Explanation Step :1 Let w is the width of rectangular hole and l is the length of rectangular hole, Perimeter of the rectangular hole is given by 2 ( l + w ) = 48 . Solving for w . 2 ( l + w ) = 48 l + w = 48 2 l + w = 24 w = 24 − l Step :2 Length of diagonal D of rectangular hole, Note that, by Pythagoras Theorem, ( Diagonal ) 2 = ( length ) 2 + ( width ) 2 D 2 = l 2 + w 2 D = ( l 2 + w 2 ) Substitute, w = 24 − l in D = ( l 2 + w 2 ) . D = ( l 2 + w 2 ) = ( l 2 + ( 24 − l ) 2 ) = ( l 2 + l 2 − 48 l + 576 ) = ( 2 l 2 − 48 l + 576 ) Step 3: Differentiate D with respect to l . D = ( 2 l 2 − 48 l + 576 ) d D d l = 4 l − 48 ( 2 l 2 − 48 l + 576 ) ( 1 2 ) = 2 l − 24 ( 2 l 2 − 48 l + 576 ) Step 4: For maxima and minima d D d l = 0 . That is, 2 l − 24 ( 2 l 2 − 48 l + 576 ) = 0 2 l − 24 = 0 l = 24 2 l = 12 Differentiate d D d l with respect to l

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ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 24.7, Problem 15E

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The dimensions rectangular hole so that the length of diagonal is minimum and the perimeter of the rectangular hole is 48 inches.

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Chapter 24.7, Problem 14E

Chapter 24.7, Problem 16E

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Chapter 24 Solutions

Basic Technical Mathematics

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