SOLUTION: Need help ASAP! A box is to be formed from a rectangular piece of sheet metal by cutting squares measuring 5 inches on a side and then folding the sides. The piece of sheet meta

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Need help ASAP! A box is to be formed from a rectangular piece of sheet metal by cutting squares measuring 5 inches on a side and then folding the sides. The piece of sheet meta      Log On


   

Question 934190: Need help ASAP!
A box is to be formed from a rectangular piece of sheet metal by cutting squares measuring 5 inches on a side and then folding the sides. The piece of sheet metal is twice as long as it is wide. If the volume of the box is to be 1760 inches cubed, what are the dimensions of the original piece of sheet metal?
Thx
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

if you cut away the 5 inches on each corner, the new dimensions are %282x-10%29 and %28x-10%29.
If you were to fold the corners up, the height is 5.
The equation for the volume is:
+5%282x-10%29%28x-10%29+=1760
+%282x-10%29%28x-10%29+=1760%2F5
+2x%5E2-20x-10x%2B100+=352...all terms divide by 2
+x%5E2-10x-5x%2B50+=176
+x%5E2-15x-126=0...factor
+x%5E2-21x%2B6x-126=0
+x%28x-21%29%2B6%28x-21%29=0
+%28x-21%29%28x%2B6%29+=+0
solutions:
if +%28x-21%29+=+0=> x=21
if +%28x%2B6%29+=+0=> x=-6...we don't need negative solution
so, x=+21
the dimensions of the metal are 21in by 42in