calculus optimization problem money?

Question

A rectangular storage container with an open top is to have a volume of 10m^3. The length of its base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container.

Please round the answer to the nearest cent.

Answer 1

v = 10 cubic meters

w= 1
l = 2(w)
h = 5

30 sq m * 6
2 sq m * 10
cost = $200

Answer 2

The size of the base is w * 2w.

The volume is l * w * h = 2w * w * h = 10 m^3

Solving for h, the volume equation is
h = (10)/(w)(2w) = 5/w^2

The cost of the materials is $10 per square meter for the base, and $6 per square meter for the sides.

The cost of the base is 10*w*2w (20w^2).
The cost of the sides are (6*w*h) for each of the two smaller sides, and (6*2w*h) for each of the two wider ones.

The total cost is the cost of the base plus the cost of the four sides:

c = 20w^2 + 2*6wh + 2*12wh
c = 20w^2 + 36wh

Remembering that h=5/w^2, and substituting, the cost is:

c = 20w^2 + 36w(5/w^2)
c = 20w^2 + 180/w

To minimize the cost, find dc/dw and set it equal to zero:

dc/dw = 0
d(20w^2 + 180/w)/dw = 0
40w -180/w^2 = 0
40w^3 = 180
w^3 = 9/2
w = cube root of( 9/2 )

To find the cost just substitute that value into:

c = 20w^2 + 180/w
c = 54.5136 + 109.0272
c = $163.54

PS - The previous commenter's answer (1x2x5) is incorrect. His material cost is 30*$6 ($180) for the sides and 2*$10 ($20) for the base, or $200.00, which is not the minimum possible cost within the constraints.

Answer 3

H*W*L = 10
L = 2W
Cost = LW(10) + (2HL+2HW)*6
Cost = LW(10) +2*6*H(L+W)
.........L=2W

Cost = 20W^2 + 12H(3W)

H * W * L = 10
2W^2 * H = 10
H = 10/(2W^2) = 5*w^(-1)

Cost = 20w^2 + 12(5*3)w^(-1)
Cost = 20w^2 + 180 w^(-1)

Set cost/dw = 0 for min or max

0 = 40W - 180 w^(-2)

Solve for W
then solve for other