Maximization Problem: Double Int. w/ C not Dependent on Integrals

In summary, you are asking for a function that "always" gives you the value of a function when you give it the values of two variables.
  • #1
JBD
15
1
Consider a double integral

$$K= \int_{-a}^a \int_{-b}^b \frac{B}{r_1(y,z)r_2^2(y,z)} \sin(kr_1+kr_2) \,dy\,dz$$

where
$$r_1 =\sqrt{A^2+y^2+z^2}$$
$$r_2=\sqrt{B^2+(C-y)^2+z^2} $$

Now consider a function:

$$C = C(a,b,k,A,B)$$

I want to find the function C such that K is maximized. In other words, there may be a relation between C, a, b, k, A and B such that K is maximized.

In calculus of variation, we consider a set of curves
$$Y=Y(x)$$
then we seek a member Y=y(x) of this set which minimizes/maximizes the integral
$$J(Y)= \int_m^n F(x,Y,Y') dx$$
In this case the function we want to find which is y(x) is dependent on the variable of integration, i.e. F(x,Y,Y') is to be integrated with respect to x.

However in my case, the function C that I want to find is not dependent on the variables of integration which are y and z. Is there a way to change or transform this problem so that it can be solved via calculus of variations? Thanks.
 
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  • #2
JBD said:
I want to find the function C such that K is maximized.

What do mean by that? A specific function ##C_0(a,b,k,A,B)## may produce a different value of the integral ##K## at different values of its arguments. So if we have two functions ##C_0## and ##C_1##, how do we decide which of them produces "the greater" value of ##K##? - seeing that each of the functions can produce several different values of ##K##.
 
  • #3
Stephen Tashi said:
What do mean by that? A specific function ##C_0(a,b,k,A,B)## may produce a different value of the integral ##K## at different values of its arguments. So if we have two functions ##C_0## and ##C_1##, how do we decide which of them produces "the greater" value of ##K##? - seeing that each of the functions can produce several different values of ##K##.

I solved the original maximization problem with a lot of simplifications. I tried to generalize it and ended up in this problem.

In my original calculations, I found that there is a ratio between for example a and B. For instance, a/B = 2.0 which will always produce a K with a max value. The K is different though because of simplifications.
 
  • #4
JBD said:
In my original calculations, I found that there is a ratio between for example a and B. For instance, a/B = 2.0 which will always produce a K with a max value.

You described what you did, but you didn't answer my question. And your description of what you did is ambiguous.

Let take a simple case of an integral ##K(a,b) ## that depends only on the two parameters ##(a,b)##. What would it mean to say that ##a/b = 2## "will always produce a ##K## with max value". Does that imply ##K(2b,b)## has the same constant value for any choice of ##b##? Or does it mean that some particular arguments like (6.0,3.0) make ##K(6.0,3.0) > K(2.0,1.0)## and ##K(6.0, 3.0) > K( 1.2, 15.3)## etc. ?

The phrase "will always produce" suggests that you are not doing a simple maximization such as:
Find values of the parameters ##(a,b,k,A,B)## that maximize the integral ##K(a,b,k,A,B)##. The answer to that problem can be given by a specific set of values. Suppose the answer is ##(6.0,3.0, 9.7, 1.3, 18.0)##. For those particular values ##a = 2b## and ##B = 3a##. It would be true, but silly to say that ##a## "always" equals ##2b## because there is only one set of values.

It's going to be a slight struggle to state what you want to do. I think the general idea is that you have a set of maximization problems and you want a function that always gives you the formula for the answer.
 
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  • #5
Consider
$$ \int_{y_0}^{y_1} cos (ay^2) dy = \sqrt{\frac{\pi}{2a}} [C(\frac{2ay_1}{\sqrt{2a\pi}})-C(\frac{2ay_0}{\sqrt{2a\pi}})]$$

where
C(x) is the C Fresnel integral

This is maximum when (for a given a=a_0)
$$\frac{2ay_1}{\sqrt{2a\pi}} =1$$
and
$$\frac{2ay_0}{\sqrt{2a\pi}} =-1$$
So
$$y_1=\frac{\sqrt{2a\pi}}{2a}$$
and
$$y_0=-\frac{\sqrt{2a\pi}}{2a}$$
So there is a ratio between y_1 and a, y_0 and a

Anyway, thank you for all the help. There are a number of assumptions and simplifications in order to arrive at these ratios so maybe it can't be done generally. Once again, thank you.
 
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  • #6
We can think about the general type of problem.

Thinking in terms of two variables and some "variable constants":

Let ## F(V_1,V_2,...V_n)## be a function of n-variables. Define the set ##R(X_1,X_2,V_2,V_3,..V_n.)## to be the set:: ##(x_1,x_2,v_3,v_4,...v_n) \in R## iff there exists numbers ##v_3,v_4,...v_n## such that ##V_1 = x_1, V_2 = x2## is a solution to the problem: Maximize ##F(V_1,V_2,v_3,v_4,...v_n)## with respect to picking values for ##V_1,V_2##.

If the set ##R## turns out to be a curve (as opposed to a surface or volume) we can parameterize ##R## by ##P_R(t) = ( X_1(t),X_2(t),V_3(t),V_4(t)...V_n(t)) )##. This can be interpreted as a trajectory. It would be nice if the line integral of ##F(V_1,V_2,...V_n)## over the path ##P_R## between ##t_0## and ##t_1## was the maximum attained over all possible paths between those points. But we must worry about paths that make the line integral large by taking a very long route through a set of moderate values of ##F##.

Even if ##P_R## is not the path that maximizes the line integral (functional) , intuitively it should be produce a local maximum over the set of curves that are small perturbations from ##P_R##. The usual 2-D Calculus of Variations technique relies on finding paths that are extrema , so perhaps it could find local extrema.

Generalizing to more that two variables, you can define a surface or volume ##C(X_1,X_2,...X_k,V_{k+1},...V_n) )## by the set of all solutions to maximizing ##F(V_1,V_2,...V_K,.v_{k+},...v_n)## with respect to ##V_1,V_2,...V_k## for given values ##v_{k+1},...v_n##. (Just keep in mind that if you let ##k = n## then ##C## may become a single point in n-dimensional space.) There is such a thing as the multivariable Calculus of Variations where the problem is to find extrema of integrals over surfaces or volumes.
 
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  • #7
Stephen Tashi said:
Generalizing to more that two variables, you can define a surface or volume ##C(X_1,X_2,...X_k,V_{k+1},...V_n) )## by the set of all solutions to maximizing ##F(V_1,V_2,...V_K,.v_{k+},...v_n)## with respect to ##V_1,V_2,...V_k## for given values ##v_{k+1},...v_n##. (Just keep in mind that if you let ##k = n## then ##C## may become a single point in n-dimensional space.) There is such a thing as the multivariable Calculus of Variations where the problem is to find extrema of integrals over surfaces or volumes.

Thank you very much.
 

What is a maximization problem?

A maximization problem is a type of mathematical problem in which an objective function is optimized, or maximized, subject to certain constraints. The goal is to find the maximum value of the objective function.

What is a double integral?

A double integral is an integral with two variables, typically used to find the volume between a surface and a region in two-dimensional space. It is represented by the symbol ∫∫ f(x,y) dA, where f(x,y) is the integrand and dA is the area element.

How is C not dependent on integrals in this problem?

In this context, C refers to a constant or a fixed number. In a maximization problem, C is typically not dependent on the integrals being used to find the maximum value of the objective function. This means that the value of C remains constant throughout the problem and does not change based on the integrals being used.

What are some common applications of maximization problems?

Maximization problems have a wide range of applications in fields such as economics, engineering, and physics. They can be used to optimize production processes, minimize costs, and determine the most efficient use of resources.

What are some strategies for solving maximization problems?

There are several strategies for solving maximization problems, including using graphical methods, setting up and solving equations, and using calculus techniques such as derivatives and integrals. It is important to carefully analyze the problem and choose the most appropriate method for solving it.

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