Aruwin
Aruwin
Branch Unspecified
08 Feb 2012

Extreme values of implicit function

Please check if my solution is correct or not.
Find the extreme values of the following implicit function:
2xy^3 + y - x^2 = 0
My attempt:
2xy³ + y - x² = 0
2y³ + 6xy²(dy / dx) + dy / dx - 2x = 0
6xy²(dy / dx) + dy / dx = 2x - 2y³
(6xy² + 1)(dy / dx) = 2x - 2y³
dy / dx = (x - y³) / (6xy² + 1)

dy / dx = 0
(x - y³) / (6xy² + 1) = 0
x - y³ = 0
y³ = x

2y⁶ + y - y⁶ = 0
y⁶ + y = 0
y(y⁵ + 1) = 0
y = 0 OR y⁵ + 1 = 0
y = 0 OR y⁵ = -1
y = 0 OR y = -1

hence the extreme points:
When y = 0, x = 0
When y = -1, x = -1
The extreme points are (-1, -1) and (0, 0).

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