# Stationary points

1. Feb 27, 2014

### Retro95

1. The problem statement, all variables and given/known data
Find the stationary point(s) and also find out whether the stationary point(s) are minimum/maximum
z = 2x + 5y - 5y2 - 5x2 - 4xy - 1000

2. Relevant equations

-

3. The attempt at a solution
I'm not 100% sure if I was correct and laid the answer out properly, can someone please help me check it out and correct any mistakes

fx = 2 - 10x - 4y
fy = 5 - 10y - 4x
fxx = -10
fyy = -10
fxy = -4

2 - 10x - 4y = 0, therefore 4y = 2 and y = 0.5
Possible stationary points (0,0.5) (0,-0.5)

5 - 10y - 4x = 0, therefore 4x = 5 and x = 1.2
Possible stationary points (0,0.5) (0,-0.5)
(1.25,0) (-1.25,0)

fxxfyy-f2xy = 84 (Greater than 0, and fxx & fyy are less than 0 therefore it's a maximum)

2. Feb 27, 2014

### BvU

Don't understand the "therefore". You solve fx=0 for x=0 . There is only one solution. Where does the (0, -0.5) appear ?

10x + 4y=2 is the equation for a straight line.

Idem 5 - 10y - 4x = 0. (0, -0.5) does not satisfy that !

What is/are the condition(s) for a stationary point ?

3. Feb 27, 2014

### Staff: Mentor

4y = 2 only if x = 0, which you don't know. The equation 2 - 10x - 4y = 0 has an infinite number of solutions.
No, 4x = 5 only if y = 0, which you also don't know.

Solve the two equations simultaneously to find the point or points that are on both lines. If you're in a calculus class, you should know how to solve a system of two linear equations.

4. Feb 28, 2014

### Retro95

Solving both of them simultaneously I got..

2-10x-4y=0
5-4x-10y=0

Multiplying the top by 4 and the bottom by 10 gives..

8-40x-4y=0
50-40x-100y=0

The x's cancel out, so..

-42+84y=0
84y=42
y=0.5

Subbing y into the original equation makes x=0.12

I've done the main working out, could you please confirm how I would then present the answer

5. Feb 28, 2014

### Staff: Mentor

You really make it hard on yourself. A little work in advance saves time and minimizes the chance for errors. I would write this system as
10x + 4y = 2
4x + 10y = 5

No, but this might have been a transcription error. The first equation should be 8 - 40x - 16y = 0.
Also, it makes more sense to multiply one equation by a positive number and the other by a negative number. Then you can add the equations rather than subtract one from the other.

Correct, despite the error I noted.
No. You're looking for the solution of the system of equations. Use your y value to find the x value for which fx = fy.
When you have found the critical point, use the other partials to determine whether it's a local maximum, local minimum, or whatever.

6. Feb 28, 2014

### Retro95

Thanks for the help I really appreciate it but if you're gonna help me then you might aswell help me set out the answer since I've done the main working outs

Since I've concluded that a stationary point is at (0,0.5) and it's a maximum point since fxx & fyy are negative, also because fxxfyy-f2xy is positive. How would I actually present the answer?

7. Feb 28, 2014

### BvU

You found one point where fx=0 AND fy=0 so that is a stationary point.

Same question for criteria for maximum, minimum, saddle point.

At (0, 0.5), fxx and fyy are both negative, so it is a maximum.

You bring in fxy but I don't see much use for that.

My question: this is homework for a chapter or section in your syllabus or textbook. Don't they summarize these conditions/criteria somewhere ?

8. Feb 28, 2014

### Staff: Mentor

Pretty much like what you have above.

9. Feb 28, 2014

### Ray Vickson

The second mixed partials are crucial to overall behavior and cannot be simply dismissed. Having $f_{xx}, f_{yy} < 0$ does not guarantee a maximum; it does eliminate a minimum, but still leaves open the possibility of a saddle point. For example for both functions $f = -x^2-y^2 \pm 100 xy$ the origin is the only stationary point, but is a saddle point rather than a maximum.

10. Feb 28, 2014

### BvU

Thanks Ray, I learned something. Must have gotten used too much to less pathologic functions! Retro did just fine then.