A Solution To Second Mat(h) Problem

Sadly, we had no winners in this contest.  Here is a solution to that math problem:

There is more than one way to solve this problem, but we will be exploiting three different relationships. First, in preserving the aspect ratio, the length of the image (we’ll call L) is 1¹/₂ times the width (W).

\( L = 1.5W. \)

Then, adding up the components making up the overall width of the mat, the image width (less two overlaps of ¹/₈“) plus two mat widths (M) would equal 16 inches.

\( W – \frac{1}{4}” + 2M = 16″ \)

By the same token, the image length (less same overlaps) plus two mat widths would be 20 inches.

\( L – \frac{1}{4}” + 2M = 20″ \)

If you replace the L in the last equation with its W equivalent from the first equation, and then add ¹/₄” to both sides of both equations to combine constants, you are left with the following two equations to solve with two unknown variables:

\( \begin{array}{r c l} 1.5W & + 2M = & 20.25 \\ W & + 2M = & 16.25 \end{array} \)

From here, you can use linear algebra (matrices) or algebraic manipulation to simplify until you are left with just one variable.  For example, just subtracting the bottom equation from the top (subtracting the left sides separately from the right sides of each equation), you will wind up with

\( 0.5W = 4 \)

so the image width is eight inches, which means its length is twelve inches, and you would set the mat guide to 4¹/₈“.

What’s Next

I’ve come up with one more printing-inspired math problem, which I will share as soon as I master a new plug-in for this blog.  After that, I’m not sure.  Response has been weak, but the former teacher in me feels a need to keep pointing out opportunities to use some of this stuff you learned in school (or is it just to torment those students who were the most difficult – I’m not telling).  This isn’t really costing anything, and I give enough warning for the math-averse to stay clear.  So stay tuned.