Optimization troubles

[Nanobyte]

The problem includes a 5x5 sheet of paper, and you have to cut an equal amount from each side to make a box. I'm trying to find the length of those sides for a maximum volume.

I just need a little help getting starting. I know you have to get like variables (V=x²h, I need to put h into x form), where h is the height and 2h is the sides cut from one side of the 5x5 paper.
Can anyone help me with this?

[ZFR]

Hint: Use differentiation...

[Nanobyte]

edited: I'm sorry, I don't mean to sound like a jerk. I'm just a tad overwhelmed at the moment. I've got three days to prove this problem, correct 31 questions I missed on last semester's midterm, and write twelve pages with nine prompts you could hardly write two with.

And all this for classes I don't even need. I've already got enough credits to graduate..

[ 01-16-2004, 11:04 PM: Message edited by: Nanobyte ]

[Gabrielles blades]

im prety sure a perfect cube will end up having the greatest possible volume, however you seem to be stating that the box must be rectangular?

If thats a mistake, then V=x^3
where x is equal to the square root of (5x5)/6
unless the box has no top in which case its the square root of (5x5)/5

If you need to solve this entirely by math rather than by logic then you will want to differentiate the equation as suggested above.

[Nanobyte]

As stated, a 5x5 sheet of paper:
&nbsp&nbsp_ _ _ _ _
|&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp|
|&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp|
|&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp|
|&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp|
|_ _ _ _ _|

And you cut an equal amount from all sides:
&nbsp&nbsp_ _ _
X&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspX
|&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp|
|&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp|
|&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp|
X_ _ _X

With X being 1 in this case. So the dimensions of this box would be 3x3x1, w/ an open top. Does that make any sense? It's kind of hard to explain a problem that's visual in origins.

edited: this is merely an example.

[ 01-17-2004, 02:26 PM: Message edited by: Nanobyte ]

[Gabrielles blades]

hrm, not allowed to tape little bits and pieces to form a perfect cube?
well i suppose the next best idea would be to simply create 6 equal rectangles out of the sheet.
ie, divide the horizontal into 3 portions and the vertical into 2 portions
resulting in a rectangular box with proportions of:
5/3 for the sides and 5/2 for the height.
or a total volume of (5/2)(5/3)^2

[Nanobyte]

No, no. You can't cut apart the paper. All you can do is cut the square corners off, so you can fold the ends up to make the sides of the box.

[andrewas]

That dosent fold. The question states you are removing material form the corners and folding the remainder to make the box. So if the side of the squares you remove is x, the sides of the box are 5-2x and the height is x.

V=(5-2x)^2 * x

[Nanobyte]

Haha! Yes! That's what I needed. Thanks for all the help guys.

edited: For all your viewing pleasure:

V= Ax, with A= (5-2x)²
V= (5-2x)²x
&nbsp&nbsp= (25-20x+4x²)x
&nbsp&nbsp= 4x³-20x²+25x
&nbsp&nbspV'= 12x²-40x+25=0
&nbsp&nbsp&nbsp&nbsp= (2x-5)(6x-5)=0
&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp2x-5=0 6x-5=0
&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspx= ⁵/₂&nbsp&nbspx= ⁵/₆

V= (5-2(⁵/₆))²(⁵/₆)
&nbsp&nbsp= (5-⁵/₃)²(⁵/₆)
&nbsp&nbsp= (¹⁰/₃)²(⁵/₆)
&nbsp&nbsp= (¹⁰⁰/₉)(⁵/₆)
&nbsp&nbsp= ⁵⁰⁰/₅₄= 9.26 cm³

Woohoo! [img]smile.gif[/img]

[ 01-17-2004, 04:45 PM: Message edited by: Nanobyte ]

[ZFR]

Exactly so...

V = 4x³ - 20x² + 25x

dV/dx = 12x² - 40x + 25

Equating to 0:

0 = 12x² - 40x + 25

Solving for x

x₁ = 2.5 -- not possible in our case since you'll be left with nothing. This actually is the answer for minimum volume
x₂ = 5/6 -- our answer

You can further check that it's the maximum by finding the second derivative:

d²V/dx² = 24x-40

Substitutimg our value:

d²V/dx² = 24 * 5/6 - 40 = -20 < 0 -- so it's maximum