
  \]

Substitute this expression for \( l \) in the second equation:
  \[
  2\left( \frac{144}{w^2 + w} \right)w + \frac{144}{w^2 + w} + w^2 + w = 96
  \]
  Simplify the terms:
  \[
  \frac{288w}{w^2 + w} + \frac{144}{w^2 + w} + w^2 + w = 96
  \]
  Combine the fractions:
  \[
  \frac{288w + 144}{w^2 + w} + w^2 + w = 96
  \]
  \[
  \frac{288w + 144 + (w^2 + w)(w^2 + w)}{w^2 + w} = 96
  \]
  \[
  \frac{288w + 144 + w^4 + 2w^3 + w^2}{w^2 + w} = 96
  \]

Simplifying further and solving for \( w \):
We can solve this polynomial equation for \( w \) using numerical methods or factorization. Let's use a reasonable approximation method and trial and error within this context.

After calculations (which would be shown step by step in a detailed solution), we find:
- \( w = 3 \)
- Substitute \( w \) back to find \( l \):
  \[
  l = \frac{144}{3^2 + 3} = \frac{144}{12} = 12
  \]
- And \( h \):
  \[
  h = w + 1 = 3 + 1 = 4
  \]

The dimensions of the box are \( l = 12 \), \( w = 3 \), and \( h = 4 \).

Finally, we find the length of the space diagonal \( d \) using the formula for the diagonal of a rectangular box:
  \[
  d = \sqrt{l^2 + w^2 + h^2}
  \]
  \[
  d = \sqrt{12^2 + 3^2 + 4^2}
  \]
  \[
  d = \sqrt{144 + 9 + 16}
  \]
  \[
  d = \sqrt{169} = 13
  \]

The distance from one corner of the box to the diagonally opposite corner is 13 inches.

Would you like more details or have any questions?

Here are 8 related questions for further exploration:

1. How do you derive the formula for the diagonal of a rectangular box?
2. What are other ways to calculate the surface area of a rectangular box?
3. How can you check if your dimensions are correct for given volume and surface area?
4. How do you solve polynomial equations in general?
5. What if the height was 2 inches greater than the width? How would that change the solution?
6. Can you derive the volume and surface area formulas for a rectangular prism from basic principles?
7. How can numerical methods help in solving algebraic equations?
8. What are some real-world applications of calculating the diagonal of a box?

**Tip:** When solving algebraic equations involving multiple variables, try to express one variable in terms of another to reduce the complexity.

Learn how to calculate the distance from one corner to the diagonally opposite corner of a rectangular box given its volume, surface area, and dimensions. This problem involves geometry and algebraic equations, including the use of the Pythagorean theorem. Suitable for high school students.

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