Closed Top Box Optimization Problem

[Gwen]

This is a project for BC Calculus in which I have to make an actual box made from posterboard, using the measurements I calculated. I'm not sure I did it correctly, but as I am on Thanksgiving break, my teacher is not accepting emails, so I don't have a way to make sure I'm not about to waste all my posterboard. The size of my designated posterboard is 12 inches x 16 inches. My work is below: I found a volume formula, found its derivative, solved for x, and then plugged x into all the dimension measurements. Help would be much appreciated.

 

[Otis]

Hi Gwen. I see an issue with your measurements adding up to 16 inches.

x + 8-2x + x + 8-2x + x

That's not 16.

 

[Gwen]

Okay, this is what I tried this time. For the calculations, I changed x into z so that length could be x and width could be y.

 

[Otis]

I changed x into z so that length could be x and width could be y.

That's fine, Gwen. You may want to consider writing a cross-hash on letter z, to distinguish it from digit 2.

Everything looks good on your second attempt, with two exceptions.

1) The first two lines below are not correct. Somehow, you'd recovered, as the decimal approximation for z is correct.



2) Your reported length (5.1815) is not correctly rounded. (Correcting it will not change your rounded volume.)

I don't know whether your class follows the rules of Significant Figures. If it doesn't, then you may ignore the following comment.

If you're going to round a dimension to four decimal places, then you ought to round each dimension to four decimal places. In that case, each of the dimensions would be reported to five significant figures. Their product could not exceed five significant figures, so that would justify reporting the volume as 80.244 cubic inches (five significant figures).

Good job, overall.



PS: I don't see anything for the second box. Have you already determined that the other possibility has a smaller volume?

 

[Gwen]

Thank you so much! Yes, I tried to find the volume of the second box. My work is a bit messier, but I just switched where I placed 16 and 12 inches in the equations (ex: y=16-2z instead of y=12-2z).

 

[Otis]

I tried to find the volume of the second box.

Very good. It looks fine, but there are still issues with roundoff error (eg: width ought to be 12.619, rounding to three places).

There's also a notation error, where you'd inadvertently switched from symbol z to x.



If you were to round all dimensions to five significant figures, then you'd be justified in reporting the volume to three places. That value would be 73.902 cubic inches. In general, when rounding intermediate results, I try to carry at least two digits more than the number of decimal places I expect in the final answer. Otherwise, I don't round intermediate values at all; I use exact expressions, or I let my machine carry all its internal digits from one operation to the next.

 

[Gwen]

I see, I’ll fix those errors when I’m constructing my box. Thank you so much for all your help!