Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So I'm not much of a mathematician, and I've been trying to figure out how one would solve this real world problem I have.

I have 2"x4" wood cut into squares (2x4x4).

I am trying to figure out at what distance to place each board on top of each other so that the top edges would follow the curve (-x^2)+25 from -5 to 0 where each increment of 1 is 1 foot.

I hope this makes sense, please let me know if i need to clarify.

share|improve this question
    
When you say "on top of", are you stacking them so that the square faces meet, or so that the "edges" (i.e. the 2 inch sides) meet? –  Zev Chonoles Nov 29 '11 at 18:23
    
@Zev: so that the square faces meet. –  kylex Nov 29 '11 at 18:27
add comment

2 Answers

up vote 2 down vote accepted

It's a bit easier to solve this problem by flipping the curve on its side; i.e., instead of looking at $y=-x^2+25$:
$\hskip2.7in $ enter image description here
looking at $y=\sqrt{25-x}$:

$\hskip1.4in $ enter image description here

Using the latter setup, what you want to find first (if I have understood your question correctly) are the non-negative values of $x$ where $$y=\sqrt{25-x}=\frac{n}{3}$$ where $n$ is an integer (all my numbers will be in feet, hence $4$ inches is $\frac{1}{3}$). These are just the numbers $$25-\frac{n^2}{9}$$ as $n$ ranges from $1$ to $15$, i.e.

224/9, 221/9, 24, 209/9, 200/9, 21, 176/9, 161/9, 16, 125/9, 104/9, 9, 56/9, 29/9, 0

or, to use approximate values,

24.89, 24.56, 24.00, 23.22, 22.22, 21.00, 19.56, 17.89, 16.00, 13.89, 11.56, 9.000, 6.222, 3.222, 0

This shows how the number of wood pieces you can fit within the boundary increases by one at each of these values of $x$: enter image description here

As you can see, for $x\leq25-\frac{n^2}{9}$, we can fit in $n$ wood pieces.

Now, the thickness of each wood piece is 2 inches, or $\frac{1}{6}$, so the top of the $m$th row (counting the row whose bottom is the $y$-axis as row 1) is at $x=\frac{m}{6}$. Thus, you will be able to fit $n$ wood pieces in on the $m$th row if and only if $$\frac{m}{6}\leq25-\frac{n^2}{9},$$ or $$n\leq \sqrt{225-\frac{3m}{2}},$$ so the number of wood pieces you can fit in on the $m$th row is $$\left\lfloor\sqrt{225-\frac{3m}{2}}\right\rfloor.$$ Using this formula, we can generate a side-on view of what the final result of your project will look like: enter image description here

Is this what you had in mind?


Mathematica code for that final image:

f[m_] := Floor[Sqrt[225 - (3 m/2)]]

Show[ParametricPlot[Table[{m/6, u*f[m]/3}, {m, 1, 150}], {u, 0, 1}], 
Plot[Sqrt[25 - x], {x, 0, 25}, AspectRatio -> 1/5], 
ParametricPlot[Table[{u Floor[150 - 2 n^2/3]/6, n/3}, {n, 1, 15}], {u, 0, 1}], 
PlotRange -> {0, 5}]
share|improve this answer
    
Thank you! Unfortunately I did not specify that the 2" side would be facing the curve, not the 4" side, but that's a simple substitution. What you've provided solves my dilemma. –  kylex Nov 29 '11 at 19:27
    
@kylex: Ah, I see - glad I could help! –  Zev Chonoles Nov 29 '11 at 19:40
    
+1 for the nice pictures. –  TonyK Nov 29 '11 at 20:57
    
Thanks, @TonyK :) –  Zev Chonoles Nov 30 '11 at 4:22
add comment

You can make a table using your equation: $\begin {array} {c c c}x&h&boards\\ -5 &50&25\\ -4& 41&20.5\\ -3&34&17\\ -2&29&14.5\\ -1&26&13\\ 0 &25&12.5 \end {array}$

Then at each $x$ position pile up enough boards to get the height you want: $25$ of them at $5$ feet. As your boards are only $1/3$ foot in size, maybe you need more lines in the table. Is this what you were looking for?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.