Finding max value of this expression

[utkarshakash]

Homework Statement

In a triangle ABC let A,B,C denote the angles of the triangle. Then maximum value of the expression
sinA/A + sinB/B + sinC/C is

Homework Equations

The Attempt at a Solution

Applying Cauchy Shwarz inequality
\dfrac{sinA}{A} + \dfrac{sinB}{B} + \dfrac{sinC}{C} \leq \sqrt{sin^2A + sin^2 B + sin^2 C} \sqrt{\frac{1}{A^2}+\frac{1}{B^2}+\frac{1}{C^2}}

 

[Ray Vickson]

Homework Statement

In a triangle ABC let A,B,C denote the angles of the triangle. Then maximum value of the expression
sinA/A + sinB/B + sinC/C is

Homework Equations

The Attempt at a Solution

Applying Cauchy Shwarz inequality
\dfrac{sinA}{A} + \dfrac{sinB}{B} + \dfrac{sinC}{C} \leq \sqrt{sin^2A + sin^2 B + sin^2 C} \sqrt{\frac{1}{A^2}+\frac{1}{B^2}+\frac{1}{C^2}}

This problem can be set up as a Calculus problem in two variables. The resulting optimality conditions are extremely complicated---absolutely calling for use of a symbolic computer algebra/calculus package (I used Maple). The resulting equations are extremely complicated, but can be solve numerically (after first performing a surface plot to gain insight into the location of the solution). The final numerical solution can be converted back into symbolic form and then verified explicitly.

I am sure there must be an easier, more insightful way to solve the problem, but it is late and I am tired and just don't see it.

 

[utkarshakash]

This problem can be set up as a Calculus problem in two variables. The resulting optimality conditions are extremely complicated---absolutely calling for use of a symbolic computer algebra/calculus package (I used Maple). The resulting equations are extremely complicated, but can be solve numerically (after first performing a surface plot to gain insight into the location of the solution). The final numerical solution can be converted back into symbolic form and then verified explicitly.

I am sure there must be an easier, more insightful way to solve the problem, but it is late and I am tired and just don't see it.

I'm still in my high school and the method which you told me is way ahead of my current syllabus. Nonetheless, I was trying Lagrange's Multiplier Method but no success.

 

[Millennial]

From the Law of Sines, \displaystyle \frac{\sin(A)}{A} = \frac{\sin(B)}{B} = \frac{\sin(C)}{C}.

Does that help?

 

[Ray Vickson]

I'm still in my high school and the method which you told me is way ahead of my current syllabus. Nonetheless, I was trying Lagrange's Multiplier Method but no success.

Congratulations! Your way is hundreds of times simpler than mine. It gives you all the information you need to solve the problem.

 

[Ray Vickson]

From the Law of Sines, \displaystyle \frac{\sin(A)}{A} = \frac{\sin(B)}{B} = \frac{\sin(C)}{C}.

Does that help?

It would help if it were true---but it isn't. The Law of Sines says
\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}
where $A,B,C$ are the angles and $a,b,c$ are the opposite sides.

 

[Millennial]

It would help if it were true---but it isn't. The Law of Sines says
\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}
where $A,B,C$ are the angles and $a,b,c$ are the opposite sides.

Sorry, I misread the question.

Using Wolfram, I can see that the maximum is reached in an equilateral triangle. However, I don't have an idea on why this is so. Using Lagrange multipliers, it is not hard to see that this is a critical point, but one needs to evaluate the other critical points before coming to the conclusion that this is the maximum.

Any ideas? I am sure there is a simpler solution to this.

 

[utkarshakash]

Sorry, I misread the question.

Using Wolfram, I can see that the maximum is reached in an equilateral triangle. However, I don't have an idea on why this is so. Using Lagrange multipliers, it is not hard to see that this is a critical point, but one needs to evaluate the other critical points before coming to the conclusion that this is the maximum.

Any ideas? I am sure there is a simpler solution to this.

I'm really sorry. The original question was to find the minimum value of this expression. But I don't think this will make the question easier than before. Finding minimum is as difficult as finding the maximum.

 

[Millennial]

I'm really sorry. The original question was to find the minimum value of this expression. But I don't think this will make the question easier than before. Finding minimum is as difficult as finding the maximum.

Really? The infimum on the range of this function with the given constraint must be quite obvious.

 

[Ray Vickson]

I'm really sorry. The original question was to find the minimum value of this expression. But I don't think this will make the question easier than before. Finding minimum is as difficult as finding the maximum.

Finding the minimum is an ill-posed problem: there is no true minimum of your expression f = f(A,B,C). The "optimal" solution is to make A = 180 degrees (π radians) and B = C = 0 degrees (0 radians); this makes the triangle collapse down into a line segment.

The smallest possible value is f = 2. However, this minimum is not achievable by any actual triangle; that is, it is an 'infimum', rather than a minimum. You can find triangles giving f = 2.0001 or f = 2.0000001, etc., but no actual triangle giving f = 2. Certainly, you cannot set derivatives to zero or use simple Lagrange multipliers on this problem: you need to use the more complex so-called Karush-Kuhn-Tucker conditions to solve the problem, because you need to impose inequality constraints $A, B, C \geq 0$ on the variables. (These must be imposed as well on the maximization problem, but they turn out to be ignorable, because the Lagrange conditions automatically satisfy them.)

 

[utkarshakash]

Finding the minimum is an ill-posed problem: there is no true minimum of your expression f = f(A,B,C). The "optimal" solution is to make A = 180 degrees (π radians) and B = C = 0 degrees (0 radians); this makes the triangle collapse down into a line segment.

The smallest possible value is f = 2. However, this minimum is not achievable by any actual triangle; that is, it is an 'infimum', rather than a minimum. You can find triangles giving f = 2.0001 or f = 2.0000001, etc., but no actual triangle giving f = 2. Certainly, you cannot set derivatives to zero or use simple Lagrange multipliers on this problem: you need to use the more complex so-called Karush-Kuhn-Tucker conditions to solve the problem, because you need to impose inequality constraints $A, B, C \geq 0$ on the variables. (These must be imposed as well on the maximization problem, but they turn out to be ignorable, because the Lagrange conditions automatically satisfy them.)

Here's the solution for this question given in my book

Let f(x)=sinx/x
We know that sin x<x for all x in (0,pi/2) and limx-->0 (sinx/x)=1.
Hence sinx/x is decreasing function in (0,pi/2).
f(A)>f(π/2)
f(B)>f(π/2)
f(C)>f(π/2)

Hence the answer is 6/π.

But I have confusion from 4th line onwards. Why they chose f(π/2) for comparison?They could have chosen any other angle as well.

 

[Ray Vickson]

Here's the solution for this question given in my book

Let f(x)=sinx/x
We know that sin x<x for all x in (0,pi/2) and limx-->0 (sinx/x)=1.
Hence sinx/x is decreasing function in (0,pi/2).
f(A)>f(π/2)
f(B)>f(π/2)
f(C)>f(π/2)

Hence the answer is 6/π.

But I have confusion from 4th line onwards. Why they chose f(π/2) for comparison?They could have chosen any other angle as well.

Their answer is incorrect, at least as regards the minimum. What they give is a lower bound, not a best lower bound.

Your original message did not mention that angles were to be ≤ π/2; that changes things a lot!

So, let us look at two possible versions of the problem (with $f(x) \equiv \sin(x)/x$).

(I) The unrestricted-angle problem:
\min F = f(A) + f(B) + f(C)\\<br /> \text{subject to}\\<br /> A+B+C=\pi\\<br /> A,B,C \geq 0
(II) The restricted-angle problem:
\min F = f(A) + f(B) + f(C)\\<br /> \text{subject to}\\<br /> A + B + C = \pi\\<br /> 0 \leq A,B,C \leq \pi/2

There are three solutions of (I): (A,B,C) = (π,0,0) or (0,π,0) or (0,0,π). These all give F = 2, which is strictly larger than your textbook's lower bound of 6/π = 1.909859317. As I said before, these three solutions do not correspond to actual triangles; they are 'degenerate' triangles that have been collapsed down to a line segment. However, for any F > 2 you can find actual triangles giving that value of F.

There are three solutions of (II): (A,B,C) = (π/2,π/2,0) or (π/2,0,π/2) or (0,π/2,π/2), all giving F = 1 + 4/π = 2.273239544. Again, these do not correspond to any real triangles, because no real triangle can have two 90-degree internal angles. However, we can find real triangles that give any value > 1 + 4/π. For example, the triangle with corners (0,0), (1,0) and (0,n) in the Cartesian plane will have internal angles π/2, arctan(n) and arctan(1/n). For very large finite n we have nearly (but not quite) two right angles; in the limit n → ∞ we get our 'solution'.

 

[utkarshakash]

Their answer is incorrect, at least as regards the minimum. What they give is a lower bound, not a best lower bound.

Your original message did not mention that angles were to be ≤ π/2; that changes things a lot!

So, let us look at two possible versions of the problem (with $f(x) \equiv \sin(x)/x$).

(I) The unrestricted-angle problem:
\min F = f(A) + f(B) + f(C)\\<br /> \text{subject to}\\<br /> A+B+C=\pi\\<br /> A,B,C \geq 0
(II) The restricted-angle problem:
\min F = f(A) + f(B) + f(C)\\<br /> \text{subject to}\\<br /> A + B + C = \pi\\<br /> 0 \leq A,B,C \leq \pi/2

There are three solutions of (I): (A,B,C) = (π,0,0) or (0,π,0) or (0,0,π). These all give F = 2, which is strictly larger than your textbook's lower bound of 6/π = 1.909859317. As I said before, these three solutions do not correspond to actual triangles; they are 'degenerate' triangles that have been collapsed down to a line segment. However, for any F > 2 you can find actual triangles giving that value of F.

There are three solutions of (II): (A,B,C) = (π/2,π/2,0) or (π/2,0,π/2) or (0,π/2,π/2), all giving F = 1 + 4/π = 2.273239544. Again, these do not correspond to any real triangles, because no real triangle can have two 90-degree internal angles. However, we can find real triangles that give any value > 1 + 4/π. For example, the triangle with corners (0,0), (1,0) and (0,n) in the Cartesian plane will have internal angles π/2, arctan(n) and arctan(1/n). For very large finite n we have nearly (but not quite) two right angles; in the limit n → ∞ we get our 'solution'.

So my original question corresponds to your problem (II) and according to you the answer should be 1+4/π. Am I right?

 

[ehild]

A+B+C=pi defines a plane. 0<A≤pi, 0<B<pi,0<C<pi, the allowed points P(A,B,C) lay on the yellow triangle with vertices (pi,0,0), (0,pi,0), (0,0,pi). In case 0<A≤pi/2,0<B<pi/2, 0<C<pi/2, the points belong to the triangle with vertices (0, pi/2, pi/2), (pi/2, 0, pi/2), (pi/2, pi/2,0)
The function can have either local extrema inside its domain or absolute extrema on the boundaries of its domain. The function is symmetric for the permutations A,B,C. The critical points lay on symmetry elements. In the first case,f(A,B,C)=2 at the vertices. At the centre of the edges (pi/2, pi/2, 0) (and permutations) f=1+4/pi>2. But these places do not belong to either domain of the function. They can be infima, but not minima. There is a special point inside the triangle: its centre. What is f(pi/3, pi/3, pi/3)? The value is greater than at the previous points. We can expect a maximum there.


ehild