Arectangular bin with an open top and volume of 38.72 cubic feet is to be built. the length of its base must be twice the width and th bin must be at least 3 feet high. material for the base of the bin costs $12 per square foot and material for the sides costs $8 per square foot. if it costs $538.56 to build the bin, what are it's dimensions?

Data:
lenght=2x
width=x
height=y≥3

Volume=lenght x width x heigth
Therefore:
(2x)(x)(y)=38.72
2x²y=38.72

Area of the base: lenght x width
Area of the base=(2x)(x)=2x²
Cost of the base=12(2x²)=24x²

Area of the sides=2(length x height) + 2(width x height)
Area of the sides=2(2xy)+2(xy)=4xy+2xy=6xy
Cost of the sides=8(6xy)=48xy

Therefore:
Cost of the base + cost of the sides=538.56
24x²+48xy=538.56

We have this system of equations:
2x²y=38.72
24x²+48xy=538.56

We can solve this system of equations by substitution method:
2x²y=38.72    ⇒y=38.72 / 2x²=19.36 /x²

24x²+48x(19.36 / x²)=538.56
24x²+929.28/x=538.56
24x³+929.28=538.56x

I have solved this equation graphycally (by fooPlot) and the result would be:
x≈3.237  This value is not correct , because "y" would be < 3ft
And
x≈2.2   This value is correct, if "y" is at least 3 ft high.
Therefore:
width=x=2.2
lenght=2x=2(2.2)=4.4
y=19.36 / x²=19.36 / (2.2)²=4

it´s dimensions would be:
width=2.2 ft
lenght=4.4 ft
height=4 ft