A rectangular piece of metal is 30 in longer than it is wide. Squares with sides 6 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 3354 incubed , what were the original dimensions of the piece of metal?

Question

Hana Roman

Mathematics

31 May, 10:33

A rectangular piece of metal is 30 in longer than it is wide. Squares with sides 6 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 3354 incubed , what were the original dimensions of the piece of metal?

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Answers (1)

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Quincy · Answer 1

Answer = 55,25 inches

Solution -

let's take x as length and y as width of the metal piece. As per the question x is 30 more than y,

⇒ x = y + 30

Then four square pieces of side 6 are cut from each corner,

so the new length and width are

x-12, y-12

Then the volume of the new box created will be

(x-12) (y-12) 6

in the question the volume of the given figure is given to be 3354

so (x-12) (y-12) 6 = 3354

putting the value of x in the the above equation

⇒ (y+30 - 12) (x-12) = 3354/6 = 559

⇒ (y+18) (y-12) = 559

⇒ y² + 6y - 775 = 0

⇒ y² + 31y - 25y - 775 = 0

⇒ (y+31) (y-25) = 0

⇒ y = - 31, 25

as length can not be - ve, so y = 25

then x = 25+30 = 55

Hence the dimensions of the metal piece are 55, 25 inches