SOLUTION: A piece of machinery is capable of producing rectangular sheets of metal such that the width is three inches less than two-thirds the length. Furthermore, equal-sized squares meas

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Question 823428: A piece of machinery is capable of producing rectangular sheets of metal such that the width is three inches less than two-thirds the length. Furthermore, equal-sized squares measuring 6 inches on a side can be cut from the corners so that the resulting piece of metal can be shaped into an open box by folding up the flaps. If specifications call for the volume of the box to be 4446 cubic inches, what should the dimensions of the original piece of metal be?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A piece of machinery is capable of producing rectangular sheets of metal such that the width is three inches less than two-thirds the length.
Furthermore, equal-sized squares measuring 6 inches on a side can be cut from the corners so that the resulting piece of metal can be shaped into an open box by folding up the flaps.
If specifications call for the volume of the box to be 4446 cubic inches, what should the dimensions of the original piece of metal be?
:
Let L = the length of the sheet of metal
Let W = the width
"the width is three inches less than two-thirds the length."
W = 2%2F3L-3 or (.67L-3)
:
Removing the 6" from the metal dimensions give a box dimension of:
(L-12)*(W-12) * 6 = 4446
Divide both sides by 6
(L-12)*(W-12) = 741
FOIL
LW - 12L - 12W + 144 = 741
LW - 12L - 12W + 144 - 741 = 0
LW - 12L - 12W - 597 = 0
Replace W with (.67L-3)
L(.67L-3) - 12L - 12(.67L-3) - 597 = 0
.67L^2 - 3L - 12L - 8L + 36 - 597 = 0
Combine like terms
.67L^2 - 23L - 561 = 0
Multiply by 3 to get rid of the annoying decimal
2L^2 - 69L - 1683 = 0
You can use the quadratic formula, but this will factor to
(2L + 33)(L - 51) = 0
the positive solution is all we want here
L = 51 inches is the length of the metal sheet
Find the Width
2%2F3*51 - 3 = 31 inches is the width
:
;
Check this
(51-12)*(31-12)* 6 = 4446 cu/in