Explanation 1) Concept: Solving an expression is to find the value of unknown in the expression, which on substitution must satisfy the expression. 2) Calculation: Given, Let ‘ x ’ be the width of the rectangular piece of metal. Hence the dimensions of the box is, l = x + 6 w = x − 4 h = 2 Hence the value is given by, ( x + 6 ) ( x − 4 ) ( 2 ) = 832 [ ( x + 6 ) ( x − 4 ) ] ( 2 ) = 832 Multiplying with each term of the first factor with each term of the second factor, [ x ⋅ x − x ⋅ 4 + 6 ⋅ x − 6 ⋅ 4 ] ( 2 ) = 832 [ x 2 − 4 x + 6 x − 24 ] ( 2 ) = 832 . Combining the like terms, [ x 2 + ( − 4 x + 6 x ) − 24 ] ( 2 ) = 832 [ x 2 + 2 x − 24 ] ( 2 ) = 832 Divide 2 on both sides, [ x 2 + 2 x − 24 ] ( 2 ) 2 = 832 2 x 2 + 2 x − 24 = 416 Add − 416 on both sides, x 2 + 2 x − 24 − 416 = 416 − 416 Combining the like terms, x 2 + 2 x + ( − 24 − 416 ) = 0 x 2 − 2 x − 440 = 0 Solving by the middle term splitting method: Product of the coefficient of x 2 and constant = − 440. Coefficient of the middle term = 2. Now, − 440 is factorized as − 20 and 22 so as to get the sum 2

12th Edition

ISBN: 9780135902578

Author: Lial

Publisher: PEARSON

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Chapter 1.5, Problem 27E

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The original dimensions of the piece of metal such that a rectangular piece of metal is 10 in. longer than it is wide. Squares with sides 2 in. long are cut from the four comers, and the flaps are folded upward to form an open box. If the volume of the box is 832 in.³.

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Chapter 1.5, Problem 26E

Chapter 1.5, Problem 28E

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ESSENTIALS OF COLL...-MYLAB ACCESS

Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E

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The original dimensions of the piece of metal such that a rectangular piece of metal is 10 in. longer than it is wide. Squares with sides 2 in. long are cut from the four comers, and the flaps are folded upward to form an open box. If the volume of the box is 832 in. 3 .

To find: The original dimensions of the piece of metal such that a rectangular piece of metal is 10 in. longer than it is wide. Squares with sides 2 in. long are cut from the four comers, and the flaps are folded upward to form an open box. If the volume of the box is 832 in.3.

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To find:

The original dimensions of the piece of metal such that a rectangular piece of metal is 10 in. longer than it is wide. Squares with sides 2 in. long are cut from the four comers, and the flaps are folded upward to form an open box. If the volume of the box is 832 in.³.