Optimization problem - defining domain

otownsend

Hi,

Can someone help me formally define what the domain would be for this question?

I was fortunately able to reach the correct solution for this question without a well defined range, but for future sake, I would like to be able to know how to derive it.

I know that the domain would consider the variables dealing with the maximum passenger capacity of 15,000 people and also the minimum fare sales of $130,000 ... but I'm struggling to construct a domain in the context of these restrictions. - Olivia JeffM How did you set up the problem? To do optimization you must have defined a function. The argument of that function is what will determine the domain. I see what your problem is here. You are thinking about constraints on two different variables. That can be translated into the language of range and domain, but may not be the most intuitive way to think about the problem of these constraints. So let's start with what you have done. HallsofIvy MHF Helper You say, at one point "range" and, at another, "domain". I presume you are thinking of the number of passengers per day as a function of the fare so I would say that, since they have "space to serve up to 15,000 passengers per day" the range will be 0 to 15,000. At the same time, you are told that, at$20 fare, they serve 10,000 passengers per day and "if the fare increases by $0.50 200 fewer people will ride the bus". (This problem does not say "per day" but I assume that is intended.) That is, taking "P" to be the number of passengers and "f" the fare, P= 10000- 200((f- 20)/0.50= 10000- 400(f- 20). There cannot be less than 0 passengers so the fare cannot be more than f such that 10000- 400(f- 20)= 0. That is, 400(f- 20)= 10000, f- 20= 25 or f=$45.00. On the other hand, since there cannot be more than 15000 passengers, the fare cannot be les than f such that 10000- 400(f- 20)= 15000. 400(f- 20)= -5000, f- 20= -12.5, f= $7.50. The domain is$7.50 to $45.00. But you really don't need to know either domain or range to answer this question as you appear to have discovered. otownsend Yea it seems slightly more tedious than usual to find the domain and range just by briefly looking at your response. Thanks for letting me know otownsend How did you set up the problem? To do optimization you must have defined a function. The argument of that function is what will determine the domain. I see what your problem is here. You are thinking about constraints on two different variables. That can be translated into the language of range and domain, but may not be the most intuitive way to think about the problem of these constraints. So let's start with what you have done. Basically I just created a revenue function... r(x) = (10,000 - 200x)(20 + 0.5x) "x" represents the number of times the fare increased by$0.5

I'm curious what the domain restriction would be for this.

otownsend

How did you set up the problem? To do optimization you must have defined a function. The argument of that function is what will determine the domain. I see what your problem is here. You are thinking about constraints on two different variables. That can be translated into the language of range and domain, but may not be the most intuitive way to think about the problem of these constraints. So let's start with what you have done.
Basically I just created a revenue function... r(x) = (10,000 - 200x)(20 + 0.5x)

"x" represents the number of times the fare increased by $0.5 I'm curious what the domain restriction would be for this. JeffM Basically I just created a revenue function... r(x) = (10,000 - 200x)(20 + 0.5x) "x" represents the number of times the fare increased by$0.5

I'm curious what the domain restriction would be for this.
Well x must clearly be an integer so the domain is the set of integers. Now you could also put some reasonableness bounds on x, which would further restrict the domain.

I reiterate that I think putting constraints in as explicit functions is a more straightforward way to go and leads to understanding LaGrangian multipliers.

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otownsend

Okay, sounds good. Thanks for your help!