Arectangular 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?

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

cosine

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