Question

A rectangular piece of metal is 5 in longer than it is wide. Squares with sides 1 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 500 in³​, what were the original dimensions of the piece of metal?

What is the original​ width?

Answer 1

The original dimensions of the piece of metal were 15 inches by 20 inches.

Step-by-step explanation:

To find the original dimensions of the metal piece, we start by defining the width of the metal as \(x\) inches. Given that the length is 5 inches longer than the width, the length can be represented as
`\(x + 5\)` inches.

After cutting squares with sides 1 inch from each corner and folding the flaps to form an open box, the resulting box's height becomes 1 inch. Therefore, the dimensions of the box base become \
`((x - 2)`\times
`(x + 3)\)` inches after the squares are cut.

The volume of the box is calculated by multiplying the base area by the height:
`\((x - 2) \`times
`(x + 3) \`times 1 = 500\) in³. Solving this equation yields
`\(x^2 + x - 506 = 0\)`. Factoring or using the quadratic formula gives us
`\(x = 20\)` or
`\(x = -25\)`. Since the width cannot be negative, the width of the original metal piece is 20 inches.

Therefore, the original dimensions of the metal piece were 20 inches by
`\((20 + 5) = 25\)`inches, resulting in a piece measuring 15 inches by 20 inches after cutting and folding.