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57. The dimensions will be x in. by 10 2 2 x in. by 16 2 2 x in., so V ( x ) 5 x (10 2 2 x )(16 2 2 x ) 5 4 x 3 2 52 x 2 1 160 x for 0 , x , 5. Then V 9 ( x ) 5 12 x 2 2 104 x 1 160 5 4( x 2 2)(3 x 2 20), so the critical point in the correct domain is x 5 2. This critical point corresponds to the maximum possible volume because V 9 ( x ) . 0 for 0 , x , 2 and V 9 ( x ) , 0 for 2 , x , 5. The box of largest volume has a height of 2 in. and a base measuring 6 in. by 12 in., and its volume is 144 in 3 . Graphical support: [0, 5] by [ 2 40, 160] 58. Step 1: r 5 radius of circle A 5 area of circle Step 2: At the instant in question, } d d r t }52} p 2 } m/sec and r 5 10 m. Step 3: We want to find } d d A t } . Step 4: A 5 p r 2 Step 5: } d d A t } 5 2 p r } d d r t } Step 6: } d d A t } 5 2 p (10) 1 2} p 2 } 2 52 40 The area is changing at the rate of 2 40 m 2 /sec. 59. Step 1: x 5 x -coordinate of particle y 5 y -coordinate of particle D 5 distance from origin to particle Step 2: At the instant in question, x 5 5 m, y 5 12 m, } d d x t }52 1 m/sec, and } d d y t 5 m/sec. Step 3: We want to find } d d D t } . Step 4: D 5 ˇ x 2 w 1 w y w 2 w Step 5: } d d D t } 5 } 2 ˇ x 2 w 1 1 w y w 2 w } 1 2 x } d d x t } 1 2 y } d d y t } 2 5 Step 6: } d d D t } 55 2 5 m/sec Since } d d D t } is negative, the particle is approaching the origin at the positive rate of 5 m/sec. 60. Step 1: x 5 edge of length of cube V 5 volume of cube Step 2: At the instant in question, } d d V t } 5 1200 cm 3 /min and x 5 20 cm. Step 3: We want to find } d d x t } . Step 4: V 5 x 3 Step 5: } d d V t } 5 3 x 2 } d d x t } Step 6: 1200 5 3(20) 2 } d d x t } } d d x t } 5 1 cm/min The edge length is increasing at the rate of 1 cm/min. (5)( 2 1) 1 (12)( 2 5) }}} ˇ 5 w 2 w 1 w 1 w 2 w 2 w x } d d x t } 1 y } d d y t } }} ˇ x 2 w 1 w y w 2 w Chapter 4 Review 201

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61. Step 1: x 5 x -coordinate of point y 5 y -coordinate of point D 5 distance from origin to point
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