extreme values of an implicit function

[aruwin]

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).​

 

[SlipEternal]

You need to check second derivatives to make sure you don't have points of inflection, but other than that, it looks good.

 

[aruwin]

You need to check second derivatives to make sure you don't have points of inflection, but other than that, it looks good.

I just realized something. My friend did the same exercise and he got a different answer because he used a different method. He said that to find the extreme values of an implicit function is different than the usual way.I went through my reference book and it seems like he's right. Here's the work:

 

[SlipEternal]

I just realized something. My friend did the same exercise and he got a different answer because he used a different method. He said that to find the extreme values of an implicit function is different than the usual way.I went through my reference book and it seems like he's right. Here's the work:

View attachment 1700

What you did assumes that you have \(\displaystyle f(x,y) = 2xy^3+y-x^2\). You do not. You have \(\displaystyle 2xy^3+y-x^2=0\). Examining partials is only useful if the function can exist for all \(\displaystyle x\) and \(\displaystyle y\). It cannot. Your actual equation can be expressed in the form \(\displaystyle y=f(x)\). There is no \(\displaystyle f(x,y)\) there. So, finding extreme values, you must assume that \(\displaystyle y=f(x)\) or \(\displaystyle x=f(y)\) because if \(\displaystyle f(x,y)=2xy^3+y-x^2=0\), then this is a constant function, and it has no extreme values.

Edit: In other words, implicit differentiation is correct. Taking partials is not.

 

[Deleted member 4993]

I have a "hiccup" about the use of word "extreme" - without specifying domain.

At x = 8 ± √(72) we have y = 2 → thus y = 1 is clearly not an "extreme" value in global sense.

 

[SlipEternal]

True. Those are local extreme values.