Lesson Minimize the cost of an aquarium with the given volume

Minimize the cost of an aquarium with the given volume

Problem 1

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.