Maths Problem That I Really Should Know!

3x^2 = -y = x/2
=> x(6x - 1) = 0
=> x = 0 or x = 1/6

Case 1: x = 0.
y = 0.

Case 2: x = 1/6.
y = - x/2 = -1/12.

ok, thanks.

so how do i find the co-ordinates?

Surely those are the co-ords?

ummm... shouldn't there be more?

There are stationary points at (0, 0) and (1/6, -1/12).

ok, thanks, i thought there might be 4.

Any help much apprieciated with this differential equation problem:

Find the stationary points of the following functions determining in each case whether they are maxima, minima or saddle points:
iii) f = x^3 + xy + y^2

i said the derivitives with respect to...
fx = 3x^2 +y
fy = x + 2y

they must both equal zero for a stationary point. therefore are equal. Therefore should be able to work it out from there but my brain's gone dead!

I just need help to work out the co-ordinates. I can work out if it is max etc.

now we know why you're not a moderator here

Yeah i know not all have 4 stationary points, but i couldn't see why the other one did.

Okay. Let's set the partial derivatives to zero and find the corresponding x and y values:

fx = 3x^2 - 3 + y^2 = 0 (1)
fy = 2xy = 0 (2)

As before, we want to solve these equations simultaneously. Now (2) is true (i.e. LHS = RHS) if x = 0 OR y = 0. In the case where x = 0, we can find the corresponding y value by substituting x into (1):

fx = -3 + y^2 = 0 => y^2 = 3 => y = +- sqrt3. In other words we have TWO values of y associated with this value of x, and thus two stationary points when x = 0: (0,sqrt3) AND (0, -sqrt3).

In the case where y=0, we again substitute into (1) to get:

fx = 3x^2 - 3 = 0 => x^2 - 1 = 0 => x^2 = 1 => x = +-sqrt1 = +-1. Therefore this value of y is associated with TWO x values, so we have a further TWO stationary points: (1,0) AND (-1,0).

HTH

~Darkness~

f(x,y) = x^2y - y + 3x^2 + 4y^2

fx = 2xy + 6x = 2x(y+3) = 0 (1)
fy = x^2 -1 + 8y = 0 (2)

hey guys, come back round to this topic, finding it still hard to get the coordinates. I've got an answer but don't think it's right.

for example:
f(x,y) = x^2y - y + 3x^2 + 4y^2

fx = 2xy + 6x = 2x(y+3) = 0
fy = x^2 -1 + 8y = 0

now, I reckon there are 3 stationary points, where do you make them to be?

Cheat! Bouncing your thread back to the top by deleting and reposting the same thing!

I've never done partial differentiation, but:
From (2): 2y = (1 - x^2)/4
Sub into (1): x(1 - x^2)/4 + 6x = 0
x(1 - x^2) + 24x = 0
x^3 - 25x = 0
x(x+5)(x-5) = 0 so stationary points occur at x = -5,0,5

thanks, (and fermat), this agrees with my answer so have some rep for being 1st.

Always the bridesmaid