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A rectangular piece of metal is 5in longer than it is wide. Squares with sides 1in 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 594in, what were the original dimensions of

the piece of metal?
What is the original width ____ inches?
What is the original length ____ inches?

User Jsinglet
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1 Answer

5 votes

Final answer:

The original width of the piece of metal is 25 inches and the original length is 30 inches.

Step-by-step explanation:

To find the original dimensions of the piece of metal, we need to find the length and width before the squares were cut. Let's let the width be x inches. According to the problem, the length is 5 inches longer than the width, so the length would be x+5 inches.



After the squares with sides 1 inch long are cut from the corners, the dimensions of the resulting box would be (x-2) inches (length), (x-2) inches (width), and 1 inch (height). The volume of the box would be (x-2)(x-2)(1) = 594 cubic inches.



Simplifying this equation and solving for x, we have (x-2)(x-2) = 594. Expanding and rearranging, we get x^2-4x+4 = 594. Subtracting 594 from both sides, we have x^2-4x-590 = 0. Using the quadratic formula, we find that x = 25 or x = -23. Since the width cannot be negative, the original width of the piece of metal is 25 inches.



To find the original length, we can substitute the value of x = 25 into the equation for the length, which is x+5. Therefore, the original length of the piece of metal is 25+5 = 30 inches.

User Jack Nutkins
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