Acompany wishes to manufacture a box with a volume of 2424 cubic feet that is open on top and is twice as long as it is wide. find the width of the box that can be produced using the minimum amount of material

 Let x = the width 
 Let 2x = the length
 Let h = the height
 then vol = x*2x*h. So we have 2x^2*h = 24
 h=24/(2*x^2)=12/x^2
 Surface area: two ends + 1 bottom + 2 sides (no top)
 S.A. = 2(x*h) + 1(2x*x) + 2(2x*h)
 S.A. = 2xh + 2x^2 + 4xh S.A. = 2x^2 + 6xh
 Replace h with  12/x^2
 S.A = 2x^2 + 6x(12/x^2)
 S.A = 2x^2 + 6(12/x)
 S.A = 2x^2 + (72/x)
 Graph this equation to find the value of x for minimum material
 Min surface area when x = 3.0 is the width then
 2(3) = 6 is the length
 Find the height:
 h=12/(3.0)^2
 h=1.33
 Box dimensions for min surface area: 3.0 by 6 by 1.33; much better numbers
 Check the vol of these dimensions: 3.0*6*1.33 ~ 24
 graphic attachment