+322 A rectangular storage container with an open top is to have a volume of 10 m^3. The length of this base is twice the width. Material for the base costs $5 per square meter. Material for the sides costs $3 per square meter. Find the cost of materials for the cheapest such container.
+129845 The area of the base is W * 2W = 2W^2
The volume of the container = Base area * height....which implies that
10 = 2W^2 * H ⇒ H = 10 / [ 2W^2 ] = 5 /W^2
So....the total surface area is given by.....area of the base + side area =
2W^2 + 2(5/W^2) [ W + 2W] =
2W^2 + (10/W^2}[ 3W] =
2^W^2 + 30/W
So....the cost, C, to be minimized is this :
C = (base cost of materials)(base area) + (side cost of materials)(side area)
C = 5(2W^2) + 3(30/W)
C = 10W^2 + 90/W
I can do this in 2 ways
(1) Calculus.....take the derivative of the cost....set to 0 and solve
So we have
C' = 20W - 90/W^2 = 0
20W = 90/W^2
W^3 = 90/20
W^3 = 9/2
W^3 = 4.5
W = (4.5)^(1/3) ≈ 1.651
Subbing this back into the cost function, the minimized cost is
C = 10[(4.5)^(1/3)] ^2 + 90/[4.5]^(1/3) ≈ $81.77
(2) If you haven't had Calculus...we can use a graph here : https://www.desmos.com/calculator/e0v6iwu5bo
Note that the function's minimized cost is ≈ $81.77
