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)