Posted by ibranian on Friday, August 10, 2012 at 10:57am.
with width x, length y, height z, in m,
volume v = xyz = 1
bottom area is xy
side areas are xz and yz
cost
c = 5.6xy + 2*3.5(xz+yz)
c = 5.6xy + 7xz + 7yz
Using Lagrange multipliers, let
f(x,y,z) = 5.6xy + 7xz + 7yz
g(x,y,z) = xyz
we have to solve
∇f = λ∇g
g(x,y,z) = 1
5.6y+7z = λyz
5.6x+7z = λxz
7x+7y = λxy
xyz = 1
5.6xy + 7xz = λxyz
5.6xy + 7yz = λxyz
7xz + 7yz = λxyz
xyz=1
dividing out the xyz, we get
5.6xy + 7xz = λ
5.6xy + 7yz = λ
7xz + 7yz = λ
7xz + 7yz = 5.6xy + 7xz
now, since z = 1/xy,
7x/y + 7y/x = 5.6xy + 7x/y
7y/x = 5.6xy
7/x = 5.6x
x^2 = 7/5.6
x = 1.25
Now, going back to our original equations,
c = 5.6(1.25)y + 7z(1.25+y)
z = 1/(1.25y)
c = 7y + 5.6 + 7/y
dc/dy = 7 - 7/y^2
dc/dy = 0 when y = 1
so, the aquarium is
1.25 x 1 x 0.8
cost of materials is
c(1.25,1,0.8) = 5.6(1.25) + 7(0.8)(1.25+1)
= 7 + 5.6(2.25)
= 19.60
Hmmm. According to wolframalpha, minimum c occurs at
(x,y,z) = (∛(5/4),∛(5/4),∛(16/25))
c = 21∛(4/5) = 19.4947
That is, in fact, less than my value using Lagrange multipliers. Better check my math (as usual)