Pre-Calc Homework Solutions 198

# Pre-Calc Homework Solutions 198 - 198 Chapter 4 Review r 6...

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47. If the dimensions are x ft by x ft by h ft, then the total amount of steel used is x 2 1 4 xh ft 2 . Therefore, x 2 1 4 xh 5 108 and so h 5 } 108 4 2 x x 2 } . The volume is given by V ( x ) 5 x 2 h 5 } 108 x 4 2 x 3 }5 27 x 2 0.25 x 3 . Then V 9 ( x ) 5 27 2 0.75 x 2 5 0.75(6 1 x )(6 2 x ) and V 0 ( x ) 52 1.5 x . The critical point occurs at x 5 6, and it corresponds to the maximum volume because V 0 ( x ) , 0 for x . 0. The corresponding height is } 108 4( 2 6) 6 2 3 ft. The base measures 6 ft by 6 ft, and the height is 3 ft. 48. If the dimensions are x ft by x ft by h ft, then we have x 2 h 5 32 and so h 5 } 3 x 2 2 } . Neglecting the quarter-inch thickness of the steel, the area of the steel used is A ( x ) 5 x 2 1 4 xh 5 x 2 1 } 12 x 8 } . We can minimize the weight of the vat by minimizing this quantity. Now A 9 ( x ) 5 2 x 2 128 x 2 2 5 } x 2 2 } ( x 3 2 4 3 ) and A 0 ( x ) 5 2 1 256 x 2 3 . The critical point occurs at x 5 4 and corresponds to the minimum possible area because A 0 ( x ) . 0 for x . 0. The corresponding height is } 3 4 2 2 } 5 2 ft. The base should measure 4 ft by 4 ft, and the height should be 2 ft.
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