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Calculating Stationary Points of a Function

  • Thread starter Hart
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  • #1
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Homework Statement



Finding the stationary point(s) of the function:

[tex]f(x,y) = xy - \frac{y^{3}}{3}[/tex]

.. on the line defined by [itex]x+y = -1[/itex].

For each point, state whether it is a minimum or maximum.

Homework Equations



.. within the problem statement and solutions.

The Attempt at a Solution



This is what I have so far:

[tex]f(x,y) = xy - \frac{y^{3}}{3}[/tex]

[tex]g(x,y) = x+y-1 = 0[/tex]

Therefore need to extemise:

[tex]F(x,y,\lambda) = f + \lambda g = xy - \frac{y^{3}}{3} + \lambda(x+y-1)[/tex]

So calculating the partial derivatives:

[tex]\frac{\partial F}{\partial x} = y + \lambda = 0[/tex]

[tex]\frac{\partial F}{\partial y} = x - 3\left(\frac{y^{2}}{3}\right) + \lambda = x - y^{2} + \lambda = 0[/tex]

[tex]\frac{\partial F}{\partial \lambda} = x + y - 1 = 0[/tex]

Then need to look for all consistent solutions:

[tex]1. y = \lambda[/tex]

.. but now I'm stuck on what to do now, seemto have done something wrong because I can't get more consistent soluations and then nice simultaneous equations to equate :confused:
 

Answers and Replies

  • #2
tiny-tim
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Hi Hart! :smile:

(have a lambda: and a curly d: ∂ :wink:)

(Actually, it's y = -λ, isn't it? :wink:)

ok, now substitute that in the other two equations, and you'll get a quadratic equation in λ. :smile:
 
  • #3
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OK.. So from the first partial derivative equation I get that:

[tex]y = -\lambda[/tex]

.. then put this into the second partial derivative equation, which gives:

[tex]x - \lambda^{2} + \lambda = 0[/tex]

.. hence:

[tex]x = \lambda^{2} - \lambda[/tex]

.. then put this into the third partial derivative equation, which gives:

[tex]\lambda^{2} - \lambda - \lambda - 1 = \lambda^{2} - 2\lambda - 1 = 0[/tex]

.. then solve quadratic equation to get values of [itex]\lambda[/itex]:

[tex]\lambda = \frac{2 \pm \sqrt{(4) + (4)}}{2} = 1 \pm \sqrt{2}[/tex]

.. hence:

[tex]\lambda = 1 + \sqrt{2} , 1 - \sqrt{2}[/tex]

.. yes??!

.. note: but:

[tex]\lambda = 1 \pm \sqrt{2} \neq - y[/tex]

:confused:
 
  • #4
tiny-tim
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What's wrong with that? :confused:

y = -1 - √2, x = 2 + √2;

y = -1 + √2, x = 2 - √2.
 
  • #5
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um.. nothing! I don't know why I thought there was an issue there earlier! :redface: .. maybe just going a bit mad! :wink:

Right, so since now have:

[tex]y = -1 - \sqrt{2}, x = 2 + \sqrt{2}[/tex]

[tex]y = -1 + \sqrt{2}, x = 2 - \sqrt{2}[/tex]

.. then there are just 2 stationary points of the function, which are those 2 above?

Also, to find if the points are maxima or minima, just need to put each set of [itex](x,y)[/itex] values into this equation:

[tex]f(x,y) = xy - \frac{y^{3}}{3}[/tex]

.. and if [itex]f(x,y) > 0[/itex] then maxima point, and if [itex]f(x,y) < 0[/itex] then minima point?
 
Last edited:
  • #6
tiny-tim
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.. and if [itex]f(x,y) > 0[/itex] then maxima point, and if [itex]f(x,y) < 0[/itex] then minima point?
(why do you keep going into LaTeX?!)

No, that only tells you the value of f(x,y) …

you'll need more than that to check whether the value is a local maximum or minimum or stationary point.
 
  • #7
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.. calculate second derivatives? negative = max point, positive = min point?
 
  • #8
tiny-tim
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Hi Hart! :smile:

(just got up :zzz: …)
.. calculate second derivatives? negative = max point, positive = min point?
Yes, except you'll need the second derivatives "dot" the line. :wink:
 
  • #9
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erm.. so for the first set of values I have:

[tex]f(x,y) = xy - \left(\frac{y^{3}}{3}\right) = \left(\left(2+\sqrt{2}\right)\left(-1-\sqrt{2}\right)\right) - \left(\frac{\left(-1 - \sqrt{2}\right)}{3}\right) = \left(-4 - 3\sqrt{2}\right) - \left(-7 + 5\sqrt{2}\right) = 3 + 2\sqrt{2} [/tex]

which is positive, so this is a minium stationary point.

.. correct?!
 
  • #10
tiny-tim
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what happened to the second derivatives? :confused:
 
  • #11
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OH.. I just forgot them! :redface:

ok, so:

[tex]f(x,y) = xy - \left(\frac{y^{3}}{3}\right)[/tex]

[tex]\frac{\partial f(x,y)}{\partial xy} = 1 - y^{2}[/tex]

[tex]\frac{\partial f(x,y)}{\partial xy} = 2y[/tex]

.. correct? (I'm not too good on partial derivations :frown:)

So then:

[tex]2y = 2\left(-1 - \sqrt{2}\right) = -2 - 2\sqrt{2}[/tex]

.. which is positive.

[tex]2y = 2\left(-1 + \sqrt{2}\right) = -2 + 2\sqrt{2}[/tex]

.. which is negative.

Hopefully somewhat getting there! :smile:
 
  • #12
tiny-tim
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hmm … this has become needlessly complicated.

It would have been easier at the beginning, instead of using the λ method (i know it has a name, but i've forgotten it :redface:), to simply use the directional derivative.

Since g(x,y) in this case is a straight line, in the direction (1,-1), all you needed to do was to find the directional derivative, ∇gf, in that direction, ie ∂f/dx - ∂f/∂y, which was y + y2 - x, and put that = 0 (subject to x + y = 1).

Substituting x = 1 - y gives ∇gf = 2y + y2 - 1 = 0 …

this is the same final equation as before, but without using λ ! :wink:

Now, to distinguish between maxima and minima, you need to know whether ∇gf is increasing or decreasing, so just consider d/dy (∇gf). :smile:
 
  • #13
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.. I was doing in the [itex]\lambda[/itex] way (um.. lagrange multiplier?!) as this is how I was taught a similar problem and hence should really use this way.

Looking back through my notes on this, once the values of x and y have been found (as they have been) then just put these into the original equation for f and then can deduce if the stationary point is a min or a max.

Basically, for the first point:

[tex]x = 2 + \sqrt{2}, y = -1 - \sqrt{2}[/tex]

.. input the values into:

[tex]f(x,y) = xy - \frac{y^{3}}{3}[/tex]

and consider whether is positive or negative, and hence max or min point.

Then repeat for second point.
 
  • #14
tiny-tim
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(oh yes, Lagrange multiplier! :rolleyes:)

But the problem with that method is that f(x,y) isn't necessarily positive at a maximum or negative at a minimum.

You can change it slightly, and say that if there are only two turning-points, at A and B, then either f(A) > f(B) or f(B) > f(A) …

in either case, the first one must be a maximum, and the second a minimum! :smile:

(of course, that doesn't rule out points of inflexion, so perhaps for completeness the values towards infinity in both directions should also be considered)
 

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