# Thread: Saddle points = sore points

1. ## Saddle points = sore points

Hi, I'm trying to solve the following maths problem and have fallen at the first hurdle... which is funny as it relates to a maxima minima "Saddle" Point question

Find the stationary points of the surface (variables x and y) and saddle points

f(x,y) = 3x² + 6xy + 2y³ + 12x - 24y

Ive got it to:

i) dz/dx = 6x +6y +12 = 0

ii) dz/dy = 6x + 6y² - 24 = 0

I've tried substituting i into ii but all i get is a mess which doesn't make sense, im getting confused as there's three numbers that equal 0 , any ideas 2. Originally Posted by thegull Hi, I'm trying to solve the following maths problem and have fallen at the first hurdle... which is funny as it relates to a maxima minima "Saddle" Point question

Find the stationary points of the surface (variables x and y) and saddle points

f(x,y) = 3x² + 6xy + 2y³ + 12x - 24y

Ive got it to:

i) dz/dx = 6x +6y +12 = 0

ii) dz/dy = 6x + 6y² - 24 = 0

I've tried substituting i into ii but all i get is a mess which doesn't make sense, im getting confused as there's three numbers that equal 0 , any ideas Divide the equations by 6 to simplify them:
i) $\displaystyle x + y + 2 = 0$,
ii) $\displaystyle x + y^2 - 4 = 0$.
Subtract i) from ii): $\displaystyle y^2-y-6=0$. Solve that for y, substitute into i) to find the corresponding values of x.

Next, differentiate i) and ii) to find the second partial derivatives of z, and use the test $\displaystyle \tfrac{\partial^2 z}{\partial x^2}\tfrac{\partial^2 z}{\partial y^2} - \bigl(\tfrac{\partial^2 z}{\partial x\partial y}\bigr)^2<0$ on the two stationary points to see whether they are saddle points.

3. Thanks fot that

Ok, ive got y= (6/y²)

so substituted into i) x + (6/y²) + 2 = 0

im unable to simplify this to get the stationary points from here, I must be wrong?!

4. Originally Posted by thegull Thanks fot that

Ok, ive got y= (6/y²)

so substituted into i) x + (6/y²) + 2 = 0

im unable to simplify this to get two stationary points from here, I must be wrong?!
Go back and re-read the reply given in post #2. It doesn't look to me like you've followed and done what was said.

5. mmmmm, I see ive missed the -out from y.

so: y² - y - 6 = 0

is y = y² - 6

substituted into i) x + y + 2 = 0
becomes x + (y²-6) + 2 = 0 ????

stationary points

x = 0 or (y²-6) + 2. = 0
y² = 8
y = 4

where x = 0 using 3 y = 0
where x = 4 using 3 y = 4²/4 = 2

stationary points (0,0) and (4,4)... not very convinced myself. !

which I think are wrong for sure !

6. Originally Posted by thegull mmmmm, I see ive missed the -out from y.

so: y² - y - 6 = 0
[snip]
This is a quadratic equation. Solve it for y. (I suggest factorise and use null factor law).

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