Jamal wants to make a box with no top out of a $24$ $inch^2$ piece of cardboard. She plans to cut smaller squares of equal size from the corners of the cardboard and fold up the resulting sides. By rounding to the nearest inch, what size of the square cut outs will create a box with the largest volume? I was thinking of using calculator, but I don't think that will work How do I solve this?

So the things i got are that the length

L=24-x or is 24-2x? To find a volume you do lwh But i dont think i did it right

Sorry for not editing the post nicely im doing this in my phone

  • $\begingroup$ The bottom of the box will be a square which is $24-2x$ by $24-2x$. The volume will be $x(24-2x)^2$. $\endgroup$ – André Nicolas Jan 6 '16 at 16:04
  • $\begingroup$ The bottom of the square? Like the length and width and x is the hieght? So how would i find the maximum $\endgroup$ – MATH ASKER Jan 6 '16 at 16:10
  • $\begingroup$ Yes, the length of the bottom is $24-2x$ and the width is $24-2x$. The height of the box is $x$. The volume is $x(24-2x)^2$, as I wrote earlier. We want to maximize $x(24-2x)^2$, with the constraint that $0\le x\le 12$. The standard way to maximize is to use the derivative. There are also non-calculus approaches, but calculus is simplest. $\endgroup$ – André Nicolas Jan 6 '16 at 16:21
  • $\begingroup$ Isnt (24-x)(24-x) gonna be 576-24x-24x+x^2 $\endgroup$ – MATH ASKER Jan 6 '16 at 16:25
  • $\begingroup$ I would like to solve it with the calculator if possible $\endgroup$ – MATH ASKER Jan 6 '16 at 16:30

Imagine cutting out the four little squares and folding the four flaps upward. Then we create a box whose bottom is a $24-2x$ by $24-2x$ square.

The volume of the box is then $x(24-2x)^2$. Note that we must have $0\lt x\lt 12$.

We want to maximize $x(24-2x)^2$, where $x$ ranges over the interval $(0,12)$. The standard technique for this uses the calculus. But you are asking for a procedure that uses a calculator, presumably a graphing calculator. Have the machine graph $x(24-2x)^2$ (over the interval $0\lt x\lt 12$). You should be able to see for what $x$ in this interval the function attains a maximum.

  • $\begingroup$ thanks a lot for answering, I forgot that it was (24-2x), so basically all i have to do is LWH and that would give me the equation to maximize? $\endgroup$ – MATH ASKER Jan 6 '16 at 22:35
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    $\begingroup$ You are welcome. Yes, and as I mentioned earlier it is not worthwhile to expand $x(24-2x)^2$. Your calculator will be perfectly happy to graph it without you going to the trouble of expanding. Saves a lot of time, and minimizes the probability of an arithmetical slipup. The answer by the way is that the maximum is reached at exactly $x=4$. (I do not have a graphing calculator, so used calculus.) $\endgroup$ – André Nicolas Jan 6 '16 at 22:39
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    $\begingroup$ I hope you can see from your picture why it is $24-2x$ rather than $24-x$. We are removing a strip of width $x$ from both ends. So the side of the "inner" square in your picture is $24-2x$. $\endgroup$ – André Nicolas Jan 6 '16 at 22:42
  • $\begingroup$ Yes, I understood that thank you for answering my question $\endgroup$ – MATH ASKER Jan 6 '16 at 23:21

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