The Biggest Box Fedex Will Take

March 10, 2009 at 5:12 pm · Filed under Math, Uncategorized

My friend Megan Goering asks what is the largest volume box you can send through Fedex given that length + twice the width + twice the height must be less than 165 inches. It might seem odd to care about only the volume instead of the dimensions, but it makes sense if you’re shipping walnuts, balloons or bows for Christmas presents. Anything pourable, that is.

Here’s my solution. Say that the dimensions are $x$, $y$ and $z$. Then we can write the problem as

\begin{eqnarray*}\textrm{max } & & xyz \\\textrm{subject to} & &x + 2y + 2z \le 165 \\& & x,y,z \ge 0 \\\end{eqnarray*}

First, observe that if $x + 2y + 2z < 165$, then the product $xyz$ can be increased by increasing any one of the dimensions. Therefore, the maximum will occur when $x + 2y + 2z = 165$. Thus, we can eliminate $z$ because

$\displaystyle z$ $\textstyle =$ $\displaystyle \frac{165 - 2y -x}{2}$ (1)

Here is the problem

\begin{eqnarray*}\textrm{max } & & xy\left(\frac{165 - 2y -x}{2} \right)\\\textrm{subject to} & &x + 2y \le 165 \\& & x,y \ge 0 \\\end{eqnarray*}

The constraints impose that the maximum be achieved in a triangular area with one vertex at the origin and the other two each on one of the $x$ and $y$ axes. Notice that if we use a point $(x,y)$ on the border of this triangular region, the product $xyz$ will be equal to 0, so this is clearly not the maximum. Therefore, the maximum occurs on the inside of the triangle. This is important, because it means that the maximum is some sort of “bump” which means the problem can be solved directly with calculus.

We want to maximize $xy\left(\frac{165 - 2y -x}{2} \right)$. The choice of $x$ and $y$ that achieves the maximum is unaffected by multiplying that product by a positive constant, so use this opportunity to get rid of the $\frac{1}{2}$. Let $f = xy(165 - 2y -x)$. Then

\begin{eqnarray*}f & = & 165xy - 2xy^{2} -x^{2}y \\\frac{\partial f}{\partial...... -2xy \\\frac{\partial f}{\partial y} & = & 165x - 4xy -x^{2}\end{eqnarray*}

Thus, in order for the maximum to be achieved, we need both derivatives to be equal to 0. Thus, we need

\begin{eqnarray*}165y - 2y^{2} -2xy & = & 0 \\165x - 4xy -x^{2} & = & 0\end{eqnarray*}

Which is, conveniently enough, two equations in two variables, so we can probably solve this. First simplify as

\begin{eqnarray*}165 - 2y -2x & = & 0 \\165 - 4y -x & = & 0\end{eqnarray*}

Then as

\begin{eqnarray*}2y + 2x & = & 165 \\4y + x & = & 165\end{eqnarray*}

And finally

\begin{eqnarray*}x & = & 55 \\y & = & \frac{55}{2}\end{eqnarray*}

and we get

\begin{eqnarray*}z & = & \frac{55}{2}\end{eqnarray*}

from Equation (1).

Thus the maximum volume you can send using this Fedex rule is $55 \times \frac{55}{2} \times \frac{55}{2} = \frac{166375}{4}$ cubic inches, or about 24 cubic feet. This many walnuts weighs over 300 kg, so presumably the problem was motivated by something lighter.

1 Comment »

  1. Daniel Grossberg said

    April 14, 2009 @ 10:29 pm

    You should submit this one to Dr. Math (the leading blog for such things). I think he’d like it.

    Reply

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