Double integral volume question
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Double integral volume question

[From: ] [author: ] [Date: 12-04-14] [Hit: ]
are made from wood costing only £3 per square metre.thickness of the wood is negligible).to minimize the cost of the wood?-This is not an integral problem!Let x, y be the dimensions for the base and z be the length of the height (in metres).......
I know how to work out double integrals but how would i go about solving this problem? I have the numerical answer i'm just looking for the method thanks:

Consider a rectangular wooden box. The base and the two shaded ends are made
from wood costing £4 per square metre while the lid and the other two sides
are made from wood costing only £3 per square metre.
The box is to have a volume of one cubic metre (you may assume that the
thickness of the wood is negligible). What should its dimensions be in order
to minimize the cost of the wood?

-
This is not an integral problem!

Let x, y be the dimensions for the base and z be the length of the height (in metres).
(Sketching this may be useful.)

We are given that V = xyz = 1 cubic metre, and the cost C can be written as (thinking about the areas of each face) as C = 4(xy + 2 * xz) + 3(xy + 2 * yz) = 7xy + 8xz + 6yz.

Hence, we need to minimize C = 7xy + 8xz + 6yz, subject to V = xyz = 1.
------------------------
Using Lagrange Multipliers,
∇C = λ∇V ==> <7y + 8z, 7x + 6z, 8x + 6y> = λ.

Equate like entries:
7y + 8z = λyz
7x + 6z = λxz
8x + 6y = λxy

So, λxyz = x(7y + 8z) = y(7x + 6z) = z(8x + 6y)
==> x(7y + 8z) = y(7x + 6z) and y(7x + 6z) = z(8x + 6y)
==> x = 3y/4 and z = 7y/8.

Substitute this into V:
(3y/4) * y * (7y/8) = 1
==> y = (32/21)^(1/3).

So, the dimensions are
(3/4)(32/21)^(1/3) metres x (32/21)^(1/3) metres x (7/8)(32/21)^(1/3) metres.

I hope this helps!

-
If you have not seen this method before, you can still rewrite C in terms of two variables (like solving for z = 1/(xy), and plugging this into the equation for C), and find the critical points by setting the first partials equal to 0. Then, check it's a minimum with 2nd derivative test.

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