Math Problem Statement

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?

Solution

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:

  • ll as the length,
  • ww as the width,
  • hh as the height.

From the problem statement, we have:

  1. l×w×h=144l \times w \times h = 144 (volume)
  2. 2(lw+lh+wh)=1922(lw + lh + wh) = 192 (surface area)
  3. h=w+1h = w + 1 (height is one inch greater than width)

First, let's express everything in terms of ll and ww:

  • Substitute hh in the volume equation: l×w×(w+1)=144l \times w \times (w + 1) = 144 lw(w+1)=144lw(w + 1) = 144 lw2+lw=144lw^2 + lw = 144

  • Next, substitute hh in the surface area equation: 2(lw+l(w+1)+w(w+1))=1922(lw + l(w + 1) + w(w + 1)) = 192 Simplify: 2(lw+lw+l+w2+w)=1922(lw + lw + l + w^2 + w) = 192 2(2lw+l+w2+w)=1922(2lw + l + w^2 + w) = 192 4lw+2l+2w2+2w=1924lw + 2l + 2w^2 + 2w = 192 Divide by 2: 2lw+l+w2+w=962lw + l + w^2 + w = 96

Now we have two equations:

  1. lw2+lw=144lw^2 + lw = 144
  2. 2lw+l+w2+w=962lw + l + w^2 + w = 96

Let's express ll in terms of ww using the first equation: l(w2+w)=144l(w^2 + w) = 144 l=144w2+wl = \frac{144}{w^2 + w}

Substitute this expression for ll in the second equation: 2(144w2+w)w+144w2+w+w2+w=962\left( \frac{144}{w^2 + w} \right)w + \frac{144}{w^2 + w} + w^2 + w = 96 Simplify the terms: 288ww2+w+144w2+w+w2+w=96\frac{288w}{w^2 + w} + \frac{144}{w^2 + w} + w^2 + w = 96 Combine the fractions: 288w+144w2+w+w2+w=96\frac{288w + 144}{w^2 + w} + w^2 + w = 96 288w+144+(w2+w)(w2+w)w2+w=96\frac{288w + 144 + (w^2 + w)(w^2 + w)}{w^2 + w} = 96 288w+144+w4+2w3+w2w2+w=96\frac{288w + 144 + w^4 + 2w^3 + w^2}{w^2 + w} = 96

Simplifying further and solving for ww: We can solve this polynomial equation for ww 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=3w = 3
  • Substitute ww back to find ll: l=14432+3=14412=12l = \frac{144}{3^2 + 3} = \frac{144}{12} = 12
  • And hh: h=w+1=3+1=4h = w + 1 = 3 + 1 = 4

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

Finally, we find the length of the space diagonal dd using the formula for the diagonal of a rectangular box: d=l2+w2+h2d = \sqrt{l^2 + w^2 + h^2} d=122+32+42d = \sqrt{12^2 + 3^2 + 4^2} d=144+9+16d = \sqrt{144 + 9 + 16} d=169=13d = \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.

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Math Problem Analysis

Mathematical Concepts

Geometry
Algebra
Quadratic Equations

Formulas

Volume of a rectangular prism
Surface area of a rectangular prism
Distance formula

Theorems

Pythagorean theorem

Suitable Grade Level

Grades 9-12