Maximizing a Function with a Constraint: The Lagrangian Approach

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Homework Help Overview

The discussion revolves around maximizing the function x^-1 + y^-1 under the constraint m = x + y, utilizing the Lagrangian optimization method. Participants explore the implications of the optimization process and the nature of the function's extrema.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the initial assumption of finding a maximum at x*=y*=m/2 and the subsequent realization that this point corresponds to a local minimum. Questions arise regarding the existence of global extrema, particularly in light of vertical asymptotes and singular points.

Discussion Status

The conversation is ongoing, with participants sharing insights about the nature of the function and its behavior at various points. Some express confusion about the problem's request for maximization, given the identified local minimum and the absence of global extrema. There is acknowledgment of the potential for extreme values at endpoints and singular points, but no consensus has been reached.

Contextual Notes

Participants note the possibility of a typo in the problem statement, which may have led to the confusion regarding the maximization request. The discussion also highlights the constraints of the problem and the implications of the function's behavior near asymptotes.

naturalnumbas
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Homework Statement


Seems straightforward enough, Lagrangian optimization


Homework Equations


Find the max of x^-1 + y^-1 subject to the constraint m=x+y



The Attempt at a Solution



At first I thought no problems, x*=y*=m/2, however:

Using the Lagrangian formula yields derivatives as follows:
wrt x: -x^-2 - lambda
wrt y: -y^-2 - lambda
lambda: m-x-y

Putting the coefficients into a bordered Hessian seems to give a positive def. matrix implying a minimum? Is this a trick question or is it possible to maximize?
 
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Yes, x=y=(1/2) is a local minimum. You can also just use your constraint to eliminate y in 1/x+1/y and graph it as a function of x. As x ranges over all possible values you can see there is no global max or global min. There are vertical asymptotes.
 
Dick said:
Yes, x=y=(1/2) is a local minimum. You can also just use your constraint to eliminate y in 1/x+1/y and graph it as a function of x. As x ranges over all possible values you can see there is no global max or global min. There are vertical asymptotes.

Thanks for the reply, you are making sense to me. I'm still confused as to why I was asked to maximize it, when there are no maximum points?
 
naturalnumbas said:
Thanks for the reply, you are making sense to me. I'm still confused as to why I was asked to maximize it, when there are no maximum points?

Typo? Always a possibility.
 
Dick said:
Typo? Always a possibility.

I'm thinking you are probably right.

On an exam no less :S.

Guess I should have known better. Thanks again.
 
Although, I've been reading that extreme values can occur at endpoints, stationary points, and singular points. So obviously this function has some singular points, but I do not know enough to go any further.

Could there be a max at the asymptotes?
 
naturalnumbas said:
Although, I've been reading that extreme values can occur at endpoints, stationary points, and singular points. So obviously this function has some singular points, but I do not know enough to go any further.

Could there be a max at the asymptotes?

As I said, graph 1/x+1/(m-x) to see what the function looks like. Yes, it's singular at x=0 and x=m. No maxes there.
 

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