This is a question from a precalculus class that I'm a TA for.

You build a box that has a volume of 100 cubic feet. It takes two people ten minutes to spray paint this box. How many minutes does it take three people to spray paint a similar box that has a volume of 500 cubic feet? Assume that the area of surface to be covered is proportionate to both the number of people working and the time spent spray painting.

A student just asked me how to answer it, so I figured I'd write up the correct calculations and post it online to help anyone else who may wander across it.

  • $\begingroup$ Does the box have any flaps? $\endgroup$
    – LinAlg
    Nov 5, 2018 at 18:45
  • 1
    $\begingroup$ I'd assume no flaps or anything. It's just a precal questions, so let's keep it simple. Besides, it shouldn't matter so long as the two boxes are similar. $\endgroup$ Nov 5, 2018 at 18:50
  • $\begingroup$ Wrong assumptions can have horrible consequences. Where I come from, boxes have two sets of flaps, and the inner flaps often do not cover the entire top of the box. Assuming you'd like those painted too, the area may not scale linearly with $k^2$. $\endgroup$
    – LinAlg
    Nov 5, 2018 at 19:00
  • $\begingroup$ Mike, I have to ask: why don't you use website for the course you are teaching to post problems and solutions for your students? $\endgroup$
    – Ennar
    Nov 5, 2018 at 19:16
  • 2
    $\begingroup$ I find it peculiar that there is no mention of the shape of the two boxes. Perhaps that is the whole point of the exercise: if you don't know the shape, you can not conclude anything about the relation between volume and surface area. $\endgroup$
    – M. Wind
    Nov 5, 2018 at 20:03

3 Answers 3


Suppose that the linear measurements of the large box are $k$ times the linear measurements of the small box. This would mean $3$-dimensional measurements, like volume, of the large box will be $k^3$ times those of the small box. So $100k^3 = 500$, and we see that $k = \sqrt[3]{5}$. The amount of surface area of a box to be spray painted is a $2$-dimensional measurement, so it will scale by $k^2 = \sqrt[3]{25}$. Since there is this much more surface area to spray paint, it will take this much more time to paint it, and so it'll take two people $10\sqrt[3]{5}$ minutes to paint the larger box. Dividing the work among three people instead of two, we see that to paint the large box it'll take $$ \frac{2}{3}10\sqrt[3]{25} \text{ minutes.} $$


For similar rectangular boxes, area ratio is the square of length ratio and volume ratio the cube of length ratio. So $5$ times the volume is $5^{\frac{2}{3}}$ times the area which is equivalent to $5^{\frac{2}{3}}$ boxes.

If it takes two people $10$ minutes to spray paint a box then the rate is $20$ minutes per person per box

For three people, it will take $\frac{20}{3} \cdot 5^{\frac{2}{3}}$ minutes to paint the $500$ cubic foot box.

$T = 19.4934$ minutes or $19$ min $29.6$ sec.


Refer to the table: $$\begin{array}{c|c|c|c} \text{People} & \text{Time} & \text{Surface}\\ \hline \color{blue}2 & \color{blue}{10} & \color{blue}{6\sqrt[3]{100^2}}&\text{divide by $2$}\\ \color{blue}1 & \color{blue}{10} & \color{blue}{3\sqrt[3]{100^2}}&\text{multuply by $3$}\\ \color{blue}3 & \color{blue}{10} & \color{blue}{9\sqrt[3]{100^2}}\\ \hline \color{red}3 & \color{red}x & \color{red}{6\sqrt[3]{500^2}}\\ \end{array} \Rightarrow \color{red}x=\frac{\color{red}{6\sqrt[3]{500^2}}\cdot \color{blue}{10}}{\color{blue}{9\sqrt[3]{100^2}}}=\frac{20\sqrt[3]{25}}{3}.$$


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