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An aquarium that holds 40 cubic meters of water is to be made
such that the length of its base is twice the width.
If material for the base costs $20 per square meter, and the material for the sides
costs $16 per square meter, find the cost of the materials for the cheapest such aquarium.
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Let w be the width of the aquarium;
then its length is 2w, according to the problem.
If the height is h, then the volume is
V = w*(2w)*h = 2w^2*h,
so
2w^2*h = 40 cubic meters, or
w^2*h = 20 cubic meters.
It gives h = . (1)
The base area is w*(2w) = 2w^2; the base cost is 20*2w^2 = 40w^2 dollars.
The lateral area is (w + 2w + w + 2w)*h = 6wh. The lateral sides cost is 16*6wh = 96wh dollars.
The total cost is
C = 40w^2 + 96wh = substitute h from (1) = = + = + .
So, we want to minimize this function C(w) = + . (2)
To find the minimum, take the derivative and equate it to zero.
Doing it, you will get, step by step
80w =
80w^3 = 1920
w^3 = 1920/80 = 24
w = = 2.88.
Thus the width is 2.885 m; the length is twice of it 2*2.885 = 5.77 m.
the height is = = 2.41 m.
The cheapest cost is C = formula (2) = + = 998.43 dollars.
Solved.