if y^3=6xy-x^3-1 show that the maximum value of y occurs where x^3=8+2sqrt(14)

Answer 1

find dy/dx

3y^2 dy/dx = 6x dy/dx + 6y - 3x^2
(dy/dx)(3y^2 - 6x) = 6y - 3x^2
dy/dx = (6y - 3x^2)/(3y^2 - 6x)
= 0 for a max/min

6y - 3x^2 = 0
y = x^2/2

sub that back into the original:
(x^2/2)^3 = 6x(x^2/2) - x^3 - 1
x^6/8 = 3x^3 - x^3 - 1
x^6/8 = 2x^3 - 1
x^6 = 16x^3 - 8
let x^3 = t , then
t^2 - 16t = -8
complete the square:
t^2 - 16t + 64 = -8+64
(t-8)^2 = 56
t-8 = ± √56 = ± 2√14
t = 8 ± 2√14

thus x^3 = 8 ± 2√14

I will leave it up to you to determine which of the two answers represents the maximum, and which produces the minimum.