Pine costs twice as much as fibreboard

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Homework Help Overview

The problem involves determining the dimensions of a rectangular wooden toy chest, where the length is twice the width. The chest must have a volume of 0.3 m³, with pine used for certain sides and fibreboard for others. The cost of pine is twice that of fibreboard, and the goal is to minimize the overall cost.

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  • Mixed

Approaches and Questions Raised

  • Participants discuss their calculations and assumptions regarding the dimensions and costs associated with the materials. There is mention of variables representing width, length, and height, with suggestions to clarify these definitions for better understanding. Some participants express confusion over the area calculations for the back of the chest, questioning whether it should be width times height or length times height.

Discussion Status

There are multiple interpretations of the problem setup and the calculations involved. Some participants have offered alternative formulations of the cost function based on their assumptions. The discussion remains open, with no explicit consensus reached on the correct approach or final dimensions.

Contextual Notes

The problem is noted to be vague in its specifications, particularly regarding the dimensions and areas involved, which has led to confusion among participants.

danizh
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I've been stuck on this question for more than an hour and I still can't seem to get the right answer (:mad::mad:). So here goes:

Question:Sindy is building a rectangular wooden btoy chest so that its length is two times its width. She has picked pine for the top, front, and two sides of the chest. She will use fibreboard for the back and bottom. The chest much have volume of 0.3m^3. Pine costs twice as much as fibreboard. Find the dimensions that will minimize the cost of the chest.

My wrong calculations:
Let a be the price of fibreboard, and 2a be price of pine.
2x^2y = 0.3
so 7 = 0.3/(2x^2)
Cost = 2a (2xy + 2x^2 + 2xy) + a (2xy+2x^2)
= 10axy + 6ax^2
= 10ax(0.3/2x^2) + 6ax^2
= (1.5a/x) + 6ax^2

Now if I take the derivative of this, I get:

Cost' = -1.5x^(-2) + 12x
1.5 = 12x^3
0.125 = x^3
x = 0.5

And so, y=0.6

The correct answer shold be 0.585m x 1.170m x 0.438m, so any help would be great as to what I'm doing wrong!
 
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danizh said:
I've been stuck on this question for more than an hour and I still can't seem to get the right answer (:mad::mad:). So here goes:
Question:Sindy is building a rectangular wooden btoy chest so that its length is two times its width. She has picked pine for the top, front, and two sides of the chest. She will use fibreboard for the back and bottom. The chest much have volume of 0.3m^3. Pine costs twice as much as fibreboard. Find the dimensions that will minimize the cost of the chest.
My wrong calculations:
Let a be the price of fibreboard, and 2a be price of pine.
2x^2y = 0.3
so 7 = 0.3/(2x^2)
Cost = 2a (2xy + 2x^2 + 2xy) + a (2xy+2x^2)
= 10axy + 6ax^2
= 10ax(0.3/2x^2) + 6ax^2
= (1.5a/x) + 6ax^2
I guess x is the width, so 2x is the length and y is the height of the box. Still, it'd be helpful if you stated what the variables you introduce mean and you should give them symbols that are 'obvious' (i.e. l,w,h for length width, height) for several reasons:
(1) Your work becomes easier to follow so other can help you more easily.
(2) It becomes easier to check you work yourself
(3) It will prevent confusion, which I think is the reason it went wrong here.

The area of pine is: 2lh+wh+lw (2 sides + front + top)
the area of board is: lw+wh (bottom + back)
So the cost is: 2a(2lh+wh+lw)+a(lw+wh). Or, since l=2w:
Cost = 2a(5wh+2w^2)+a(2w^2+wh)
Which is already different from your result:
Cost = 2a (4xy + 2x^2) + a (2x^2+2xy)

EDIT: Wait a sec. I don't have a picture of the box. Is the area of the back of the box length times height or width times height? I assumed it was wh, but if it's lh your expression is correct.
 
Last edited:
It isn't specified, that is what's causing the confusion. In a situation like this, would you consider "width x height" the back (the shorter length multiplied by height)?
 
The question is pretty vague on that. I thought it was obvious the length (longer side) would correspond to a side and the width to the back.
Anyway, I get that the dimensions are 0.516 x 1.032 x 0.563.
So I guess the answer in your book is wrong. w=0.516 is definitely the value which minimizes 2(pine area)+(board area) under the given constraints.
 
Makes sense, thanks a lot Galileo! :)
 

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