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F(x, y) Min Max problem with boundaries

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  • #1
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Homework Statement:

Find max or min value of function $$f(x,y)=\frac{x+y^2}{1+x^2+y^2} $$ which is defined by $$x^2+y^2=4$$

Homework Equations:

$$f(x,y)=\frac{x+y^2}{1+x^2+y^2} $$
First I took the partial derivatives (f'x=0 and f'y=0) and got x=-1 or 1 and y=0 but these values are outside our definition right?

I did a parameter change to trigs and got result f(0,2) is max and f(-2,0) is min. is this correct?

edit: maybe its f(1, 3) is max?
 

Answers and Replies

  • #2
WWGD
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The statement seems unclear. Do you mean your ##f(x,y)## is subject to the constraint that ##x,y## lie on the EDIT: circle ##C(0,2)##? If so, you can simplify your ##f## and sub in ##4## for ##x^2+y^2## in the denominator.
 
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  • #3
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Ok so i get 4/5 as max value and -2/5 as min value. is this correct?
 
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  • #4
WWGD
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If you simplify with the sub in then ##f_x=1, f_y=2y##. Have you tried that?
 
  • #5
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If you simplify with the sub in then ##f_x=1, f_y=2y##. Have you tried that?
interesting. thanks.
 
  • #6
WWGD
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interesting. thanks.
Sorry, forgot to include that in the resulting expression after the sub in , you can also turn ## x+ y^2## into a function of 1 variable ( seemingly easier in terms of ##x##) , which simplifies things even further.
 
  • #7
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Sorry, forgot to include that in the resulting expression after the sub in , you can also turn ## x+ y^2## into a function of 1 variable ( seemingly easier in terms of ##x##) , which simplifies things even further.
since the circle itself is only defined is there any point in taking derivatives? since the max min have to be on the circle.
 
  • #8
WWGD
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since the circle itself is only defined is there any point in taking derivatives? since the max min have to be on the circle.
If you can figure it out, no need to do so. But how would you prove it's the maximum? Edit: never mind. Once you do the sub ins the max will be pretty clear. Edit2: No, you're right, the equation is defined on the circle so that the max will also be.
 
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  • #9
WWGD
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Ok so i get 4/5 as max value and -2/5 as min value. is this correct?
Max is 17/4 at x=1/2 , no global minimum. Plot the subbed-in curve. EDIT: As pointed out, there is a min. within the circle. The curve is a parabola, which we can intersect with the circle.
 
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  • #10
StoneTemplePython
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no global minimum. Plot the subbed-in curve.
continuous function applied on a compact domain...
 
  • #11
WWGD
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continuous function applied on a compact domain...
Ah, my bad, I was considering the global min of 4+x -x^2 , not the one restricted to the circle. We get a parabola in the global one before restricting. Edited. I thought subbing -in the equation of the circle on the objective function would take care of it. Clearly not.
 
  • #12
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Max is 17/4 at x=1/2 , no global minimum. Plot the subbed-in curve. EDIT: As pointed out, there is a min. within the circle. The curve is a parabola, which we can intersect with the circle.
how can x=1/2 when its not part of the domain?
 
  • #13
WWGD
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how can x=1/2 when its not part of the domain?
Try the point in the circle with coordinate ##x =1/2## Should be ## \sqrt {15} /2 ##
 
  • #14
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Try the point in the circle with coordinate ##x =1/2## Should be ## \sqrt {15} /2 ##
how does inserting x=1/2 into the function yield 17/4 when our boundries are x^2 +y^2=4 ?????
 
  • #15
WWGD
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how does inserting x=1/2 into the function yield 17/4 when our boundries are x^2 +y^2=4 ?????
Edit: You're right ,my bad. Objective function is ##f(x,y)= \frac {x+ y^2}{5} ## for points##(x,y) ##on the circle. ##(1/2, \sqrt 15/4)## is on the circle. So max is 17/20, not 17/4, sorry.
 
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  • #16
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Edit: You're right ,my bad. Objective function is ##f(x,y)= \frac {x+ y^2}{5} ## for points##(x,y) ##on the circle. ##(1/2, \sqrt 15/4)## is on the circle. So max is 17/20, not 17/4, sorry.
im sorry but (1/2)^2 + (sqrt15 /4)^2 is not equal to 4.
 
  • #17
haruspex
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im sorry but (1/2)^2 + (sqrt15 /4)^2 is not equal to 4.
Clearly @WWGD meant ##\sqrt{15/4}##
 
  • #18
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Try the point in the circle with coordinate ##x =1/2## Should be ## \sqrt {15} /2 ##
how did you get that point?
 
  • #19
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##(1/2, \sqrt 15/4)## is on the circle.
how did you get that point?
He got it by misunderstanding how LaTeX works. If you write ##\sqrt 15/4##, it renders as ##\sqrt 15/4##, but if you write it as ##\sqrt{15/4}##, with braces around the quantity in the radical, the expression renders as ##\sqrt{15/4}##. The latter form is what I believe he intended.
 
  • #20
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He got it by misunderstanding how LaTeX works. If you write ##\sqrt 15/4##, it renders as ##\sqrt 15/4##, but if you write it as ##\sqrt{15/4}##, with braces around the quantity in the radical, the expression renders as ##\sqrt{15/4}##. The latter form is what I believe he intended.
i meant how did he find it on the circle which method did he use? derivative. trig change? how?
 
  • #21
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i meant how did he find it on the circle which method did he use? derivative. trig change? how?
Are you asking how he got the x value of 1/2? Or are you asking how he got the y value of ##\sqrt{15/4}##?
 
  • #22
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Are you asking how he got the x value of 1/2? Or are you asking how he got the y value of ##\sqrt{15/4}##?
well both. since they are related.
 
  • #23
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well both.
I'll let @WWGD explain how he came up with x = 1/2. As for the y value, just substitute x = 1/2 into the equation ##x^2 + y^2 = 4##.
 
  • #24
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I'll let @WWGD explain how he came up with x = 1/2. As for the y value, just substitute x = 1/2 into the equation ##x^2 + y^2 = 4##.
well yes, i understood that part. but where did x=0.5 come from. cause you don't get it with partial derivates and you don't get it from trig change either.
 
  • #25
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well yes, i understood that part. but where did x=0.5 come from. cause you don't get it with partial derivates and you don't get it from trig change either.
Edited to fix a typo:
Your original function is constrained to the circle ##x^2 + y^2 = 4##, so what do you get if you replace ##y^2## in your original function with ##4 - x^2##? If you do that, you get a critical point for x = 1/2.
 

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