http://freshwater-aquariums.net/what-should-be-the-length-and-width-of-the-aquarium-to-minimize-cost-of-materials-and-what-is-the-min-cost.html Freshwater Aquarium Care, Fish, Guides, and Tips Fri, 21 May 2010 15:29:02 +0000 http://wordpress.org/?v=2.9.2 hourly 1 http://freshwater-aquariums.net/what-should-be-the-length-and-width-of-the-aquarium-to-minimize-cost-of-materials-and-what-is-the-min-cost.html/comment-page-1#comment-415 Mikey Tue, 23 Jun 2009 01:01:46 +0000 http://freshwater-aquariums.net/what-should-be-the-length-and-width-of-the-aquarium-to-minimize-cost-of-materials-and-what-is-the-min-cost.html#comment-415 Side width = x Front length = y Height = 4 Volume 4xy = 88 so 4y = 88/x Now cost: 2.35*Front area + 1.35*(2*Side Area + Back Area) 2.35*4y + 1.35*(2*4x+4y) = cost substitute 4y = 88/x 2.35*88/x + 1.35 * (8x+88/x) = cost take the derivative with respect to x for the change in cost with respect to x. my calculator says it's 10.8(x^2*30.1481481481)/x^2 set that equal to zero for the min of the cost. That comes to about x=5.49. Sub that into your volume equation to get y = 4.01 and sub those both into your cost equation to get the total min cost at about $118.60. That's the idea. Side width = x
Front length = y
Height = 4
Volume
4xy = 88
so 4y = 88/x
Now cost:
2.35*Front area + 1.35*(2*Side Area + Back Area)
2.35*4y + 1.35*(2*4x+4y) = cost
substitute 4y = 88/x
2.35*88/x + 1.35 * (8x+88/x) = cost
take the derivative with respect to x for the change in cost with respect to x.
my calculator says it’s 10.8(x^2*30.1481481481)/x^2
set that equal to zero for the min of the cost. That comes to about x=5.49. Sub that into your volume equation to get y = 4.01 and sub those both into your cost equation to get the total min cost at about $118.60.
That’s the idea.

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