Question 1201504
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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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<pre>
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 = {{{20/w^2}}}.      (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) = {{{40w^2 + 96w*(20/w^2)}}} = {{{40w^2}}} + {{{(96*20)/w}}} = {{{40w^2}}} + {{{1920/w}}}.


So, we want to minimize this function C(w) = {{{40w^2}}} + {{{1920/w}}}.    (2)


To find the minimum, take the derivative and equate it to zero.


Doing it, you will get, step by step

    80w = {{{1920/w^2}}}

    80w^3 = 1920

      w^3 = 1920/80 = 24

    w = {{{root(3,24)}}} = 2.88.


Thus the width is 2.885 m;  the length is twice of it 2*2.885 = 5.77 m.

                            the height is  {{{20/w^2}}} = {{{20/2.88^2}}} = 2.41 m.


The cheapest cost is  C = formula (2) = {{{40*2.88^2}}} + {{{1920/2.88}}} = 998.43 dollars.
</pre>

Solved.