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Pre-Calc Homework Solutions 154

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

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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 9 5 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 9 5 6 x 2 1 6 x 2 12 5 6( x 1 2)( x 2 1) y 0 5 12 x 1 6 The critical points occur at x 5 2 2 or x 5 1, since y 9 5 0 at these points. Since y 0 ( 2 2) 5 2 18 , 0, the graph has a local maximum at x 5 2 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. V 5 } 1 3 } p r 2 h 5 } 1 3 } p (5) 2 (8) 5 } 20 3 0 p } cm 3 4. V 5 p r 2 h 5 1000 SA 5 2 p rh 1 2 p r 2 5 600 Solving the volume equation for h gives h 5 } 1 p 0 r 0 2 0 } .
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