{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

8_Cal_Solution of Calculus_6e

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

This preview shows page 1. Sign up to view the full content.

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 find 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 18 (because y 0). The volume of the cylinder is V = πy 2 x , and so V ( x ) = πx (18 x ) 2 , 0 x 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 = 1000. But r can be arbitrarily small positive as well as arbitrarily large positive (by making
This is the end of the preview. Sign up to access the rest of the document.

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 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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online