Carboard Box Optimization: Maximize size, Minimize Cost (not understanding "hint")
- Thread starter jimbo1323
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I use these boxes for moving and for storage. They're cheap and sturdy!
. . . . .\(\displaystyle V\, =\, LHW\, =\, 1\)
This also tells us that we'll need to measure length, width, and height in terms of feet, rather than inches. (We can always convert to inches when we're done.)
What are the dimensions (in terms of given variables) for the area of one of the double-thickness sides? How many thicknesses of cardboard are there for one of these sides? So, for the two double-thickness sides, how many times of that area do you need?
What are the dimensions (in terms of given variables) for the area of the bottom? How many thicknesses of cardboard are there for the bottom? So, for the bottom, how many times of that area do you need?
As for the box top, they gave you a drawing and an expression for the area of cardboard needed. Just use that.
If you get stuck, please reply with your work so far, starting with your answers to my questions above. Thank you!
This tells us that:
. . . . .\(\displaystyle V\, =\, LHW\, =\, 1\)
This also tells us that we'll need to measure length, width, and height in terms of feet, rather than inches. (We can always convert to inches when we're done.)
I'm not sure what you mean by the lid "taking up" 12.5% of the height. You're told that it's the lid's overlap (that is, the sides that cover the tops of all the vertical sides of the box) that is equal to 12.5% of the box's height. So, whatever "H" is, the overhang of the lid is going to be 12.5% of that. If the height is 18 inches, then the lid's overhang will be two and a quarter inches.The box is cleverly designed so that it requires no tape or staples to hold it together. A diagram showing how the box is constructed is shown below.
As you can see, the bottom is secured by two side pieces that fold up. This results in a box that has two single-thickness sides, two double-thickness sides, and a double-thickness bottom.
The box top is also self-securing. The top is designed to have a lip that fits over the box; this lip is 12.5% of the height.
View attachment 9587
The single-thickness sides are \(\displaystyle L\, \times\, H,\) the double-thickness sides are \(\displaystyle W\, \times\, H,\) and the double-thickness bottom is \(\displaystyle L\, \times\, W.\) The dimensions on the top of the box are as shown above, so its area is given by:
. . . . .\(\displaystyle (L\, +\, 4\, \times\, 0.125H)\, \times\, (W\, +\, 2\, \times\, 0.125H)\, =\, (L\, +\, 0.50H)\, \times\, (W\, +\, 0.25H)\)
(a) The wholesale price of cardboard is $0.10/ft2. What are the raw-materials cost and dimensions for the cheapest box that can be manufactured?
(b) Show that it is, in fact, a minimum by examining the Hessian matrix.
(Hint: Eliminate one of the variables, say L, using the volume constraint, so that you end up with a two-variable problem.)
The hint near the bottom and the sentence about the box lid taking up 12.5% of the height is what is currently tripping me up about the problem.
Yes.
What are the dimensions (in terms of given variables) for the area of one of the single-thickness sides? How many thicknesses of cardboard are there for one of these sides? So, for the two single-thickness sides, how many times of that area do you need?
What are the dimensions (in terms of given variables) for the area of one of the double-thickness sides? How many thicknesses of cardboard are there for one of these sides? So, for the two double-thickness sides, how many times of that area do you need?
What are the dimensions (in terms of given variables) for the area of the bottom? How many thicknesses of cardboard are there for the bottom? So, for the bottom, how many times of that area do you need?
As for the box top, they gave you a drawing and an expression for the area of cardboard needed. Just use that.
If you get stuck, please reply with your work so far, starting with your answers to my questions above. Thank you!