What are the dimensions of the metal sheet?

A box is being constructed by cutting 3-inch squares from the corners of a rectangular sheet of metal that is 4 inches longer than it is wide. If the box is to have a volume of 135 cubic inches, find the dimensions of the metal sheet?What are the dimensions of the metal sheet?
okay the first thing you need to do is find the dimensions of the box. when it is laid out flat the dimensions are x and x+4. but since 3 inch squares are being cut out of each side of the box the new dimensions are x-6 and x-2. these are your length and width. and your height is 3 because when you take the 3 inch squares out and fold it, your box will be 3 inches high. now take all of these values and plug it into the volume formula.





v=l x w x h


135= (x-2)(x-6)(3)


135=(x^2-8x+12)(3)


135=3x^2-24x+36


0=3x^2-24x-99


0=3(x^2-8x-33)


0=3(x-11)(x+3)


x=11 and -3


-3 is rejected because you cannot have a negative measurement.





now plug this into you length and width (x-2, x-6)


11-2=9


11-6= 5





therefore the length is 9 and the width is 5.What are the dimensions of the metal sheet?
Volume of a rectangular solid is given to be 135 cubic inches.





The volume formula for this shape is V = LWH.





135 = 3x(x + 4)





135 = 3x^2 + 12x





3x^2 + 12x - 135 = 0





x^2 + 4x - 45 = 0





(x + 9) (x - 5) = 0





x + 9 = 0





x = - 9...This answer is rejected because distance cannot be negative.





x - 5 = 0





x = 5





Dimensions:





width = 5 inches





length = x + 4 or 5 + 4 = 9 inches





As you know, 9 inches is bigger than 5 inches by 4 inches.
L = W + 4





135 = 3(L-6)(W-6)


135 = 3(W-2)(W-6)


45 = W^2 - 8W + 12


0 = W^2 - 8W - 33


0 = (W-11)(W+3)


W = 11, -3 (negative dimension doesn't make sense, so W= 11)


L = 15