Simple word problem, wrong solution manual?

trip20

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Nov 4, 2006
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All,

Please take a look at the following word problem, it is from the Linear Programming section of my Math131 course:

A theatre is presenting a program on drinking and driving for students and parents. The proceeds will be donated to a alcohol information center. Admission is $2 for parents and $1 for students. There are two constraints; the theater can hold no more than 150 people, and every two parents must bring at least 1 student. How many parents and students should attend to raise the most money?

I know the answer is 100 parents and 50 students however I wasn't exactly sure how to put it in an algebric equation so I checked the solution manual which shows the following:

Let x = number of students
Let y = number of parents

x+y<=150 and 2x-y>=0

Answer is x+2y (100, 50) or 100 students and 50 parents.

I believe this to be wrong, that would equal $200. It should be the other way around! 100 parents and 50 students meets all the constraints and equals $250.

Can someone please confirm and give the correct formula(s). Wouldn't x+2y<=150 meet both constraints?

Thank you
 

stapel

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Feb 4, 2004
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Since the max/min points occur at corners of the feasibility region, you know from your graph that (x, y) = (100, 50) can be neither a maximum nor a minimum.

I would guess that they typoed the answer, reversing the x- and y-coordinates. The point (x, y) = (50, 100) is a corner point, and clearly provides a more optimal solution.

You might want to point out this error to your instructor, so he can pass it along to the rest of the class.

Eliz.
 

Denis

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Feb 17, 2004
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1,467
trip20 said:
Answer is x+2y (100, 50) or 100 students and 50 parents.
That's obviously "or 50 students and 100 parents"; typo perhaps?

Easy to see that maximum is made up of "2 parents + 1 student" combos,
or 3 persons for $5; 150/3 = 50; 50 * $5 = $250

By the way, minimum would be 2 parents bringing 148 students, right? :wink:
 
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