Pre-Calc Homework Solutions 161

# Pre-Calc Homework Solutions 161 - 31. Refer to the...

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Unformatted text preview: 31. Refer to the illustration in the problem statement. Since x 2 1 y 2 5 9, we have x 5 9 w 2 w y w 2 w . Then the volume of the cone is given by V 5 } 1 3 } p r 2 h 5 } 1 3 } p x 2 ( y 1 3) 5 } 1 3 } p (9 2 y 2 )( y 1 3) 5 } p 3 } ( 2 y 3 2 3 y 2 1 9 y 1 27), for 2 3 , y , 3. Thus } d d V y } 5 } p 3 } ( 2 3 y 2 2 6 y 1 9) 5 2 p ( y 2 1 2 y 2 3) 5 2 p ( y 1 3)( y 2 1), so the critical point in the interval ( 2 3, 3) is y 5 1. Since } d d V y } . 0 for 2 3 , y , 1 and } d d V y } , 0 for 1 , y , 3, the critical point does correspond to the maximum value, which is V (1) 5 } 32 3 p } cubic units. 32. (a) Note that w 2 1 d 2 5 12 2 , so d 5 1 w 4 w 4 w 2 w w w 2 w . Then we may write S 5 kwd 2 5 kw (144 2 w 2 ) 5 144 kw 2 kw 3 for 0 , w , 12, so } d d w S } 5 144 k 2 3 kw 2 5 2 3 k ( w 2 2 48). The critical point (for 0 , w , 12) occurs at w 5 4 w 8 w 5 4 3 w . Since } d d w S } ....
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## This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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