Pre-Calc Homework Solutions 154

Pre-Calc Homework Solutions 154 - 154 Section 4.4 9. x 2 x2...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
3. The circumference of the original circle of radius 4 is 8 p . Thus, 0 # x # 8 p . [0, 8 p ] by [ 2 10, 40] 4. The maximum occurs at about x 5 4.61. The maximum volume is about V 5 25.80. 5. Start with } d d V x } 5 } 2 3 p } rh } d d x r } 1 } p 3 } r 2 } d d h x } . Compute } d d x r } and } d d h x } , substitute these values in } d d V x } , set } d d V x } 5 0, and solve for x to obtain x 5 } 8(3 2 3 ˇ 6 w ) p } < 4.61. Then V 5 } 128 2 p 7 ˇ 3 w } < 25.80. Quick Review 4.4 1. y 95 3 x 2 2 12 x 1 12 5 3( x 2 2) 2 Since y 9$ 0 for all x (and y 9. 0 for x ± 2), y is increasing on ( 2‘ , ) and there are no local extrema. 2. y 95 6 x 2 1 6 x 2 12 5 6( x 1 2)( x 2 1) y 05 12 x 1 6 The critical points occur at x 52 2 or x 5 1, since y 95 0 at these points. Since y 0 ( 2 2) 52 18 , 0, the graph has a local maximum at x 52 2. Since y 0 (1) 5 18 . 0, the graph has a local minimum at x 5 1. In summary, there is a local maximum at ( 2 2, 17) and a local minimum at (1, 2 10). 3.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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.

Ask a homework question - tutors are online