# Maths Problem That I Really Should Know!

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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.

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.

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#2

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.

=> 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.

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(Original post by

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.

**Jonny W**)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.

so how do i find the co-ordinates?

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#4

(Original post by

ok, thanks.

so how do i find the co-ordinates?

**Ollie**)ok, thanks.

so how do i find the co-ordinates?

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(Original post by

Surely those are the co-ords?

**Nylex**)Surely those are the co-ords?

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(Original post by

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

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

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(Original post by

ok, thanks, i thought there might be 4.

**Ollie**)ok, thanks, i thought there might be 4.

f = x^3 - 3x + xy^2

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

fy = 2xy (2)

2 is satisfied by x=0, y=0

x=0 into (1) give y^2 = 3 therefore y = +- root3.

therefore (0, root3) and (0, -root3) are stationery

y=0 into (1) gives x^2 =1, x = +-1

therefore (1,0) and (-1,0) are stationery

Thats what he wrote up in the example.

I'm not entirely sure whats going on!

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#10

(Original post by

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.

**Ollie**)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.

fx = 3x^2 +y = 0

fy = x + 2y = 0 => x = -2y (*)

Substitute -2y for x in fx to obtain:

fx = 3(-2y)^2 + y = 0

fx = 12(y)^2 + y = 0 => y(12y + 1) = 0 => y=0 or y=-1/12

Substitute y values in (*) to obtain corresponding x values: x = 0 or x = 1/6

Thus, the stationary points are: (0,0),(1/6,-1/12).

As for your example, not all functions have 4 stationary points!! Some have 1, some have none, some have 4, some have 3 etc...

~Darkness~

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(Original post by

now we know why you're not a moderator here

**elpaw**)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.

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#13

(Original post by

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

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

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~

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(Original post by

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~

**Darkness**)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~

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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?

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?

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#16

(Original post by

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)

**Ollie**)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)

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

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#17

(Original post by

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?

**Ollie**)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?

f'x = 2xy + 6x

f'y = x² - 1 + 8y

f'x = 2xy + 6x = 0

2x(y + 3) = 0

x=0, y= -3

=======

x=0

==

f'y = x² - 1 + 8y = 0

0 - 1 + 8y = 0

y = 1/8

====

y= - 3

====

f'y = x² - 1 + 8y = 0

x² - 1 - 24 = 0

x² = 25

x = ± 5

=====

Stationary points are

(0, 1/8), (5, -3), (-5, -3)

================

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(Original post by

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

**Bezza**)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

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#19

(Original post by

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

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

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