Question 1145631: A rectangular sheet of tin is 40cm longer than it is wide. Squares 15cm by 15cm are cut from each corner of the sheet, and the sides are folded up and soldered to make an open box with a volume of 67,500 cm^3 What were the dimensions of the original sheet of tin?
Answer by ikleyn(52778)   (Show Source):
You can put this solution on YOUR website! .
Let w be the original width of the sheet, in centimeters.
Then the original length was (w + 40) cm.
After cutting squares at the corners, folding and soldering, the base of the box has
dimensions (w-2*15) = (w-30) cm and (w+40-2*15) = (w+10) cm.
Therefore, the volume of the open box is (w-30)*(w+10)*15 cm^3.
It gives you an equation for the volume
(w-30)*(w+10)*15 = 67500 cm^3.
Cancel by the factor 15 both sides
(w-30)*(w+10) = 4500
You can transform this equation to the standard quadratic equation form and solve it using the quadratic formula.
You can also notice that the only decomposition of the number 4500 into the product of two numbers
with the difference of 40 is 4500 = 50*90,
so you can get the solution MENTALLY w-30 = 50, w = 50+30 = 80.
ANSWER. The dimensions of the original rectangular sheet were 80 cm (width) and 80+40 = 120 cm (length).
Solved.
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