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?
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!
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.
Report Abuse