March 28, 2009
What should be the length and width of the aquarium to minimize cost of materials, and what is the min cost?
Can you answer Changyeon L's question about aquariums?:
A rectangular aquarium is to be 4 ft high and have a volume of 88 cu ft. The base, ends, and back are to be made of slate which costs $1.35 sq/ft, and the front is to be made of special reinforced glass that costs $2.35 sq/ft. What should be the length and width of the aquarium to minimize the cost of materials, and what is the minimum cost to build the aquarium?
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Comments on What should be the length and width of the aquarium to minimize cost of materials, and what is the min cost? »
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.