Math Problem Statement

a rectangular box has a volume of 144 cubic inches, a surface area of 192 sq inches, and a height 1in greater than its width. what is the distance from one corner of the box to the diagonally opposite corner?

Solution

Given:

  • The volume VV of the rectangular box is 144 cubic inches.
  • The surface area SS of the rectangular box is 192 square inches.
  • The height hh is 1 inch greater than the width ww.

We are to find the distance from one corner of the box to the diagonally opposite corner.

  1. Let's denote:

    • Width =w= w
    • Height =w+1= w + 1
    • Length =l= l
  2. Using the volume formula for a rectangular box: V=lwh=144V = l \cdot w \cdot h = 144 Substituting h=w+1h = w + 1: lw(w+1)=144l \cdot w \cdot (w + 1) = 144

  3. Using the surface area formula for a rectangular box: S=2lw+2lh+2wh=192S = 2lw + 2lh + 2wh = 192 Substituting h=w+1h = w + 1: 2lw+2l(w+1)+2w(w+1)=1922lw + 2l(w + 1) + 2w(w + 1) = 192 Simplifying: 2lw+2lw+2l+2w2+2w=1922lw + 2lw + 2l + 2w^2 + 2w = 192 4lw+2l+2w2+2w=1924lw + 2l + 2w^2 + 2w = 192 2lw+l+w2+w=962lw + l + w^2 + w = 96

  4. Solving for ll in terms of ww: From lw(w+1)=144l \cdot w \cdot (w + 1) = 144: l=144w(w+1)l = \frac{144}{w(w + 1)} Substituting ll into the surface area equation: 2(144w(w+1))w+144w(w+1)+w2+w=962 \left( \frac{144}{w(w + 1)} \right) w + \frac{144}{w(w + 1)} + w^2 + w = 96 Simplifying: 2144/(w+1)+144/(w2+w)+w2+w=962 \cdot 144 / (w + 1) + 144 / (w^2 + w) + w^2 + w = 96

  5. Solve for ww by simplifying and solving the quadratic equation derived from the volume and surface area constraints. After simplification and solving, we get: w=3and thush=4,l=12/1=4w = 3 \quad \text{and thus} \quad h = 4, \quad l = 12 / 1 = 4

  6. Calculate the diagonal distance dd using the Pythagorean theorem in 3D: d=l2+w2+h2=42+32+42=16+9+16=41d = \sqrt{l^2 + w^2 + h^2} = \sqrt{4^2 + 3^2 + 4^2} = \sqrt{16 + 9 + 16} = \sqrt{41}

Therefore, the distance from one corner of the box to the diagonally opposite corner is: 41 inches\sqrt{41} \text{ inches}

Would you like any details or have any questions?

Here are 8 related questions for further exploration:

  1. How would you find the volume of a rectangular prism if given different dimensions?
  2. Can the Pythagorean theorem be applied in 3D for any rectangular box?
  3. How do you derive the formula for the surface area of a rectangular prism?
  4. How does changing one dimension of the box affect its volume and surface area?
  5. What other shapes can you use the Pythagorean theorem in 3D for?
  6. How do you solve a quadratic equation by factoring?
  7. What is the significance of the diagonal in the context of a rectangular prism?
  8. How do you calculate the surface area if one dimension is unknown?

Tip: When dealing with 3D problems, visualizing the shape and drawing diagrams can significantly aid in understanding and solving the problem effectively.

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Math Problem Analysis

Mathematical Concepts

Geometry
Pythagorean Theorem
Quadratic Equations

Formulas

Volume of a rectangular box
Surface area of a rectangular box
Pythagorean theorem

Theorems

Pythagorean theorem

Suitable Grade Level

Grades 7-9