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

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I don't see how the given solution set is correct, as there is no ##x## which satisfies the first equation when we solve the second for ##y##. The only stationary point is the critical point ##(1/6, 1/2)##, which, under the conditions for a stationary point, is a saddle point.
  • #1
Retro95
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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
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)
 
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  • #2
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 ?
 
  • #3
Retro95 said:

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.
Retro95 said:
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.
Retro95 said:
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)
 
  • #4
Mark44 said:
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
 
  • #5
Retro95 said:
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

Retro95 said:
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.

Retro95 said:
The x's cancel out, so..

-42+84y=0
84y=42
y=0.5
Correct, despite the error I noted.
Retro95 said:
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 fx = fy.
Retro95 said:
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.
 
  • #6
Mark44 said:
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?
 
  • #7
Post #2: What is/are the condition(s) for a stationary point ?
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
Retro95 said:
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.
 
  • #9
BvU said:
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 ?

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

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

What are stationary points?

Stationary points are points on a graph where the derivative (slope) is equal to zero. This means that the function is neither increasing nor decreasing at that point.

What is the significance of the coordinates (0,0.5), (0,-0.5), (1.25,0), and (-1.25,0) as stationary points?

These coordinates represent the x and y values of points on a graph where the derivative is equal to zero. This means that the function is not changing at these points, and they can be either maximum or minimum points depending on the shape of the graph around them.

What is a maximum point?

A maximum point is a point on a graph where the function reaches its highest value. This means that the slope of the function changes from positive to negative at this point, indicating a local maximum value.

What is the difference between a local maximum and a global maximum?

A local maximum is a point on a graph where the function reaches its highest value within a specific interval. A global maximum is a point where the function reaches its highest value over the entire domain of the function.

How can stationary points be used in real-life applications?

In real-life applications, stationary points can be used to find maximum or minimum values of a function, which can be helpful in optimizing processes or predicting outcomes. For example, in economics, stationary points can be used to find the maximum profit of a company or the minimum cost of production. In physics, stationary points can be used to find the maximum height of a projectile or the minimum energy required for a system to function.

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