What box would be best to ship the most packing peanuts?

Question

So I struggle with optimization and we were given a problem and I have no idea on how to do it and I could really use some help please.
So a person wants to ship a lot of packing peanuts but they want to choose the best type of box so they can send the most packing peanuts (so basically biggest volume?). The three choices are a tube, a box with a square base, or a box with triangular bases. The maximum girth and length of the box together are 108 inches. So I have to find which dimensions of these types of boxes would allow for the most room to store the packing peanuts and which box would allow most space.
Hopefully, this makes sense. If you need me to clarify something, just ask.

Thanks in advance. The help really means a lot.

Answer 1

Basically the strategy will be to determine the dimensions of each box that will generate the largest volume, determine what that volume is, and then compare the max volumes generated by each box type.

 

The fact that the sum of the height of the box with its "girth" (think of this as circumference/perimeter) equals 108 is used to establish a relationship between the dimensions. Then write a formula for the volume of the box in terms of one variable.

 

Let's take the box with square base first. Let x = the length of side of the square base and h = the height. 

This means 4x + h = 108, or h = 108 - 4x.

 

The volume of this box is V = x² (108 - 4x)   --area of the base times height in terms of x

V = 108x²-4x³

 

dV/dx = 216x - 12x²

 

That equals zero when x = 18 (or 0 which we can ignore). When x = 18, h = 36.

You might want to convince yourself that these values generate a maximum volume by using 2nd derivative test.

So we have V = 11664 cubic inches.

 

Here is a start for the tube.

Let r = radius of the base and h equal its height.

 

The girth is 2(pi)r. So 2(pi)r + h = 108. Or h = 108 - 2(pi)r

The volume of the tube is V = (pi)r²h = (pi)r²(108 - 2(pi)r) = 108(pi)r² - 2(pi)²r³

 

Now find dV/dr, set equal to zero to find critical number(s), then use that value to determine the max volume.

 

I happened to get r= 36/pi  as the value of r that generates the max volume. That volume was greater than the volume of the box with the square base.

 

For the box with a triangular base you might need more details on the type of triangle (equilateral?).