Arectangular storage container with an open top is to have a volume of 20 cubic meters. the length of its base is twice the width. material for the base costs 20 dollars per square meter. material for the sides costs 7 dollars per square meter. find the cost of materials for the cheapest such container.

The volume of a rectangular box is calculated by multiplying the length, width and the height of the box. For this case, it is written as:

20 = lwh

From the problem statement, the length of its base is twice the width. Making the expression for volume.

20 = (2w)wh = 2(w^2)(h)
h = 10/(w^2)

Side A: l *h = (2w)(10/(w^2)) = 20/w
Side B: w*h = w*(10/(w^2)) = 10/w

Base = l*w = 2w*w = 2w^2
Total Area of Side A: 2 * (20/w) = 40/w
Total Area of Side B: 2 * (10/w) = 20/w
Total Area = (40/w)+(20/w) + 2w^2 = 60/w + 2w^2
 
Total area weighted cost = 7*(60/w) + 20*(2w^2) = 420/w + 40w^2
 
Find the critical point: 1. Take derivative. 2. Set to zero. 3. Solve for w.
   
Derivative = 80w - (420/(w^2)) = 0
80w = 420/w^2 --> 80w^3 = 420
w^3 = 420/80
w = apprx (1.738)
 
Re-input into cost formula: 420/1.738 + 40*(1.738)^2 = apprx (362.48)