Gift Wrapping Math

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Charles Hamilton Houston, NAACP Litigation Director.


Introduction

Figure 1: Examples of Christmas Gift Wrapping.

Figure 1: Examples of Christmas Gift
Wrapping (Source).

My sister works as an event planner/wedding planner. She wrote me an email today with the following gift wrapping question.

I need to wrap 375 boxes. 16x14x6. The wrapping paper is 30 inches wide. I say I need about 1100 feet. People are telling me I only need half that.

She then referred me to a web site that gives  a formula for the amount of wrapping paper for a given prismatic box. Her team was not familiar with evaluating formulas, and she asked me if I could do the calculations.

This formula was supposedly created by Sara Santos, a well-known applied mathematician. However, when I looked at the equation (Equation 1), I knew something was wrong. I am sure Sara derived it correctly – the problem is one of getting it on the page correctly. I still remember a conversation I had years ago with a typesetter who complained about setting mathematical type because it was "fussy" – he charged extra for it because it had to be done EXACTLY right.

Here is a video that provides a good background on the formula and how it is used.

For those who don't want the joy of computing, I include an Excel spreadsheet here to help you. No macros, just a simple calculator.

Background

Equation as Stated

Equation 1 is supposed to the give the area of wrapping paper required for a single box.

Eq. 1 GiftWrap

The presence of the equality is what confused me. However, I have figured out what is going on and the equality should not be there. The first term \frac{1}{2}\cdot {{\left( {d+2\cdot h+w} \right)}^{2}}  is for rectangular boxes and the second term \displaystyle 2\cdot {{\left( {w+h} \right)}^{2}} is for square boxes – the square box term is a special case of the rectangular term with d = w.

The gift wrapping formula does not care numbers are assigned to d, w, or h. If you have two dimensions that are equal, then make those d and w because you have a box with a square side.

Analysis

Derivation

The derivation is straightforward given two cases: (1) square box, and (2) rectangular box. Figure 3 provides my drawings of these cases. The required wrapping paper area follows directly from these drawings.

DerivationSquare Rectangular
Figure 3(a): Square-Base Box. Figure 3(b): Rectangular-Base Box.

Now I understand that Equation 1 should not contain an equality. Instead, there are two separate cases covered. Really, the square-base box case is just a special case of the rectangular-box case.

Wrapping Paper Constraint

The web page did not mention that the derivation makes an assumption that the wrapping paper is wide enough to support this approach to wrapping the box. To wrap a box this way requires that the wrapping paper meet the following width constraint (Equation 2).

Eq. 2 \displaystyle {{w}_{{Wrap}}}=\sqrt{2}\cdot \left( {w+h} \right) square-base box
\displaystyle {{w}_{{Wrap}}}=\frac{{w+d+2\cdot h}}{{\sqrt{2}}} rectangular-base box

Solution to My Sister's Problem

My sister needed to know how much 30-inch wide wrapping paper to buy. Figure 4 shows my answer.

Figure 4: My Answer to My Sister's Problem.

Figure 4: My Answer to My Sister's Problem.

1100 feet of wrapping paper is a lot of paper.

Conclusion

I love to work geometric problems, and my sister gave me a practical one – everyone has to wrap packages. This problem also fits in nicely with my interests in origami and paper engineering.

Save

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7 Responses to Gift Wrapping Math

  1. Pingback: Yet More Gift Wrapping | Math Encounters Blog

  2. TJ says:

    Your a genius , you made my day with this equation. Merry Christmas .

     
  3. Beatrice Kondo says:

    Thank you! I kept looking at the original equation and wondering how it could be an equality. The fact that it is not was tripping me up!

     
    • mathscinotes says:

      That was what threw my sister for a loop. I see all sorts of formulas on the web that contain typographic errors.

      Thanks for reading my post.

      mark

       
  4. Simon Wigzell says:

    I am terrible with maths so this is really useful 🙂 If I a box that was 82mm x 112mm x 43mm and had to wrap 200 of them - how much paper would I require? (presuming it is 700mm wide and 3000mm long)

     
  5. alok says:

    It doesn't seem that the width constraint is correct above. It is essentially taken as a square root of total area. So if width has to be as much, symmetrically the height has to be as much (since you could rotate the paper 90 degrees), and hence all we are saying is that the desired square paper must be embedded in any rectangular paper we choose. One should be able to do with lesser width (and greater length to make up for the area) paper.

    Thoughts?

     

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