Question 1201504
An aquarium that holds 40 cubic meters of water
 L*w*h = 40
is to be made such that the length of its base is twice the width.
 L = 2w
therefore if we replace L with 2w
 2w^2h = 40
divide eq by 2
 w^2h = 20
or
h = 20/(w^2)
If material for the base costs $20 per square meter,
 20(L*w)
replace L with 2w
 20(2w^2 = 40w^2 is cost of the base
and the material for the sides costs $16 per square meter,
 16(2(L*h)
 32Lh
replace L with 2w
 32(2wh) = 64wh is the cost of the long sides
and
 16(2(w*h))
 32wh is the cost of the shorter sides
cost of the sides: 
 64wh + 32wh = 96wh the total cost of the sides
find the cost of the materials for the cheapest such aquarium.
 40w^2 + 96wh = total cost
replace h with 20/w^2
 40w^2 + 96w(20/w^2) 
multiply, cancel w
 40w^2 + 1920/w = total cost
we can simplify to solve, divide by 40
 w^2 + 48/w
find the minimum on your graphing calc, w=3 is minimum cost
then L = 6 is the length and h = 20/3^2 = 2.22 meters
Check: 6 * 3 * 2.22 = 40 cubic meters
:
find the cost of the materials for the cheapest such aquarium.
Base cost: 40*3^2 = 360
Long sides 64*3*2.22 = 426.24
Short sides 32*3*2.22 = 213.12
----------------------------------
total cost of aquarium 999.36 ~ $1000
:
Check using total equation
40(3^2) + 1920/3 = 1000