Pine costs twice as much as fibreboard

[danizh]

I've been stuck on this question for more than an hour and I still can't seem to get the right answer (). 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!

 

[Galileo]

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

 

[danizh]

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)?

 

[Galileo]

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.

 

[danizh]

Makes sense, thanks a lot Galileo! :)