math

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Donald wants to build an aquarium with internal volume of 1 m^3 that made of flat glass sheet. The bottom and side walls of the aquarium have to be made of 7 and 5 mm-thick glass sheet, respectively. Current price of 5 mm and 7 mm galss sheet are \$3.5 and \$5.6 per m^2, respectively. Determine the dimensions (wide x length x depth) of the aquarium such that the price of required glass sheets is minimum.

• math - ,

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)