Find Stationary Points of u(x,y) Function & Determine Max/Min/Saddle

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Thanks for the help!In summary, we are tasked with finding the four stationary points of the function u(x, y) = 4x^3 − 18(x^2)y + 24x(y^2) − 120y and determining whether they are maxima, minima, or saddle points. To do so, we use the equation E= (d^2u/dxy)^2 - [(d^2u/dx^2) * (d^2u/dy^2)]. After deriving the equations du/dx = 12x^2 - 36xy +24y^2 and du/dy = -18x^2 + 48xy-120, we can factor the first
  • #1
Jenkz
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Homework Statement



Find the four stationary points of the function:
u(x, y) = 4x^3 − 18(x^2)y + 24x(y^2) − 120y

Determine whether they are maxima, minima or saddle points.

Homework Equations



To find stationary points use:

E= (d^2u/dxy)^2 - [(d^2u/dx^2) * (d^2u/dy^2)]

E>0 saddle
E < 0 Maximum -> d^2u/dx^2 < 0
Minimum -> d^2u/dx^2 > 0

The Attempt at a Solution



du/dx = 12x^2 - 36xy +24y^2 (1)
du/dy = -18x^2 + 48xy-120 (2)

Stationary points mean both du/dx and du/dy are equal to 0. Here I should find simultaneous solution of equations (1) and (2). This is where I get stuck and I am not sure how to find them.

I have done the next part though:

E= (-36x + 48y)^2 - (1152x^2 - 1728xy)

But I need the stationary points to find the min/max/saddle points.
 
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  • #2
Try factoring one of those equations.
 
  • #3
Ok I've tried:

du/dx = 12x^2 - 36xy +24y^2 (1)
du/dy = -18x^2 + 48xy-120 (2)

(1) x^2 - 3xy + 2y^2 = 0
(2)-3x^2 + 8xy -20 = 0

Eliminate x^2 ; xy +6y^2 - 20= 0 ; x = 20/y - 6y
Or
Eliminate xy ; x^2 - 16y^2 + 60 = 0 (not sure how to factorise)

Do I sub that back into the previous equtions?

I am still not sure where to go form here.
 
  • #4
The first equation can be written in factored form as
(x - y)(x - 2y) = 0

BTW, you should post problems like this in the Calculus & Beyond section, not this section.
 
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  • #5
Thank you! I think I know what I need to do now with the other equation now.

Noted, i'll post in the other section.
 

What is the purpose of finding stationary points in a u(x,y) function?

The purpose of finding stationary points in a u(x,y) function is to determine the critical points where the function's gradient is equal to zero. These points can indicate the location of maximum, minimum, or saddle points in the function.

How do I find stationary points in a u(x,y) function?

To find stationary points in a u(x,y) function, you can use the method of partial derivatives. Take the partial derivatives of the function with respect to both x and y, set them equal to zero, and solve the resulting system of equations to find the stationary points.

What does it mean when a stationary point is a maximum, minimum, or saddle point?

If the second derivative test at a stationary point is positive, the point is a local minimum. If the second derivative test is negative, the point is a local maximum. If the second derivative test is zero, the point is a saddle point.

How can I determine the nature of a stationary point in a u(x,y) function?

To determine the nature of a stationary point in a u(x,y) function, you can use the second derivative test. Evaluate the second partial derivatives of the function at the stationary point, and use their signs to determine if the point is a maximum, minimum, or saddle point.

What are the practical applications of finding stationary points in a u(x,y) function?

Finding stationary points in a u(x,y) function can be used in many practical applications, such as optimizing functions in economics, physics, and engineering. It can also be used to find the optimal solution in mathematical models and to analyze the behavior of systems in science and engineering.

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