Arectangular storage container with an open top is to have a volume of 10 m3. the length of this base is twice the width. material for the base costs $20 per square meter. material for the sides costs $12 per square meter. find the cost of materials for the cheapest such container. (round your answer to the nearest cent.)

L=2W, V=LWH using L=2W in the Volume equation we get:

V=2W^2H and V=10 so

10=2W^2H  now we can solve this for H

H=5/W^2 and L=2W  we'll need these later :)

C=20LW+12*2LH+12*2WH

C=20LW+24LH+24WH  using our H and L found earlier...

C=20(2W^2)+24(2W*5/W^2)+24(W*5/W^2)

C=40W^2+240/W+120/W  making a common denominator...

C=(40W^3+240+120)/W

C=(40W^3+360)/W

dC/dW=(120W^3-40W^3-360)/W^2

dC/dW=(80W^3-360)/W^2

d2C/dW2=(240W^4-160W^4+720W)/W^4

d2C/dW2=(80W^3+720)/W^3

Since d2C/dW2 is positive for all possible values of W (as W>0), when dC/dW=0, C(W) will be at an absolute minimum value...

dC/dW=0 only when 80W^3-360=0

80W^3=360

W^3=45

So our minimum cost is:

C(45^(1/3))=(40W^3+360)/45^(1/3)

C(45^(1/3))=(40*45+360)/45^(1/3)

C(45^(1/3))=2160/45^(1/3)

C≈$607.27  (to the nearest cent)

The answer would be choice b