SOLUTION: A rectangular piece of metal is 15 inches longer that it is wide. Squares with sides 3 inches long are cut from the four corners and the flaps are folded upward to form an open box
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Question 582522: A rectangular piece of metal is 15 inches longer that it is wide. Squares with sides 3 inches 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 1938 in^3, what were the original dimensions of the piece of metal?
What is the original width?
What is the original width?
I have tried this question many times and my professor said the answers are 25,40 but no matter what I do I can't get that. Help. Thank you Answer by Bill M Wilcox(2) (Show Source):
You can put this solution on YOUR website! First, imagine doing this in real life. You have a sheet with length x and width x-15. When you cut out the corners, you change the dimensions, but you don't change the difference between them, so you're still looking at something like length y and width y-15, and now you have a third dimension (height) from folding those three-inch-wide flaps upward. This gives you a volume of
This can be reworked into the quadratic equation
The quadratic formula then gives the solution y = 34, the length of your box. Then y - 15 = 19 provides the width, and of course the height is 3 inches.
How do you find the original dimensions? Remember that each edge of your sheet lost 3 inches on both sides, for a total of 6 inches lost, so y = x - 6. All you need to do in order to find the original dimensions is add those 6 inches back:
There you have the original length and width of your sheet. I hope this explanation has been clear and helpful.
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