This preview shows page 1. Sign up to view the full content.
Unformatted text preview: h suﬃciently close to zero). Answer: A ( r ) = 2 πr 2 + 2000 r , < r < + ∞ . 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 > 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 , < 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 opentopped 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...
View
Full
Document
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.
 Spring '11
 CHENG
 Calculus

Click to edit the document details