Help forming equation. Constrained extrema.

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Homework Statement



show that for any triangle inscribed in a circle with radius r , the equilateral triangle has the largest perimeter. I'm supposed to use the larangian method.

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The Attempt at a Solution



ok so my problem is forming an equation for the perimeter in terms of x and y.

basically the constraint is to maximize perimeter with respect to the circle x^2 + y^2 = r

so if a b c are the sides of the triangle, how can I form an equation for each side with respect to x and y? The distance from the origin to each vertex is r. But I'm not sure how to get an equation for the distance of each side. I just want some hints.
 

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  • #2
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Pick point 1 at (r,0), point 2 at an angle theta1, and point 3 at an angle theta2.
Now you can calculate the length of the sides as a function of theta1 and theta2...
 
  • #3
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Hi, thanks for the reply. I'm not sure if this is what you meant but this is what i did.

If i drew a triangle in the circle i noticed that the length of the origin to any vertex of the triangle is R.

So applying that this is what I did, I can basically split the triangle into 3 parts to find each side of the triangle from the law of cosines:

c^2 = a^2 + b^2 - 2ab cos C

For example is c was a side of the triangle I can get its length through R. Here is my bad mspaint interpretation.

http://i.imgur.com/7NMne.png

c^2 = R^2 + R^2 - 2R^2 cos C

and R = x^2+y^2

so i can do that for the other sides as well, and form the perimeter equation

P(a,b,c) = a + b + c

I can now sub in R as x^2 + y^2 in the perimeter equation to get it in terms of x and y, but after doing that and simplifying, then solving it with the larangian with the constraint as x^y + y^2 = R, I dont get an expression for lamda in terms of x and y. All i get is a constant which is in terms of the angles.

Help?
 
  • #4
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Oh sorry.
What you got is an expression for which you don't need Lagrangian multipliers, since it does not have a boundary condition any more.
It will work to take the derivative to theta1 and to theta2 of course which both should be zero.

To use Lagrangian multipliers, I think you would need to define for instance (x1, y1) and (x2, y2) as the 2nd and 3rd point of the triangle.
Both need to satisfy your boundary condition.
So you would need to solve for 4 variables and 2 multipliers.
 
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  • #5
Ray Vickson
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If I were doing it, I would use angles. For inscribed triangle ABC, let t1 be the angle subtended by AB at the center O of the circle. In terms of r and t1, what is the expression for the length of AB? You have angles t2 and t3 associated with sides BC and CA. What is the constraint on angles t1, t2, and t3?

RGV
 
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  • #6
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Oh sorry.
What you got is an expression for which you don't need Lagrangian multipliers, since it does not have a boundary condition any more.
It will work to take the derivative to theta1 and to theta2 of course which both should be zero.

To use Lagrangian multipliers, I think you would need to define for instance (x1, y1) and (x2, y2) as the 2nd and 3rd point of the triangle.
Both need to satisfy your boundary condition.
So you would need to solve for 4 variables and 2 multipliers.
Hmm, why would (r,0) be the first point of my triangle? There doesn't necessarily have to be a vertex at (r,0) like in my mspaint drawing. I get what you mean for 2 multipliers, I will try that.
 
  • #7
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Hmm, why would (r,0) be the first point of my triangle? There doesn't necessarily have to be a vertex at (r,0) like in my mspaint drawing. I get what you mean for 2 multipliers, I will try that.
The first point is an arbitrary choice.
Any point on the circle will do.
It seems smart to pick the simplest point there is, which is (r,0).
 
  • #8
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Okay so the perimeter equation would not involve the angles then right? Using 2 arbitrary points x1,y1 and x2,y2 i got something like this

P(x1,x2,y1,y2,R) = sqrt (x1 - R)^2 + (y1)^2 + sqrt (x1-x2)^2 + (y1-y2)^2 + sqrt (x2-R)^2 + (y2)^2

Edit:

Ok im a bit confused now. When you said two multipliers, how would that work? I'm constraining the perimeter to the circle.
 
  • #9
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If I were doing it, I would use angles. For inscribed triangle ABC, let t1 be the angle subtended by AB at the center O of the circle. In terms of r and t1, what is the expression for the length of AB? You have angles t2 and t3 associated with sides BC and CA. What is the constraint on angles t1, t2, and t3?

RGV
Just noticed this post, isn't that the way i was doing it earlier? the constraint on angles t1,t2,and t3 is that they have to add up to 360 or 2pi?
 
  • #10
Ray Vickson
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Just noticed this post, isn't that the way i was doing it earlier? the constraint on angles t1,t2,and t3 is that they have to add up to 360 or 2pi?
Well, you did something like it, but I could not understand your sentence about getting a "constant in terms of the angles". I don't know whether you think you have solved the problem, or not. The Lagrangian method, using angles, solves the problem completely. You don't get expressions for x and y (at least, not immediately), but you do get the angles, which is all you need to prove your required result.

RGV
 
  • #11
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Okay so the perimeter equation would not involve the angles then right? Using 2 arbitrary points x1,y1 and x2,y2 i got something like this

P(x1,x2,y1,y2,R) = sqrt (x1 - R)^2 + (y1)^2 + sqrt (x1-x2)^2 + (y1-y2)^2 + sqrt (x2-R)^2 + (y2)^2

Edit:

Ok im a bit confused now. When you said two multipliers, how would that work? I'm constraining the perimeter to the circle.
You have 2 constraints, one for (x1,y1) and one for (x2,y2).

The Lagrange multiplier would be:
[tex]\Lambda(x_1,x_2,y_1,y_2,\lambda_1,\lambda_2) = P(x_1,x_2,y_1,y_2;R) - \lambda_1 (x_1^2+y_1^2-R^2) - \lambda_2 (x_2^2+y_2^2-R^2)[/tex]
 
  • #12
Ray Vickson
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You have 2 constraints, one for (x1,y1) and one for (x2,y2).

The Lagrange multiplier would be:
[tex]\Lambda(x_1,x_2,y_1,y_2,\lambda_1,\lambda_2) = P(x_1,x_2,y_1,y_2;R) - \lambda_1 (x_1^2+y_1^2-R^2) - \lambda_2 (x_2^2+y_2^2-R^2)[/tex]
I would very much prefer to work with the Lagrangian
[tex] L = 2r(\sin(\theta_1/2) + \sin(\theta_2/2) + \sin(\theta_3/2)) - \lambda(\theta_1 + \theta_2 +\theta_3-2 \pi). [/tex]

RGV
 
  • #13
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Well, you did something like it, but I could not understand your sentence about getting a "constant in terms of the angles". I don't know whether you think you have solved the problem, or not. The Lagrangian method, using angles, solves the problem completely. You don't get expressions for x and y (at least, not immediately), but you do get the angles, which is all you need to prove your required result.

RGV
I noticed you did this with sin. My larangian involved the angles as well as x and y so I think its wrong because I got stuck with a huge mess of substitutions. I'm wondering how you formed that equation of the larangian. Law of sines?
 
  • #14
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RGV used just the definition of the sine, applied to a half-angle.
The resulting r sin(theta/2) is half of an edge.

The resulting formula is nice and simple.
However,
[tex]L = 2r(\sin(\theta_1/2) + \sin(\theta_2/2) + \sin((2 \pi - \theta_1 - \theta_2)/2))[/tex]
will work even better.
It makes use of the Lagrangian multiplier completely redundant though.
 
  • #15
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I haven't tried the sine method yet because I was trying the perimeter larangian with two constraints but I got to a weird point now.

so first off when I did the partials I noticed that

fx1 = fy1
fx2 = fy2

that results in solving for g1 from fx1 and fy1 to be the same thing. The same goes for g2 which you can solve for from fx2 and fy2.

when you take the partial fxR, you end up with an equation that includes g1 and g2. If I subsitute those, everything cancels to 0. I'm kind of stuck here.
 
  • #16
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Don't take the partial with respect to R.
R is not one of your coordinates, but a constant.

If you're still stuck, perhaps you can show the equations that you got?
 
  • #17
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right!

but then what do I do here to make a substitution? I have a value for g1 and g2 in terms of the length of the sides (which are in terms of the co ordinates and R). For simplicity I will call a b c the length of the sides. So I got

g1 = 1/a - 1/c
g2 = 1/b + 1/c

how can I use this when I take the partials fg1 and fg2?

I'm trying to show that a = b = c as a maximum.
 
  • #18
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Huh? :confused:
I do not recognize these equations.
What are g1 and g2?

What is your set of equations?
 
  • #19
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g1 is lamda1 from your equation. Same for g2.

so when I took the partials and setting them to 0 I get

g1 =[1/2sqrt (x1-R)² + y1²] - [1/2sqrt (x2-x1)² + (y2-y1)²]

g2 = [1/2sqrt (x2-R)² + (y2)²] + [1/2sqrt (x2-x1)² + (y2-y1)²]
 
  • #20
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Sorry, I still don't understand what you did, or which partials you took, or what the placement of parentheses should be.
 
  • #21
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ok so in detail.

first off I'll just use

a = sqrt [(x1-R)² + y1²]

c = sqrt [(x2-x1)² + (y2-y1)²]

b = sqrt[ (x2-R)² + (y2)²]

so a b and c are the lengths for each side of the triangle.

then P = a + b + c

so the larangian function

F = P - g1(x1²+y1² -R²) - g2(x2² + y2² - R²)

now

Fx1 = x1/a - x1/c - 2x1g1 = 0

solve for g1

g1 = 1/2a - 1/2c

Fx2 = x2/b + x2/c - 2x2g2 = 0

solve for g2

g2 = 1/2b + 1/2c

hopefully that makes sense
 
  • #22
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ok so in detail.

first off I'll just use

a = sqrt [(x1-R)² + y1²]

c = sqrt [(x2-x1)² + (y2-y1)²]

b = sqrt[ (x2-R)² + (y2)²]

so a b and c are the lengths for each side of the triangle.

then P = a + b + c

so the larangian function

F = P - g1(x1²+y1² -R²) - g2(x2² + y2² - R²)

now

Fx1 = x1/a - x1/c - 2x1g1 = 0

solve for g1

g1 = 1/2a - 1/2c

Fx2 = x2/b + x2/c - 2x2g2 = 0

solve for g2

g2 = 1/2b + 1/2c

hopefully that makes sense
 
  • #23
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Okay, this helps.

Let's start with Fx1, or rather da/dx1.
I get a different derivative...

Btw, usually you don't really want to solve g1 or g2. You just want to eliminate them.
 
  • #24
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oh crap I see why haha. I was using chain rule for the inside of the root so I forgot that it was 2(x1-R) rather than just 2x1. I'll fix this and get back.
 
  • #25
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well yup. Ok so I fixed that mistake and now I have an even bigger mess. I got two values for g1 qnd g2.


g1 = x1-R/2x1a - x2-x1/2x1c

g1 = 1/2a - y2-y1/2y1c

g2 = x2-R/2x2b + x2-x1/2x2c

g2 = 1/2b + y2-y1/2y2c


I'm lost now. How is this supposed to show a = b = c just through co ordinates? Maybe I should go the angle route but I still don't understand how the sin larangian was made without using sin r at all.
 
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