getting dy/dx implicitly , ....
2x -2x dy/dx - 2y + 8y^3 dy/dx = 0
dy/dx( 8y^3 - 2x) = 2y - 2x
dy/dx = (y-x)/(4y^3 - x)
= 0 for a max/min
so y-x=0 --> y = x
sub back into original
x^2 - 2x^2 + 2x^4 - 1 = 0
2x^4 - x^2 - 1 = 0
(x^2 - 1)(2x^2 + 1) = 0
x^2 = -1/2 ----> no real solution
or
x^2= 1
x = ± 1
if x = 1
1 - 2y + 2y^4 - 1=0
by observation , y = 1 and y = 0
if x = -1
1 + 2y + 2y^4 - 1 = 0
by observation y = -1 and y = 0
when x = 1, y has a max of 1
check and verification, see
http://www.wolframalpha.com/input/?i=plot+x%5E2+-2xy+%2B2y%5E4%3D1
let y be the function of x defined implicitly by the equation x^2 -2xy +2y^4=1. find the value of x where y obtains its maximum value.
1 answer
1 answer