SOLUTION: PROBLEM: VOLUME OF A BOX:
A RECTANGULAR PIECE OF METAL IS 10 IN. LONGER THAN IT IS WIDE. SQUARES WITH SIDES 2 IN LONG ARE CUT FROM THE FOUR CORNERS, AND THE FLAPS ARE FOLDED UPWAR
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Question 143854This question is from textbook College Algebra
: PROBLEM: VOLUME OF A BOX:
A RECTANGULAR PIECE OF METAL IS 10 IN. LONGER THAN IT IS WIDE. SQUARES WITH SIDES 2 IN LONG ARE CUT FROM THE FOUR CORNERS, AND THE FLAPS ARE FOLDED UPWARD TO FORM AN OPEN BOX. IF THE VOLUME OF THE BOX IS 832 IN.^3, WHAT WERE THE ORIGINAL DIMENSIONS OF THE PIECE OF METAL?
This question is from textbook College Algebra
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Start by letting the width of the metal sheet be x inches, then the length is x+10 inches.
The volume of the box (832 cu.in.) will be calculated by the length (less 4 inches) times width (less 4 inches) times the height of the box (2 inches).
We can write the equation:
or...
Divide both sides by 2 to simplify a bit.
Perform the multiplication.
Subtract 416 from both sides.
Factor this quadratic equation.
so that...
or Discard the negative solution as the length cannot be a negative quantity.
The original dimensions of the metal sheet are:
Width = 20 inches and the length = 30 inches.
Check:
The volumes is:
(x-4)(x+6)(2)(2) = 832 Substitute x = 20.
OK!
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