HW 11 Solutions - Mathematics 408K: Homework 11 Gagan Tara...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Mathematics 408 K: Homework 11 Gagan Tara Nanda 02 nd November, 2004 HW 11 : # s 6 , 8 , 9 6. Let the side of the squares be a . Then the canvas covers four separate pieces. The sides surfaces are squares, each of side length a . The top is a rectangle, of width a and length x . The back is also a rectangle, again with length x and width a . So the total surface area is S = a 2 + a 2 + ax + ax =2 ¡ a 2 + ax ¢ . We are told that a total of 294 ft 2 of canvas is available. So S = 294 a 2 + ax = 147 .T h i sg i v e s ax = 147 a 2 . We wish to maximize the space inside the shelter. This is the same as the volume of aboxo fs ides a, a,x .S o V = a 2 x =( ax ) a .W r i t i n g V in terms of a only, we get V = ¡ 147 a 2 ¢ a = 147 a a 3 . We wish to maximize this quantity. So V 0 =147 3 a 2 . This has critical points when 147 3 a 2 =0 a 2 =49 a = ± 7 .C l e a
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/28/2010 for the course M 56200 taught by Professor Radin during the Fall '09 term at University of Texas at Austin.

Page1 / 2

HW 11 Solutions - Mathematics 408K: Homework 11 Gagan Tara...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online