8_Cal_Solution of Calculus_6e

8_Cal_Solution of Calculus_6e - h sufficiently close to...

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Because x> 0 and y> 0 (the box has positive volume), but because y can be arbitrarily close to zero (as well as x ), we see also that 0 <x< + . We use the equation x 2 y = 324 to eliminate y from Eq. (1) and thereby Fnd that C ( x )=3 x 2 + 1296 x , 0 <x< + . C01S01.044: If the rectangle is rotated around its side S of length x to produce a cylinder, then x will also be the height of the cylinder. Let y denote the length of the two sides perpendicular to S ; then y will be the radius of the cylinder; moreover, the perimeter of the original rectangle is 2 x +2 y = 36. Hence y =18 x . Note also that x = 0 and that x 5 18 (because y = 0). The volume of the cylinder is V = πy 2 x , and so V ( x )= πx (18 x ) 2 , 0 5 x 5 18 . C01S01.045: Let h denote the height of the cylinder. Its radius is r , so its volume is πr 2 h = 1000. The total surface area of the cylinder is A =2 πr 2 +2 πrh (look inside the front cover of the book); h = 1000 πr 2 , so A =2 πr 2 +2 πr · 1000 πr 2 =2 πr 2 + 2000 r . Now r cannot be negative; r cannot be zero, else πr 2 h 6 = 1000. But r can be arbitrarily small positive as well as arbitrarily large positive (by making
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Unformatted text preview: h sufficiently close to zero). Answer: A ( r ) = 2 πr 2 + 2000 r , &lt; r &lt; + ∞ . C01S01.046: Let y denote the height of the box (in centimeters). Then 2 x 2 + 4 xy = 600 , so that y = 600 − 2 x 2 4 x . (1) The volume of the box is V = x 2 y = (600 − 2 x 2 ) · x 2 4 x = 1 4 (600 x − 2 x 3 ) = 1 2 (300 x − x 3 ) by Eq. (1). Also x &gt; 0 by Eq. (1), but the maximum value of x is attained when Eq. (1) forces y to be zero, at which point x = √ 300 = 10 √ 3. Answer: V ( x ) = 300 x − x 3 2 , &lt; x 5 10 √ 3 . C01S01.047: The base of the box will be a square measuring 50 − 2 x in. on each side, so the open-topped box will have that square as its base and four rectangular sides each measuring 50 − 2 x by x (the height of the box). Clearly 0 5 x and 2 x 5 50. So the volume of the box will be V ( x ) = x (50 − 2 x ) 2 , 5 x 5 25 . 8...
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This note was uploaded on 09/14/2011 for the course MATH 101 taught by Professor Cheng during the Spring '11 term at Ecole Hôtelière de Lausanne.

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