Pre-Calc Homework Solutions 201

Pre-Calc Homework Solutions 201 - Chapter 4 Review 57. The...

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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
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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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