Stationary Points: (0,0.5) (0,-0.5) (1.25,0) (-1.25,0) Maximum Point

[Retro95]

Homework Statement

Find the stationary point(s) and also find out whether the stationary point(s) are minimum/maximum
z = 2x + 5y - 5y² - 5x² - 4xy - 1000

Homework Equations

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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)

 

[BvU]

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)

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 ?

 

[Mark44]

Homework Statement

Find the stationary point(s) and also find out whether the stationary point(s) are minimum/maximum
z = 2x + 5y - 5y² - 5x² - 4xy - 1000

Homework Equations

-

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

4y = 2 only if x = 0, which you don't know. The equation 2 - 10x - 4y = 0 has an infinite number of solutions.

Possible stationary points (0,0.5) (0,-0.5)

5 - 10y - 4x = 0, therefore 4x = 5 and x = 1.2

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.

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)

 

[Retro95]

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.

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

 

[Mark44]

Solving both of them simultaneously I got..

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

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

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

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

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.

The x's cancel out, so..

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

Correct, despite the error I noted.

Subbing y into the original equation makes x=0.12

No. You're looking for the solution of the system of equations. Use your y value to find the x value for which f_x = f_y.

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

When you have found the critical point, use the other partials to determine whether it's a local maximum, local minimum, or whatever.

 

[Retro95]

When you have found the critical point, use the other partials to determine whether it's a local maximum, local minimum, or whatever.

Thanks for the help I really appreciate it but if you're going to help me then you might as well 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?

 

[BvU]

Post #2: What is/are the condition(s) for a stationary point ?

You found one point where f_x=0 AND f_y=0 so that is a stationary point.

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

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

You bring in f_xy 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 ?

 

[Mark44]

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?

Pretty much like what you have above.

 

[Ray Vickson]

You found one point where f_x=0 AND f_y=0 so that is a stationary point.

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

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

You bring in f_xy 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 ?

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.

 

[BvU]

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