Need help with this PreCalculus problem?

Question

 An open top box is made from a 12 cm x 22 cm piece of cardboard. Find the dimensions of the box if its volume is to be maximized.

Answer 1

Fascinating question.  You are trying to make a box, length times width times height.  

 

You will have one bottom piece l x w

You will have two side pieces l x h

You will have two end pieces w x h

 

It really doesn't matter which direction you call width and which you call length, because the problem is the same, so let's assume the length is greater or equal to the width.  

 

Hmm.  I'm not sure there's an obvious method here, so I'll go with my gut.  

 

Whatever way you draw this, the "length" will need to be along one side of the cardboard.  There are two other pieces that use the same length, so they need to be next.  That means that we have a chunk L x (W + 2H) accounted for, and the remainder of the box sides need to come from the remainder of the cardboard.  

 

If you draw a rectangle 12 by 22, the longest "length" that you can make is 22.  By inspection, an open-ended box with a length of 22, where you only have two sides, is going to at most have twice 3 height and 6 width that fit into that 12-inch dimension. With ends, you have three copies of the height and it drops to 2.  That gives us a theoretical maximum of 264 ccs.  (I was able to prove that with calculus that you don't have yet.)

 

On the other hand, if you assume that L is on the 12 side, then there are two configurations possible.  The top part looks like this -

L x W

L x H

L x H

 

And the last slice gets split in half for either 

>     W x H    &   W x H 

> or H x W   &   H x W 

 

Basically, if you're not going to lose some material, then either the H has to be 6 or the W has to be 6.

 

Assume H = 6

 

L = 12

H = 6

22 = 2H + 2W

10 = 2W

5 = W

 

V = 12 * 6 * 5 = 360

 

 

Assume W = 6

 

L = 12

W = 6

22 = W + 2 H + H = 6 + 3 H 

H = 16/3 = 5-1/3

 

V = 12 * 6 * 16/3 = 384

 

That's biggest.

 

However, I've omitted the possible case where the long direction (22) is either (L + H) or (L + W).

 

In that case, it looks like you'd lay the bottom of the box along the top left of the 22 side, and have the two long sides underneath it - maximum being 6 width and 3 height to make 12. The ends of the box would be laid across the right side, with their height laid horizontal at 3 that would make the length of the box 19.  19 x 6 x 3 = 342, smaller than our other example.  

 

If you increase the width of the box beyond 6, your height drops below 3, the cuts to the right have to flip because you can't fit more than 2 of anything higher than 6 into 12, so the length drops below 16, and you end up with wasted material and a much smaller volume. 

 

Looks like 384 is our best layout.

 

This has been a fun afternoon's play time, but I'm not sure what method you were supposed to use in this chapter of precalc.