# Thread: nature of stationary points

1. ## nature of stationary points

Hello,

I would appreciate your help with discovering where I am going wrong with this.
In the following question (with the exception of i.) my problem is not with finding the stationary points (S.P.’s) but rather, with determining their nature. Note that I am instructed to do this without using the second derivative test. I am hoping (I think) that my error is not simply computational, if so excuse me…

Question: Investigate the S.P.’s of these surfaces

i. z = (x+y)(xy+1)

I have that the partial of z w.r.t. x is: z_x = 2xy + 1 + y^2
and that: z_y = 2xy + 1 + x^2

equating to zero and treating as a system of simultaneous eq’s => x = +/- y

this is the exception here as I can see that the S.P.’s are (1,-1,0) and (-1,1,0) however doesn’t x=+/- y suggest that, say, x = 5 and y = -5 is also a S.P., though by inserting these values into the partial’s it clearly isn’t as the outcome is not zero?...
I go on

evaluating the general point with co-ordinates x = 1+h and y = -1+k in the neighbourhood of (1,-1,0) I want to find the value of z and determining the nature of said point =>

z = (1+h-1+k)[(1+h)(-1+k)+1] = (h+k)(-1+k-h+hk+1) = (h+k)(k-h+hk)

similarly for the other S.P. => z = (h+k)(h-k+hk)

the book says these are saddle points and I also think that there is nothing to ensure that the z is for all small enough values of h and k either smaller or greater than zero, but I am reluctant and this shows more in

ii. z = x^2 + y + (2/x) + (4/y)

as before z_x = 2x – [2/(x^2)]
and z_y = 1 – [4/(y^2)]

and so the S.P.’s can be found (no problem) to be (1,2,7) and (1,-2,-1))

as before z = (1+h)^2 + (2+k) + [2/(1+h)] + [4/(2+k)]
and for the other S.P. z = (1+h)^2 + (-2+k) + [2/(1+h)] + [4/(-2+k)]

Now the book says that (1,2,7) is a minimum and (1,-2,-1) is a saddle point, but I can’t see this…?

iii. z = xy + ln|x| + 2y^2

as before z_x = y + 1/x
and z_y = x + 4y

and so the S.P.’s can be found (no problem) to be (2, -1/2, ln2-1/2) and (-2, 1/2, ln2-1/2)

as before z= (2+h)(k – 1/2) + ln|2+h| + 2(k – 1/2)^2
and for the other S.P. z = (-2+h)(k + 1/2) + ln|-2+h| + 2(k + 1/2)^2

the book says that these are both saddle points and me I also think that the second and third terms must be positive but the first can be either negative or positive for small enough values of h and k, though here again I have the feeling that i’m in the wrong. I should be comparing the value of z to ln2-1/2 not zero but that seems even more confusing…

pls help me I am stydying on my own and will feel quite incompetent if I can’t get a better grasp of these stuff.

2. Originally Posted by pb6883
Hello,

I would appreciate your help with discovering where I am going wrong with this.
In the following question (with the exception of i.) my problem is not with finding the stationary points (S.P.’s) but rather, with determining their nature. Note that I am instructed to do this without using the second derivative test. I am hoping (I think) that my error is not simply computational, if so excuse me…

Question: Investigate the S.P.’s of these surfaces

i. z = (x+y)(xy+1)

I have that the partial of z w.r.t. x is: z_x = 2xy + 1 + y^2
and that: z_y = 2xy + 1 + x^2

equating to zero and treating as a system of simultaneous eq’s => x = +/- y

this is the exception here as I can see that the S.P.’s are (1,-1,0) and (-1,1,0) however doesn’t x=+/- y suggest that, say, x = 5 and y = -5 is also a S.P., though by inserting these values into the partial’s it clearly isn’t as the outcome is not zero?...
You didn't complete the calculation. Yes, y= x or y= -x. Now put those back into the equations for z_x or z_y. With y= x, z_x= 2xy+ 1+y^2= 0 becomes 2x^2+ 1+ x^2= 0 or 3x^2= -1. That's impossible and gives no solution. With y= -x, z_x= 2xy+ 1+ y^2= 0 becomes -2x^2+ 1+ x^2= 0 or x^2= 1. That gives x= 1 or -1 and, since y= -x, the points are (1, -1) and (-1, 1).

I go on

evaluating the general point with co-ordinates x = 1+h and y = -1+k in the neighbourhood of (1,-1,0) I want to find the value of z and determining the nature of said point =>

z = (1+h-1+k)[(1+h)(-1+k)+1] = (h+k)(-1+k-h+hk+1) = (h+k)(k-h+hk)
and, for example, if h=0, k> 0, this is k(k)= k^2> 0. If h> 0, k= 0, it is h(-h)= -h^2< 0.

similarly for the other S.P. => z = (h+k)(h-k+hk)

the book says these are saddle points and I also think that there is nothing to ensure that the z is for all small enough values of h and k either smaller or greater than zero, but I am reluctant and this shows more in

ii. z = x^2 + y + (2/x) + (4/y)

as before z_x = 2x – [2/(x^2)]
and z_y = 1 – [4/(y^2)]

and so the S.P.’s can be found (no problem) to be (1,2,7) and (1,-2,-1))

as before z = (1+h)^2 + (2+k) + [2/(1+h)] + [4/(2+k)]
and for the other S.P. z = (1+h)^2 + (-2+k) + [2/(1+h)] + [4/(-2+k)]

Now the book says that (1,2,7) is a minimum and (1,-2,-1) is a saddle point, but I can’t see this…?
Did you consider using the "second derivative test" for this? While it requires calculating the second derivative, it only requires evaluating at a single point so, here, is easier to use.

If $\displaystyle f_{xx}f_{yy}-(f_{xy})^2$, evaluated at the critical point, is positive, then the point is either a maximum or a minimum. It is a minimum if f_{xx}> 0 (or f_{yy}> 0; they must have the same sign in this situation) and a maximum if f_{xx}< 0. If $\displaystyle f_{xx}f_{yy}- (f_{xy})^2$ is negative, the point is a saddle point.

Here, $\displaystyle f_{xx}= 2+ \frac{4}{x^3}$, $\displaystyle f_{yy}= \frac{8}{y^3}$ and $\displaystyle f_{xy}= 0$.

At (1, 2) $\displaystyle f_{xx}f_{yy}- (f_{xy})^2$ is (6)(1)- 0= 6> 0 while [tex]f_{xx}= 6> 0 so (1, 2) is a minimum.

At (1, -2) $\displaystyle f_{xx}f_{yy}- (f_{xy})^2$ is (-6)(1)- 0= -6< 0 so (1, -2) is a saddlepoint.

iii. z = xy + ln|x| + 2y^2

as before z_x = y + 1/x
and z_y = x + 4y

and so the S.P.’s can be found (no problem) to be (2, -1/2, ln2-1/2) and (-2, 1/2, ln2-1/2)

as before z= (2+h)(k – 1/2) + ln|2+h| + 2(k – 1/2)^2
and for the other S.P. z = (-2+h)(k + 1/2) + ln|-2+h| + 2(k + 1/2)^2

the book says that these are both saddle points and me I also think that the second and third terms must be positive but the first can be either negative or positive for small enough values of h and k, though here again I have the feeling that i’m in the wrong. I should be comparing the value of z to ln2-1/2 not zero but that seems even more confusing…

pls help me I am stydying on my own and will feel quite incompetent if I can’t get a better grasp of these stuff.
Again, it is simpler to use the "second derivative test". $\displaystyle f_{xx}= -\frac{1}{x^2}$, $\displaystyle f_{yy}= 4$ and $\displaystyle f_{xy}= 0$.

At (2, -1/2), $\displaystyle f_{xx}f_{yy}- (f_{xy})^2= (-\frac{1}{4})(4)- 0^2= -1< 0$ so (2, -1/2) is a saddle point.

At (-2, 1/2) $\displaystyle f_{xx}f_{yy}- (f_{xy})^2= (-\frac{1}{4})(4)- 0^2= -1< 0$ so (-2, 1/2) is also a saddle point.