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November 24, 2014

November 24, 2014

Posted by **ibranian** on Friday, August 10, 2012 at 10:57am.

- math -
**Steve**, Friday, August 10, 2012 at 11:56amwith 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)

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