This preview shows page 1. Sign up to view the full content.
Unformatted text preview: y = 1 & 2 x 2 : Its &rst derivative is dy dx = & 4 x: Finding where dy=dx = 0 requires solving dy dx = & 4 x = 0 : Doing this implies x = 0 : Now &nd where dy=dx = 0 for the following functions: (i) y = 1 & x 2 ; (ii) y = 1 & 2 x + x 2 ; y = x (1 & a & x ) & cx where a; c are any numbers. 5. By selling output q 1 ; &rm 1 makes pro&t & 1 = q 1 (1 & q 1 & q 2 ) & cq 1 where c; q 2 are any positive constants. Find q 1 where d& 1 =dq 1 = 0 : [If you have problems doing this, put c = 0 : 2 ; q 2 = 0 : 2 and answer the question ] : 1...
View Full
Document
 Winter '11
 HuyDuong

Click to edit the document details